Chapter 9 of 14

Updated 6 August 2026

NCERT Solutions

Straight Lines — NCERT Solutions

CBSE · Class 11 · Mathematics

NCERT Solutions for Straight Lines, CBSE Class 11 Mathematics: 70 textbook questions solved step by step. Part of the CBSE Class 11 Mathematics syllabus.

206 questions60 flashcards14 formulas & key relations5 concepts

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Exercise 9.1

1Draw a quadrilateral in the Cartesian plane, whose vertices are

(-4, 5)

(0, 7)

(5, -5)

and

(-4, -2)

. Also, find its area.Show solution

Given: Vertices of the quadrilateral are $A(-4, 5)$ , $B(0, 7)$ , $C(5, -5)$ and $D(-4, -2)$ .

Area of quadrilateral ABCD can be found by dividing it into two triangles: $\triangle ABC$ and $\triangle ACD$ .

Area of a triangle with vertices $(x_1,y_1)$ , $(x_2,y_2)$ , $(x_3,y_3)$ is:
$\text{Area} = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Area of $\triangle ABC$ with $A(-4,5)$ , $B(0,7)$ , $C(5,-5)$ :
$= \frac{1}{2}|(-4)(7-(-5))+0((-5)-5)+5(5-7)|$
$= \frac{1}{2}|(-4)(12)+0+5(-2)|$
$= \frac{1}{2}|-48-10| = \frac{1}{2}(58) = 29 \text{ sq. units}$

Area of $\triangle ACD$ with $A(-4,5)$ , $C(5,-5)$ , $D(-4,-2)$ :
$= \frac{1}{2}|(-4)((-5)-(-2))+5((-2)-5)+(-4)(5-(-5))|$
$= \frac{1}{2}|(-4)(-3)+5(-7)+(-4)(10)|$
$= \frac{1}{2}|12-35-40| = \frac{1}{2}(63) = \frac{63}{2} \text{ sq. units}$

Total Area of quadrilateral ABCD:
$= 29 + \frac{63}{2} = \frac{58+63}{2} = \frac{121}{2} = 60.5 \text{ sq. units}$

2The base of an equilateral triangle with side

2a

lies along the

y

-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.Show solution

Given: Equilateral triangle with side $2a$ . The base lies along the $y$ -axis and its mid-point is the origin.

Since the base lies along the $y$ -axis and its midpoint is the origin, the two endpoints of the base are:
$B(0, a) \quad \text{and} \quad C(0, -a)$

Let the third vertex be $A(x, 0)$ (it lies on the $x$ -axis by symmetry).

Since the triangle is equilateral, $AB = 2a$ :
$AB = \sqrt{(x-0)^2+(0-a)^2} = 2a$
$x^2 + a^2 = 4a^2$
$x^2 = 3a^2 \implies x = \pm\sqrt{3}\,a$

Therefore, the vertices of the equilateral triangle are:
$(0, a),\quad (0, -a),\quad \text{and} \quad (\sqrt{3}\,a,\, 0) \text{ or } (-\sqrt{3}\,a,\, 0)$

3Find the distance between

\mathrm{P}(x_1, y_1)

and

\mathrm{Q}(x_2, y_2)

when: (i) PQ is parallel to the

y

-axis, (ii) PQ is parallel to the

x

-axis.Show solution

Distance formula: $PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

(i) PQ is parallel to the $y$ -axis:

When a line is parallel to the $y$ -axis, the $x$ -coordinates of both points are equal, i.e., $x_1 = x_2$ .
$PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{0+(y_2-y_1)^2}$
$\boxed{PQ = |y_2 - y_1|}$

(ii) PQ is parallel to the $x$ -axis:

When a line is parallel to the $x$ -axis, the $y$ -coordinates of both points are equal, i.e., $y_1 = y_2$ .
$PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{(x_2-x_1)^2+0}$
$\boxed{PQ = |x_2 - x_1|}$

4Find a point on the

x

-axis, which is equidistant from the points

(7, 6)

and

(3, 4)

.Show solution

Let the required point on the $x$ -axis be $P(x, 0)$ .

Given: $P$ is equidistant from $A(7, 6)$ and $B(3, 4)$ , so $PA = PB$ .

$PA^2 = (x-7)^2 + (0-6)^2 = x^2 - 14x + 49 + 36 = x^2 - 14x + 85$

$PB^2 = (x-3)^2 + (0-4)^2 = x^2 - 6x + 9 + 16 = x^2 - 6x + 25$

Setting $PA^2 = PB^2$ :
$x^2 - 14x + 85 = x^2 - 6x + 25$
$-14x + 85 = -6x + 25$
$-8x = -60$
$x = \frac{15}{2}$

The required point is $\left(\dfrac{15}{2},\, 0\right)$ .

5Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points

\mathrm{P}(0, -4)

and

\mathrm{B}(8, 0)

.Show solution

Step 1: Find the mid-point $M$ of segment $PB$ where $P(0,-4)$ and $B(8,0)$ .
$M = \left(\frac{0+8}{2},\, \frac{-4+0}{2}\right) = (4,\,-2)$

Step 2: Find the slope of the line through the origin $O(0,0)$ and $M(4,-2)$ .
$m = \frac{-2-0}{4-0} = \frac{-2}{4} = -\frac{1}{2}$

The slope of the required line is $-\dfrac{1}{2}$ .

6Without using the Pythagoras theorem, show that the points

(4, 4)

(3, 5)

and

(-1, -1)

are the vertices of a right angled triangle.Show solution

Let $A(4,4)$ , $B(3,5)$ , $C(-1,-1)$ .

Concept: Two lines are perpendicular if the product of their slopes is $-1$ .

Slope of AB:
$m_1 = \frac{5-4}{3-4} = \frac{1}{-1} = -1$

Slope of BC:
$m_2 = \frac{-1-5}{-1-3} = \frac{-6}{-4} = \frac{3}{2}$

Slope of CA:
$m_3 = \frac{4-(-1)}{4-(-1)} = \frac{5}{5} = 1$

Check perpendicularity:
$m_1 \times m_3 = (-1)(1) = -1$

Since the product of slopes of $AB$ and $CA$ is $-1$ , lines $AB \perp CA$ .

Therefore, the angle at vertex $A$ is $90°$ , and the given points form a right angled triangle. $\hspace{2cm}\blacksquare$

7Find the slope of the line, which makes an angle of

30^{\circ}

with the positive direction of

y

-axis measured anticlockwise.Show solution

Given: The line makes an angle of $30°$ with the positive direction of the $y$ -axis (anticlockwise).

If a line makes an angle of $30°$ with the positive $y$ -axis, then it makes an angle of $90° + 30° = 120°$ with the positive $x$ -axis.

Slope:
$m = \tan(120°) = \tan(180° - 60°) = -\tan 60° = -\sqrt{3}$

The slope of the line is $-\sqrt{3}$ .

8Without using distance formula, show that points

(-2, -1)

(4, 0)

(3, 3)

and

(-3, 2)

are the vertices of a parallelogram.Show solution

Let $A(-2,-1)$ , $B(4,0)$ , $C(3,3)$ , $D(-3,2)$ .

Concept: A quadrilateral is a parallelogram if both pairs of opposite sides are parallel (i.e., have equal slopes).

Slope of AB:
$m_{AB} = \frac{0-(-1)}{4-(-2)} = \frac{1}{6}$

Slope of DC:
$m_{DC} = \frac{3-2}{3-(-3)} = \frac{1}{6}$

Since $m_{AB} = m_{DC}$ , $AB \parallel DC$ .

Slope of BC:
$m_{BC} = \frac{3-0}{3-4} = \frac{3}{-1} = -3$

Slope of AD:
$m_{AD} = \frac{2-(-1)}{-3-(-2)} = \frac{3}{-1} = -3$

Since $m_{BC} = m_{AD}$ , $BC \parallel AD$ .

Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram. $\hspace{2cm}\blacksquare$

9Find the angle between the

x

-axis and the line joining the points

(3, -1)

and

(4, -2)

.Show solution

Slope of the line joining $(3,-1)$ and $(4,-2)$ :
$m = \frac{-2-(-1)}{4-3} = \frac{-1}{1} = -1$

If $\alpha$ is the angle the line makes with the positive $x$ -axis, then:
$\tan\alpha = m = -1$
$\alpha = 135°$

The line makes an angle of $135°$ with the positive $x$ -axis.

10The slope of a line is double of the slope of another line. If tangent of the angle between them is

\frac{1}{3}

, find the slopes of the lines.Show solution

Let the slope of one line be $m$ and the slope of the other be $2m$ .

Formula for tangent of angle between two lines:
$\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$

Here $\tan\theta = \dfrac{1}{3}$ , $m_1 = m$ , $m_2 = 2m$ :
$\frac{1}{3} = \left|\frac{2m - m}{1 + m \cdot 2m}\right| = \left|\frac{m}{1+2m^2}\right|$

Case 1: $\dfrac{m}{1+2m^2} = \dfrac{1}{3}$
$3m = 1 + 2m^2 \implies 2m^2 - 3m + 1 = 0$
$(2m-1)(m-1) = 0 \implies m = \frac{1}{2} \text{ or } m = 1$

Case 2: $\dfrac{m}{1+2m^2} = -\dfrac{1}{3}$
$3m = -(1+2m^2) \implies 2m^2 + 3m + 1 = 0$
$(2m+1)(m+1) = 0 \implies m = -\frac{1}{2} \text{ or } m = -1$

Therefore, the slopes of the lines are:

$m = 1$ and $2m = 2$ , or
$m = \dfrac{1}{2}$ and $2m = 1$ , or
$m = -1$ and $2m = -2$ , or
$m = -\dfrac{1}{2}$ and $2m = -1$ .

11A line passes through

(x_1, y_1)

and

(h, k)

. If slope of the line is

m

, show that

k - y_1 = m(h - x_1)

.Show solution

Given: A line passes through the points $(x_1, y_1)$ and $(h, k)$ and has slope $m$ .

Slope of the line passing through $(x_1, y_1)$ and $(h, k)$ is:
$m = \frac{k - y_1}{h - x_1}$

(provided $h \neq x_1$ )

Multiplying both sides by $(h - x_1)$ :
$m(h - x_1) = k - y_1$

$\therefore\quad k - y_1 = m(h - x_1) \hspace{2cm}\blacksquare$

Exercise 9.2

1Write the equations for the

x

-and

y

-axes.Show solution

Equation of the $x$ -axis:
Every point on the $x$ -axis has its $y$ -coordinate equal to zero.
$\boxed{y = 0}$

Equation of the $y$ -axis:
Every point on the $y$ -axis has its $x$ -coordinate equal to zero.
$\boxed{x = 0}$

2Find the equation of the line passing through the point

(-4, 3)

with slope

\frac{1}{2}

.Show solution

Given: Point $(-4, 3)$ , slope $m = \dfrac{1}{2}$ .

Using point-slope form: $y - y_1 = m(x - x_1)$
$y - 3 = \frac{1}{2}(x - (-4))$
$y - 3 = \frac{1}{2}(x + 4)$
$2y - 6 = x + 4$
$\boxed{x - 2y + 10 = 0}$

3Find the equation of the line passing through

(0, 0)

with slope

m

.Show solution

Given: Point $(0, 0)$ , slope $m$ .

Using point-slope form: $y - y_1 = m(x - x_1)$
$y - 0 = m(x - 0)$
$\boxed{y = mx}$

4Find the equation of the line passing through

\left(2, 2\sqrt{3}\right)

and inclined with the

x

-axis at an angle of

75^{\circ}

.Show solution

Given: Point $(2, 2\sqrt{3})$ , angle with $x$ -axis $= 75°$ .

Slope: $m = \tan 75°$
$\tan 75° = \tan(45°+30°) = \frac{\tan 45°+\tan 30°}{1-\tan 45°\tan 30°} = \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$

Rationalising:
$= \frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2-1^2} = \frac{3+2\sqrt{3}+1}{2} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$

Using point-slope form:
$y - 2\sqrt{3} = (2+\sqrt{3})(x-2)$
$y - 2\sqrt{3} = (2+\sqrt{3})x - 2(2+\sqrt{3})$
$y = (2+\sqrt{3})x - 4 - 2\sqrt{3} + 2\sqrt{3}$
$y = (2+\sqrt{3})x - 4$
$\boxed{(2+\sqrt{3})x - y - 4 = 0}$

5Find the equation of the line intersecting the

x

-axis at a distance of 3 units to the left of origin with slope

-2

.Show solution

Given: The line intersects the $x$ -axis at 3 units to the left of origin, so the point is $(-3, 0)$ . Slope $m = -2$ .

Using point-slope form:
$y - 0 = -2(x - (-3))$
$y = -2(x + 3)$
$y = -2x - 6$
$\boxed{2x + y + 6 = 0}$

6Find the equation of the line intersecting the

y

-axis at a distance of 2 units above the origin and making an angle of

30^{\circ}

with positive direction of the

x

-axis.Show solution

Given: The line meets the $y$ -axis at $(0, 2)$ (2 units above origin). Angle with positive $x$ -axis $= 30°$ .

Slope: $m = \tan 30° = \dfrac{1}{\sqrt{3}}$

Using slope-intercept form ( $y$ -intercept $c = 2$ ):
$y = mx + c = \frac{1}{\sqrt{3}}x + 2$
$\sqrt{3}\,y = x + 2\sqrt{3}$
$\boxed{x - \sqrt{3}\,y + 2\sqrt{3} = 0}$

7Find the equation of the line passing through the points

(-1, 1)

and

(2, -4)

.Show solution

Given: Points $(-1, 1)$ and $(2, -4)$ .

Slope:
$m = \frac{-4-1}{2-(-1)} = \frac{-5}{3}$

Using point-slope form with $(-1, 1)$ :
$y - 1 = -\frac{5}{3}(x - (-1))$
$y - 1 = -\frac{5}{3}(x + 1)$
$3(y-1) = -5(x+1)$
$3y - 3 = -5x - 5$
$\boxed{5x + 3y + 2 = 0}$

8The vertices of

\Delta

PQR are P (2, 1), Q (-2, 3) and R (4, 5). Find equation of the median through the vertex R.Show solution

Given: $P(2,1)$ , $Q(-2,3)$ , $R(4,5)$ .

The median through $R$ goes to the mid-point $M$ of $PQ$ .

Mid-point $M$ of $PQ$ :
$M = \left(\frac{2+(-2)}{2},\, \frac{1+3}{2}\right) = (0, 2)$

Slope of $RM$ (through $R(4,5)$ and $M(0,2)$ ):
$m = \frac{2-5}{0-4} = \frac{-3}{-4} = \frac{3}{4}$

Equation of median (using point $M(0,2)$ , which is the $y$ -intercept):
$y = \frac{3}{4}x + 2$
$4y = 3x + 8$
$\boxed{3x - 4y + 8 = 0}$

9Find the equation of the line passing through

(-3, 5)

and perpendicular to the line through the points

(2, 5)

and

(-3, 6)

.Show solution

Slope of the line through $(2,5)$ and $(-3,6)$ :
$m_1 = \frac{6-5}{-3-2} = \frac{1}{-5} = -\frac{1}{5}$

Slope of the required line (perpendicular to above):
$m_2 = -\frac{1}{m_1} = -\frac{1}{-1/5} = 5$

Equation of line through $(-3, 5)$ with slope $5$ :
$y - 5 = 5(x - (-3))$
$y - 5 = 5x + 15$
$\boxed{5x - y + 20 = 0}$

10A line perpendicular to the line segment joining the points

(1, 0)

and

(2, 3)

divides it in the ratio

1: n

. Find the equation of the line.Show solution

Slope of segment joining $(1,0)$ and $(2,3)$ :
$m_1 = \frac{3-0}{2-1} = 3$

Slope of perpendicular line:
$m = -\frac{1}{3}$

Point dividing the segment in ratio $1:n$ (section formula):
$P = \left(\frac{1\cdot2 + n\cdot1}{1+n},\, \frac{1\cdot3 + n\cdot0}{1+n}\right) = \left(\frac{n+2}{n+1},\, \frac{3}{n+1}\right)$

Equation of the perpendicular line through $P$ with slope $-\dfrac{1}{3}$ :
$y - \frac{3}{n+1} = -\frac{1}{3}\left(x - \frac{n+2}{n+1}\right)$

Multiplying through by $3(n+1)$ :
$3(n+1)y - 9 = -(n+1)x + (n+2)$
$(n+1)x + 3(n+1)y = n + 2 + 9 = n + 11$

$\boxed{(1+n)x + 3(1+n)y = n+11}$

11Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).Show solution

Let the equal intercepts be $a$ (both on $x$ - and $y$ -axis).

Using intercept form:
$\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a$

Since the line passes through $(2, 3)$ :
$2 + 3 = a \implies a = 5$

Equation of the line:
$\boxed{x + y = 5}$

12Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.Show solution

Let the $x$ -intercept be $a$ and $y$ -intercept be $b$ .

Given: $a + b = 9 \implies b = 9 - a$

Intercept form:
$\frac{x}{a} + \frac{y}{9-a} = 1$

Since the line passes through $(2, 2)$ :
$\frac{2}{a} + \frac{2}{9-a} = 1$
$2(9-a) + 2a = a(9-a)$
$18 - 2a + 2a = 9a - a^2$
$18 = 9a - a^2$
$a^2 - 9a + 18 = 0$
$(a-3)(a-6) = 0$
$a = 3 \text{ or } a = 6$

Case 1: $a = 3$ , $b = 6$ : $\dfrac{x}{3} + \dfrac{y}{6} = 1 \implies 2x + y = 6$

Case 2: $a = 6$ , $b = 3$ : $\dfrac{x}{6} + \dfrac{y}{3} = 1 \implies x + 2y = 6$

The equations are $\boxed{2x + y - 6 = 0}$ and $\boxed{x + 2y - 6 = 0}$ .

13Find equation of the line through the point

(0, 2)

making an angle

\frac{2\pi}{3}

with the positive

x

-axis. Also, find the equation of line parallel to it and crossing the

y

-axis at a distance of 2 units below the origin.Show solution

Slope of the line:
$m = \tan\frac{2\pi}{3} = \tan 120° = -\sqrt{3}$

Equation of line through $(0,2)$ with slope $-\sqrt{3}$ (using slope-intercept form, $y$ -intercept $= 2$ ):
$y = -\sqrt{3}\,x + 2$
$\boxed{\sqrt{3}\,x + y - 2 = 0}$

Parallel line has the same slope $-\sqrt{3}$ and crosses the $y$ -axis at $2$ units below origin, i.e., at $(0, -2)$ :
$y = -\sqrt{3}\,x + (-2)$
$\boxed{\sqrt{3}\,x + y + 2 = 0}$

14The perpendicular from the origin to a line meets it at the point

(-2, 9)

, find the equation of the line.Show solution

Given: The perpendicular from origin $O(0,0)$ meets the line at $(-2, 9)$ .

Slope of the perpendicular $O(-2,9)$ :
$m_1 = \frac{9-0}{-2-0} = -\frac{9}{2}$

Slope of the required line (perpendicular to $OM$ ):
$m = -\frac{1}{m_1} = -\frac{1}{-9/2} = \frac{2}{9}$

Equation of line through $(-2, 9)$ with slope $\dfrac{2}{9}$ :
$y - 9 = \frac{2}{9}(x - (-2))$
$9(y-9) = 2(x+2)$
$9y - 81 = 2x + 4$
$\boxed{2x - 9y + 85 = 0}$

15The length

L

(in centimetre) of a copper rod is a linear function of its Celsius temperature

C

. In an experiment, if

L = 124.942

when

C = 20

and

L = 125.134

when

C = 110

, express

L

in terms of

C

.Show solution

Given: $L$ is a linear function of $C$ , so $L = aC + b$ for some constants $a, b$ .

Condition 1: When $C = 20$ , $L = 124.942$ :
$20a + b = 124.942 \quad \dots(1)$

Condition 2: When $C = 110$ , $L = 125.134$ :
$110a + b = 125.134 \quad \dots(2)$

Subtracting (1) from (2):
$90a = 0.192 \implies a = \frac{0.192}{90} = 0.00213\overline{3}$

More precisely: $a = \dfrac{0.192}{90} = \dfrac{192}{90000} = \dfrac{16}{7500} = \dfrac{4}{1875}$

From (1):
$b = 124.942 - 20 \times \frac{0.192}{90} = 124.942 - \frac{3.84}{90} = 124.942 - 0.04\overline{2}$
$b = 124.942 - \frac{4}{90} \times \frac{0.96}{1}$

Using the two-point form directly:
$L - 124.942 = \frac{125.134 - 124.942}{110 - 20}(C - 20)$
$L - 124.942 = \frac{0.192}{90}(C - 20)$

$\boxed{L = \frac{0.192}{90}(C-20) + 124.942}$

Simplifying: $L = 0.00\overline{213}(C - 20) + 124.942$

16The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?Show solution

Let selling price be $x$ (Rs/litre) and demand be $y$ (litres/week).

Given two points: $(14, 980)$ and $(16, 1220)$ .

Slope:
$m = \frac{1220 - 980}{16 - 14} = \frac{240}{2} = 120$

Equation of line through $(14, 980)$ :
$y - 980 = 120(x - 14)$
$y = 120x - 1680 + 980$
$y = 120x - 700$

When $x = 17$ :
$y = 120(17) - 700 = 2040 - 700 = 1340$

The owner could sell $\boxed{1340}$ litres of milk weekly at Rs 17/litre.

P(a, b)

is the mid-point of a line segment between axes. Show that equation of the line is

\frac{x}{a} + \frac{y}{b} = 2

.Show solution

Let the line segment have endpoints $A(\alpha, 0)$ on the $x$ -axis and $B(0, \beta)$ on the $y$ -axis.

Since $P(a,b)$ is the mid-point of $AB$ :
$a = \frac{\alpha + 0}{2} \implies \alpha = 2a$
$b = \frac{0 + \beta}{2} \implies \beta = 2b$

So the line passes through $(2a, 0)$ and $(0, 2b)$ .

Using intercept form with $x$ -intercept $= 2a$ and $y$ -intercept $= 2b$ :
$\frac{x}{2a} + \frac{y}{2b} = 1$

Multiplying both sides by 2:
$\frac{x}{a} + \frac{y}{b} = 2 \hspace{2cm}\blacksquare$

18Point R

(h, k)

divides a line segment between the axes in the ratio 1: 2. Find equation of the line.Show solution

Let the line segment have endpoints $A(a, 0)$ on the $x$ -axis and $B(0, b)$ on the $y$ -axis.

$R(h,k)$ divides $AB$ in ratio $1:2$ (using section formula):
$h = \frac{1 \cdot 0 + 2 \cdot a}{1+2} = \frac{2a}{3} \implies a = \frac{3h}{2}$
$k = \frac{1 \cdot b + 2 \cdot 0}{1+2} = \frac{b}{3} \implies b = 3k$

Using intercept form:
$\frac{x}{3h/2} + \frac{y}{3k} = 1$
$\frac{2x}{3h} + \frac{y}{3k} = 1$

Multiplying by $3$ :
$\frac{2x}{h} + \frac{y}{k} = 3$

$\boxed{\frac{2x}{h} + \frac{y}{k} = 3}$

19By using the concept of equation of a line, prove that the three points

(3, 0)

(-2, -2)

and

(8, 2)

are collinear.Show solution

Concept: Three points are collinear if they all lie on the same line.

Equation of line through $(3, 0)$ and $(-2, -2)$ :
$\text{Slope} = \frac{-2-0}{-2-3} = \frac{-2}{-5} = \frac{2}{5}$

Using point-slope form with $(3, 0)$ :
$y - 0 = \frac{2}{5}(x - 3)$
$5y = 2x - 6$
$2x - 5y - 6 = 0 \quad \dots(*)$

Check if $(8, 2)$ satisfies $(*)$ :
$2(8) - 5(2) - 6 = 16 - 10 - 6 = 0 \checkmark$

Since $(8, 2)$ satisfies the equation of the line through $(3,0)$ and $(-2,-2)$ , the three points are collinear. $\hspace{2cm}\blacksquare$

Exercise 9.3

1Reduce the following equations into slope-intercept form and find their slopes and the

y

-intercepts.
(i)

x + 7y = 0

(ii)

6x + 3y - 5 = 0

(iii)

y = 0

Show solution

Slope-intercept form: $y = mx + c$ , where $m$ = slope and $c$ = $y$ -intercept.

(i) $x + 7y = 0$ :
$7y = -x \implies y = -\frac{1}{7}x + 0$
Slope $m = -\dfrac{1}{7}$ , $y$ -intercept $c = 0$ .

(ii) $6x + 3y - 5 = 0$ :
$3y = -6x + 5 \implies y = -2x + \frac{5}{3}$
Slope $m = -2$ , $y$ -intercept $c = \dfrac{5}{3}$ .

(iii) $y = 0$ :
$y = 0 \cdot x + 0$
Slope $m = 0$ , $y$ -intercept $c = 0$ .

2Reduce the following equations into intercept form and find their intercepts on the axes.
(i)

3x + 2y - 12 = 0

(ii)

4x - 3y = 6

(iii)

3y + 2 = 0

Show solution

Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} = 1$ , where $a$ = $x$ -intercept and $b$ = $y$ -intercept.

(i) $3x + 2y - 12 = 0$ :
$3x + 2y = 12 \implies \frac{x}{4} + \frac{y}{6} = 1$
$x$ -intercept $a = 4$ , $y$ -intercept $b = 6$ .

(ii) $4x - 3y = 6$ :
$\frac{x}{6/4} + \frac{y}{-2} = 1 \implies \frac{x}{3/2} + \frac{y}{-2} = 1$
$x$ -intercept $a = \dfrac{3}{2}$ , $y$ -intercept $b = -2$ .

(iii) $3y + 2 = 0$ :
$3y = -2 \implies y = -\frac{2}{3}$
This is a horizontal line. It does not have an $x$ -intercept (parallel to $x$ -axis) and has $y$ -intercept $= -\dfrac{2}{3}$ . The intercept form is not defined in the usual sense since the line does not cross the $x$ -axis.

3Find the distance of the point

(-1, 1)

from the line

12(x + 6) = 5(y - 2)

.Show solution

Rewrite the line:
$12x + 72 = 5y - 10$
$12x - 5y + 82 = 0$

Distance formula: $d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2+B^2}}$

Here $A = 12$ , $B = -5$ , $C = 82$ , $(x_1, y_1) = (-1, 1)$ :
$d = \frac{|12(-1) + (-5)(1) + 82|}{\sqrt{144 + 25}} = \frac{|-12 - 5 + 82|}{\sqrt{169}} = \frac{|65|}{13} = 5$

The distance is $\boxed{5}$ units.

4Find the points on the

x

-axis, whose distances from the line

\frac{x}{3} + \frac{y}{4} = 1

are 4 units.Show solution

Rewrite the line:
$\frac{x}{3} + \frac{y}{4} = 1 \implies 4x + 3y - 12 = 0$

Let the point on the $x$ -axis be $(a, 0)$ .

Distance = 4:
$\frac{|4a + 3(0) - 12|}{\sqrt{16+9}} = 4$
$\frac{|4a - 12|}{5} = 4$
$|4a - 12| = 20$

Case 1: $4a - 12 = 20 \implies 4a = 32 \implies a = 8$

Case 2: $4a - 12 = -20 \implies 4a = -8 \implies a = -2$

The required points are $\boxed{(8, 0)}$ and $\boxed{(-2, 0)}$ .

5Find the distance between parallel lines
(i)

15x + 8y - 34 = 0

and

15x + 8y + 31 = 0

(ii)

l(x + y) + p = 0

and

l(x + y) - r = 0

Show solution

Formula for distance between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ :
$d = \frac{|C_1 - C_2|}{\sqrt{A^2+B^2}}$

(i) $A=15$ , $B=8$ , $C_1=-34$ , $C_2=31$ :
$d = \frac{|-34-31|}{\sqrt{225+64}} = \frac{65}{\sqrt{289}} = \frac{65}{17} = \frac{65}{17}$
$\boxed{d = \frac{65}{17} \text{ units}}$

(ii) Rewrite: $lx + ly + p = 0$ and $lx + ly - r = 0$ .
$A = l$ , $B = l$ , $C_1 = p$ , $C_2 = -r$ :
$d = \frac{|p-(-r)|}{\sqrt{l^2+l^2}} = \frac{|p+r|}{l\sqrt{2}}$
$\boxed{d = \frac{|p+r|}{l\sqrt{2}} \text{ units}}$

6Find equation of the line parallel to the line

3x - 4y + 2 = 0

and passing through the point

(-2, 3)

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7Find equation of the line perpendicular to the line

x - 7y + 5 = 0

and having

x

intercept 3.

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8Find angles between the lines

\sqrt{3} x + y = 1

and

x + \sqrt{3} y = 1

Slope of line 1: $\sqrt{3}x + y = 1 \implies y = -\sqrt{3}x + 1$ , so $m_1 = -\sqrt{3}$ .

Slope of line 2: $x + \sqrt{3}y = 1 \implies y = -\dfrac{1}{\sqrt{3}}x + \dfrac{1}{\sqrt{3}}$ , so $m_2 = -\dfrac{1}{\sqrt{3}}$ .

Angle between the lines:
$\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \left|\frac{-\frac{1}{\sqrt{3}} - (-\sqrt{3})}{1 + (-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)}\right|$
$= \left|\frac{-\frac{1}{\sqrt{3}} + \sqrt{3}}{1 + 1}\right| = \left|\frac{\frac{-1+3}{\sqrt{3}}}{2}\right| = \left|\frac{\frac{2}{\sqrt{3}}}{2}\right| = \frac{1}{\sqrt{3}}$

$\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30°$

The angle between the lines is $30°$ (and the supplementary angle is $150°$ ).

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9The line through the points

(h, 3)

and

(4, 1)

intersects the line

7x - 9y - 19 = 0

at right angle. Find the value of

h

Slope of line through $(h,3)$ and $(4,1)$ :
$m_1 = \frac{1-3}{4-h} = \frac{-2}{4-h}$

Slope of line $7x - 9y - 19 = 0$ :
$9y = 7x - 19 \implies m_2 = \frac{7}{9}$

Condition for perpendicularity: $m_1 \cdot m_2 = -1$
$\frac{-2}{4-h} \cdot \frac{7}{9} = -1$
$\frac{-14}{9(4-h)} = -1$
$14 = 9(4-h)$
$14 = 36 - 9h$
$9h = 22$
$\boxed{h = \frac{22}{9}}$

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10Prove that the line through the point

(x_{1},y_{1})

and parallel to the line

Ax + By + C = 0

\mathrm{A}(x - x_{1}) + \mathrm{B}(y - y_{1}) = 0

Slope of given line $Ax + By + C = 0$ :
$By = -Ax - C \implies m = -\frac{A}{B}$

A line parallel to it has the same slope $m = -\dfrac{A}{B}$ .

Equation of line through $(x_1, y_1)$ with slope $-\dfrac{A}{B}$ :
$y - y_1 = -\frac{A}{B}(x - x_1)$
$B(y - y_1) = -A(x - x_1)$
$A(x - x_1) + B(y - y_1) = 0 \hspace{2cm}\blacksquare$

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11Two lines passing through the point (2, 3) intersects each other at an angle of

60^{\circ}

. If slope of one line is 2, find equation of the other line.

Given: Both lines pass through $(2,3)$ . Slope of one line $m_1 = 2$ . Angle between them $\theta = 60°$ .

Let slope of the other line be $m$ .

$\tan 60° = \left|\frac{m - 2}{1 + 2m}\right|$
$\sqrt{3} = \left|\frac{m-2}{1+2m}\right|$

Case 1: $\dfrac{m-2}{1+2m} = \sqrt{3}$
$m - 2 = \sqrt{3}(1+2m) = \sqrt{3} + 2\sqrt{3}\,m$
$m(1 - 2\sqrt{3}) = \sqrt{3} + 2$
$m = \frac{2+\sqrt{3}}{1-2\sqrt{3}}$

Rationalising: $m = \dfrac{(2+\sqrt{3})(1+2\sqrt{3})}{(1-2\sqrt{3})(1+2\sqrt{3})} = \dfrac{2+4\sqrt{3}+\sqrt{3}+6}{1-12} = \dfrac{8+5\sqrt{3}}{-11}$

Case 2: $\dfrac{m-2}{1+2m} = -\sqrt{3}$
$m - 2 = -\sqrt{3} - 2\sqrt{3}\,m$
$m(1+2\sqrt{3}) = 2 - \sqrt{3}$
$m = \frac{2-\sqrt{3}}{1+2\sqrt{3}} = \frac{(2-\sqrt{3})(1-2\sqrt{3})}{(1+2\sqrt{3})(1-2\sqrt{3})} = \frac{2-4\sqrt{3}-\sqrt{3}+6}{1-12} = \frac{8-5\sqrt{3}}{-11}$

Equation of line through $(2,3)$ :

For Case 1: $y - 3 = \dfrac{-(8+5\sqrt{3})}{11}(x-2)$ , i.e., $\;(8+5\sqrt{3})x + 11y - (16+10\sqrt{3}+33) = 0$

For Case 2: $y - 3 = \dfrac{-(8-5\sqrt{3})}{11}(x-2)$ , i.e., $\;(8-5\sqrt{3})x + 11y - (16-10\sqrt{3}+33) = 0$

The two possible equations of the other line are:
$\boxed{(8+5\sqrt{3})x + 11y - (49+10\sqrt{3}) = 0}$
$\text{and}$
$\boxed{(8-5\sqrt{3})x + 11y - (49-10\sqrt{3}) = 0}$

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12Find the equation of the right bisector of the line segment joining the points (3, 4) and

(-1, 2)

Mid-point $M$ of segment joining $(3,4)$ and $(-1,2)$ :
$M = \left(\frac{3+(-1)}{2},\, \frac{4+2}{2}\right) = (1, 3)$

Slope of the segment:
$m_1 = \frac{2-4}{-1-3} = \frac{-2}{-4} = \frac{1}{2}$

Slope of the right bisector (perpendicular to segment):
$m = -\frac{1}{m_1} = -2$

Equation of right bisector through $M(1,3)$ with slope $-2$ :
$y - 3 = -2(x - 1)$
$y - 3 = -2x + 2$
$\boxed{2x + y - 5 = 0}$

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13Find the coordinates of the foot of perpendicular from the point

(-1, 3)

to the line

3x - 4y - 16 = 0

Let the foot of perpendicular be $F(a, b)$ on the line $3x - 4y - 16 = 0$ .

Condition 1: $F$ lies on the line:
$3a - 4b - 16 = 0 \quad \dots(1)$

Slope of given line: $m_1 = \dfrac{3}{4}$

Slope of perpendicular from $(-1,3)$ to $F(a,b)$ :
$m_2 = \frac{b-3}{a-(-1)} = \frac{b-3}{a+1}$

Condition 2: $m_1 \cdot m_2 = -1$ :
$\frac{3}{4} \cdot \frac{b-3}{a+1} = -1$
$3(b-3) = -4(a+1)$
$3b - 9 = -4a - 4$
$4a + 3b = 5 \quad \dots(2)$

Solving (1) and (2):
From (1): $3a - 4b = 16 \implies$ multiply by 3: $9a - 12b = 48$
From (2): multiply by 4: $16a + 12b = 20$

Adding: $25a = 68 \implies a = \dfrac{68}{25}$

From (2): $b = \dfrac{5 - 4(68/25)}{3} = \dfrac{5 - 272/25}{3} = \dfrac{(125-272)/25}{3} = \dfrac{-147}{75} = -\dfrac{49}{25}$

The foot of perpendicular is $\left(\dfrac{68}{25},\, -\dfrac{49}{25}\right)$ .

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14The perpendicular from the origin to the line

y = mx + c

meets it at the point

(-1, 2)

. Find the values of

m

and

c

Since $(-1, 2)$ lies on the line $y = mx + c$ :
$2 = m(-1) + c \implies c - m = 2 \quad \dots(1)$

Slope of line $= m$ .

Slope of perpendicular from origin $O(0,0)$ to $(-1,2)$ :
$m_{\perp} = \frac{2-0}{-1-0} = -2$

Condition: The line and the perpendicular are perpendicular:
$m \cdot m_{\perp} = -1$
$m \cdot (-2) = -1 \implies m = \frac{1}{2}$

From (1):
$c = 2 + m = 2 + \frac{1}{2} = \frac{5}{2}$

$\boxed{m = \frac{1}{2}, \quad c = \frac{5}{2}}$

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15If

p

and

q

are the lengths of perpendiculars from the origin to the lines

x\cos \theta - y\sin \theta = k\cos 2\theta

and

x\sec \theta + y\csc \theta = k

, respectively, prove that

p^2 + 4q^2 = k^2

Finding $p$ : Distance from origin to $x\cos\theta - y\sin\theta - k\cos 2\theta = 0$ :
$p = \frac{|0 - 0 - k\cos 2\theta|}{\sqrt{\cos^2\theta + \sin^2\theta}} = \frac{|k\cos 2\theta|}{1} = |k\cos 2\theta|$
$p^2 = k^2\cos^2 2\theta \quad \dots(1)$

Finding $q$ : Line $x\sec\theta + y\csc\theta = k$ can be written as:
$\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = k \implies x\sin\theta + y\cos\theta = k\sin\theta\cos\theta$
(Multiplying both sides by $\sin\theta\cos\theta$ )

Distance from origin:
$q = \frac{|k\sin\theta\cos\theta|}{\sqrt{\sin^2\theta+\cos^2\theta}} = |k\sin\theta\cos\theta| = \left|\frac{k\sin 2\theta}{2}\right|$
$q^2 = \frac{k^2\sin^2 2\theta}{4} \quad \dots(2)$

Now:
$p^2 + 4q^2 = k^2\cos^2 2\theta + 4 \cdot \frac{k^2\sin^2 2\theta}{4}$
$= k^2\cos^2 2\theta + k^2\sin^2 2\theta$
$= k^2(\cos^2 2\theta + \sin^2 2\theta) = k^2 \hspace{2cm}\blacksquare$

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16In the triangle ABC with vertices A (2, 3), B (4, -1) and C (1, 2), find the equation and length of altitude from the vertex A.

Altitude from A is perpendicular to $BC$ .

Slope of BC with $B(4,-1)$ and $C(1,2)$ :
$m_{BC} = \frac{2-(-1)}{1-4} = \frac{3}{-3} = -1$

Slope of altitude from A (perpendicular to $BC$ ):
$m = -\frac{1}{-1} = 1$

Equation of altitude through $A(2,3)$ :
$y - 3 = 1(x - 2)$
$\boxed{x - y + 1 = 0}$

Length of altitude = distance from $A(2,3)$ to line $BC$ .

Equation of line $BC$ : slope $= -1$ , through $B(4,-1)$ :
$y+1 = -1(x-4) \implies x + y - 3 = 0$

Distance from $A(2,3)$ to $x+y-3=0$ :
$d = \frac{|2+3-3|}{\sqrt{1+1}} = \frac{|2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

The length of the altitude from A is $\sqrt{2}$ units.

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17If

p

is the length of perpendicular from the origin to the line whose intercepts on the axes are

a

and

b

, then show that

\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}

Equation of line with intercepts $a$ and $b$ :
$\frac{x}{a} + \frac{y}{b} = 1 \implies bx + ay - ab = 0$

Distance from origin $(0,0)$ :
$p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2+a^2}} = \frac{ab}{\sqrt{a^2+b^2}}$

(taking $a, b > 0$ for simplicity)

$p^2 = \frac{a^2b^2}{a^2+b^2}$

$\frac{1}{p^2} = \frac{a^2+b^2}{a^2b^2} = \frac{1}{b^2} + \frac{1}{a^2} = \frac{1}{a^2} + \frac{1}{b^2} \hspace{2cm}\blacksquare$

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Miscellaneous Exercise on Chapter 9

1Find the values of

k

for which the line

(k - 3)x - (4 - k^2)y + k^2 - 7k + 6 = 0

is
(a) Parallel to the

x

-axis,
(b) Parallel to the

y

-axis,
(c) Passing through the origin.

Given line: $(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$

(a) Parallel to the $x$ -axis:
A line is parallel to the $x$ -axis if the coefficient of $x$ is zero and coefficient of $y$ is non-zero.
$k - 3 = 0 \implies k = 3$
Check: when $k=3$ , coefficient of $y = -(4-9) = 5 \neq 0$ . ✓
$\boxed{k = 3}$

(b) Parallel to the $y$ -axis:
A line is parallel to the $y$ -axis if the coefficient of $y$ is zero and coefficient of $x$ is non-zero.
$-(4-k^2) = 0 \implies k^2 = 4 \implies k = \pm 2$
Check $k=2$ : coefficient of $x = 2-3 = -1 \neq 0$ . ✓
Check $k=-2$ : coefficient of $x = -2-3 = -5 \neq 0$ . ✓
$\boxed{k = 2 \text{ or } k = -2}$

(c) Passing through the origin:
Substitute $(0,0)$ :
$k^2 - 7k + 6 = 0$
$(k-1)(k-6) = 0 \implies k = 1 \text{ or } k = 6$
$\boxed{k = 1 \text{ or } k = 6}$

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2Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and

-6

, respectively.

Let $x$ -intercept $= a$ and $y$ -intercept $= b$ .

Given: $a + b = 1$ and $ab = -6$ .

So $a$ and $b$ are roots of:
$t^2 - (a+b)t + ab = 0 \implies t^2 - t - 6 = 0$
$(t-3)(t+2) = 0 \implies t = 3 \text{ or } t = -2$

Case 1: $a = 3$ , $b = -2$ :
$\frac{x}{3} + \frac{y}{-2} = 1 \implies 2x - 3y = 6 \implies \boxed{2x - 3y - 6 = 0}$

Case 2: $a = -2$ , $b = 3$ :
$\frac{x}{-2} + \frac{y}{3} = 1 \implies -3x + 2y = 6 \implies \boxed{3x - 2y + 6 = 0}$

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3What are the points on the

y

-axis whose distance from the line

\frac{x}{3} + \frac{y}{4} = 1

is 4 units.

Rewrite the line:
$\frac{x}{3} + \frac{y}{4} = 1 \implies 4x + 3y - 12 = 0$

Let the point on the $y$ -axis be $(0, b)$ .

Distance = 4:
$\frac{|4(0) + 3b - 12|}{\sqrt{16+9}} = 4$
$\frac{|3b-12|}{5} = 4$
$|3b - 12| = 20$

Case 1: $3b - 12 = 20 \implies 3b = 32 \implies b = \dfrac{32}{3}$

Case 2: $3b - 12 = -20 \implies 3b = -8 \implies b = -\dfrac{8}{3}$

The required points are $\left(0,\, \dfrac{32}{3}\right)$ and $\left(0,\, -\dfrac{8}{3}\right)$ .

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4Find perpendicular distance from the origin to the line joining the points

(\cos \theta, \sin \theta)

and

(\cos \phi, \sin \phi)

Equation of line through $(\cos\theta, \sin\theta)$ and $(\cos\phi, \sin\phi)$ :

Using two-point form:
$y - \sin\theta = \frac{\sin\phi - \sin\theta}{\cos\phi - \cos\theta}(x - \cos\theta)$

Slope:
$m = \frac{\sin\phi - \sin\theta}{\cos\phi - \cos\theta} = \frac{2\cos\frac{\phi+\theta}{2}\sin\frac{\phi-\theta}{2}}{-2\sin\frac{\phi+\theta}{2}\sin\frac{\phi-\theta}{2}} = -\frac{\cos\frac{\phi+\theta}{2}}{\sin\frac{\phi+\theta}{2}} = -\cot\frac{\phi+\theta}{2}$

The equation of the line can be written as:
$x\cos\frac{\theta+\phi}{2} + y\sin\frac{\theta+\phi}{2} = \cos\frac{\theta-\phi}{2}$

(This is a standard result obtained by simplifying the two-point form.)

Distance from origin:
$d = \frac{\left|0 \cdot \cos\frac{\theta+\phi}{2} + 0 \cdot \sin\frac{\theta+\phi}{2} - \cos\frac{\theta-\phi}{2}\right|}{\sqrt{\cos^2\frac{\theta+\phi}{2}+\sin^2\frac{\theta+\phi}{2}}}$
$= \frac{\left|\cos\frac{\theta-\phi}{2}\right|}{1} = \left|\cos\frac{\theta-\phi}{2}\right|$

$\boxed{d = \left|\cos\frac{\theta-\phi}{2}\right|}$

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5Find the equation of the line parallel to

y

-axis and drawn through the point of intersection of the lines

x - 7y + 5 = 0

and

3x + y = 0

Find point of intersection of $x - 7y + 5 = 0$ and $3x + y = 0$ .

From $3x + y = 0$ : $y = -3x$ .

Substituting in $x - 7y + 5 = 0$ :
$x - 7(-3x) + 5 = 0$
$x + 21x + 5 = 0$
$22x = -5 \implies x = -\frac{5}{22}$
$y = -3 \times \left(-\frac{5}{22}\right) = \frac{15}{22}$

Point of intersection: $\left(-\dfrac{5}{22},\, \dfrac{15}{22}\right)$

Line parallel to $y$ -axis through this point has equation $x =$ constant:
$\boxed{x = -\frac{5}{22}} \quad \text{i.e.,} \quad 22x + 5 = 0$

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6Find the equation of a line drawn perpendicular to the line

\frac{x}{4} + \frac{y}{6} = 1

through the point, where it meets the

y

-axis.

Find where $\dfrac{x}{4} + \dfrac{y}{6} = 1$ meets the $y$ -axis (set $x = 0$ ):
$\frac{y}{6} = 1 \implies y = 6$
Point: $(0, 6)$ .

Slope of given line: $\dfrac{x}{4} + \dfrac{y}{6} = 1 \implies 3x + 2y = 12 \implies m_1 = -\dfrac{3}{2}$

Slope of perpendicular line:
$m = -\frac{1}{m_1} = \frac{2}{3}$

Equation of perpendicular through $(0, 6)$ :
$y - 6 = \frac{2}{3}(x - 0)$
$3y - 18 = 2x$
$\boxed{2x - 3y + 18 = 0}$

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7Find the area of the triangle formed by the lines

y - x = 0

x + y = 0

and

x - k = 0

Find vertices (intersections of the three lines):

Intersection of $y - x = 0$ and $x + y = 0$ :
$y = x$ and $y = -x \implies x = 0, y = 0$ . Vertex $A(0, 0)$ .

Intersection of $y - x = 0$ and $x = k$ :
$x = k, y = k$ . Vertex $B(k, k)$ .

Intersection of $x + y = 0$ and $x = k$ :
$x = k, y = -k$ . Vertex $C(k, -k)$ .

Area of triangle with $A(0,0)$ , $B(k,k)$ , $C(k,-k)$ :
$\text{Area} = \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$
$= \frac{1}{2}|0(k-(-k)) + k(-k-0) + k(0-k)|$
$= \frac{1}{2}|0 - k^2 - k^2| = \frac{1}{2} \cdot 2k^2 = k^2$

$\boxed{\text{Area} = k^2 \text{ sq. units}}$

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8Find the value of

p

so that the three lines

3x + y - 2 = 0

px + 2y - 3 = 0

and

2x - y - 3 = 0

may intersect at one point.

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9If three lines whose equations are

y = m_1x + c_1

y = m_2x + c_2

and

y = m_3x + c_3

are concurrent, then show that

m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0

Find intersection of lines 1 and 2:
$m_1x + c_1 = m_2x + c_2$
$x(m_1 - m_2) = c_2 - c_1$
$x = \frac{c_2 - c_1}{m_1 - m_2}$
$y = m_1 \cdot \frac{c_2-c_1}{m_1-m_2} + c_1 = \frac{m_1c_2 - m_1c_1 + c_1m_1 - c_1m_2}{m_1-m_2} = \frac{m_1c_2 - m_2c_1}{m_1-m_2}$

For concurrency, this point must lie on line 3: $y = m_3x + c_3$ :
$\frac{m_1c_2 - m_2c_1}{m_1-m_2} = m_3 \cdot \frac{c_2-c_1}{m_1-m_2} + c_3$

Multiplying both sides by $(m_1 - m_2)$ :
$m_1c_2 - m_2c_1 = m_3(c_2-c_1) + c_3(m_1-m_2)$
$m_1c_2 - m_2c_1 = m_3c_2 - m_3c_1 + c_3m_1 - c_3m_2$
$m_1c_2 - m_2c_1 - m_3c_2 + m_3c_1 - c_3m_1 + c_3m_2 = 0$
$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0 \hspace{1cm}\blacksquare$

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10Find the equation of the lines through the point (3, 2) which make an angle of

45^\circ

with the line

x - 2y = 3

Slope of given line $x - 2y = 3$ : $m_1 = \dfrac{1}{2}$ .

Let slope of required line be $m$ . Angle between them $= 45°$ :
$\tan 45° = \left|\frac{m - \frac{1}{2}}{1 + \frac{m}{2}}\right| = 1$

$\left|\frac{2m-1}{2+m}\right| = 1$

Case 1: $\dfrac{2m-1}{m+2} = 1 \implies 2m-1 = m+2 \implies m = 3$

Case 2: $\dfrac{2m-1}{m+2} = -1 \implies 2m-1 = -(m+2) \implies 3m = -1 \implies m = -\dfrac{1}{3}$

Equation with $m = 3$ through $(3,2)$ :
$y - 2 = 3(x-3) \implies 3x - y - 7 = 0$

Equation with $m = -\frac{1}{3}$ through $(3,2)$ :
$y - 2 = -\frac{1}{3}(x-3) \implies 3y - 6 = -(x-3) \implies x + 3y - 9 = 0$

The required equations are $\boxed{3x - y - 7 = 0}$ and $\boxed{x + 3y - 9 = 0}$ .

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11Find the equation of the line passing through the point of intersection of the lines

4x + 7y - 3 = 0

and

2x - 3y + 1 = 0

that has equal intercepts on the axes.

Find point of intersection of $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$ :

From second equation: $2x = 3y - 1 \implies x = \dfrac{3y-1}{2}$

Substituting in first:
$4 \cdot \frac{3y-1}{2} + 7y - 3 = 0$
$2(3y-1) + 7y - 3 = 0$
$6y - 2 + 7y - 3 = 0$
$13y = 5 \implies y = \frac{5}{13}$
$x = \frac{3(5/13)-1}{2} = \frac{15/13 - 1}{2} = \frac{2/13}{2} = \frac{1}{13}$

Point of intersection: $\left(\dfrac{1}{13},\, \dfrac{5}{13}\right)$

Line with equal intercepts ( $a = b$ ): $\dfrac{x}{a} + \dfrac{y}{a} = 1 \implies x + y = a$

Passing through $\left(\dfrac{1}{13}, \dfrac{5}{13}\right)$ :
$\frac{1}{13} + \frac{5}{13} = a \implies a = \frac{6}{13}$

$x + y = \frac{6}{13} \implies \boxed{13x + 13y = 6}$

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12Show that the equation of the line passing through the origin and making an angle

\theta

with the line

y = mx + c

\frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}

Let the slope of the required line (passing through origin) be $M$ .

The line through origin is $y = Mx$ .

Angle between $y = Mx$ and $y = mx + c$ :
$\tan\theta = \left|\frac{M - m}{1 + mM}\right|$

This gives two cases:

Case 1: $\dfrac{M-m}{1+mM} = \tan\theta$
$M - m = \tan\theta + mM\tan\theta$
$M(1 - m\tan\theta) = m + \tan\theta$
$M = \frac{m + \tan\theta}{1 - m\tan\theta}$

Case 2: $\dfrac{M-m}{1+mM} = -\tan\theta$
$M - m = -\tan\theta - mM\tan\theta$
$M(1 + m\tan\theta) = m - \tan\theta$
$M = \frac{m - \tan\theta}{1 + m\tan\theta}$

Combining both cases:
$M = \frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta} \hspace{2cm}\blacksquare$

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13In what ratio, the line joining

(-1, 1)

and

(5, 7)

is divided by the line

x + y = 4

Let the line $x + y = 4$ divide the segment joining $A(-1,1)$ and $B(5,7)$ in ratio $k:1$ .

Point of division (section formula):
$P = \left(\frac{5k + (-1)}{k+1},\, \frac{7k+1}{k+1}\right) = \left(\frac{5k-1}{k+1},\, \frac{7k+1}{k+1}\right)$

$P$ lies on $x + y = 4$ :
$\frac{5k-1}{k+1} + \frac{7k+1}{k+1} = 4$
$\frac{12k}{k+1} = 4$
$12k = 4(k+1)$
$12k = 4k + 4$
$8k = 4 \implies k = \frac{1}{2}$

The line divides the segment in ratio $\dfrac{1}{2}:1 = 1:2$ .

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14Find the distance of the line

4x + 7y + 5 = 0

from the point

(1, 2)

along the line

2x - y = 0

Find the point where line $2x - y = 0$ (i.e., $y = 2x$ ) meets $4x + 7y + 5 = 0$ .

Substitute $y = 2x$ :
$4x + 7(2x) + 5 = 0$
$4x + 14x + 5 = 0$
$18x = -5 \implies x = -\frac{5}{18}$
$y = 2 \times \left(-\frac{5}{18}\right) = -\frac{5}{9}$

Point of intersection: $Q\left(-\dfrac{5}{18},\, -\dfrac{5}{9}\right)$

Distance from $(1,2)$ to $Q$ :
$d = \sqrt{\left(1-\left(-\frac{5}{18}\right)\right)^2 + \left(2-\left(-\frac{5}{9}\right)\right)^2}$
$= \sqrt{\left(\frac{23}{18}\right)^2 + \left(\frac{23}{9}\right)^2}$
$= \sqrt{\frac{529}{324} + \frac{529}{81}}$
$= \sqrt{\frac{529}{324} + \frac{2116}{324}}$
$= \sqrt{\frac{2645}{324}} = \frac{\sqrt{2645}}{18} = \frac{23\sqrt{5}}{18}$

$\boxed{d = \frac{23\sqrt{5}}{18} \text{ units}}$

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15Find the direction in which a straight line must be drawn through the point

(-1, 2)

so that its point of intersection with the line

x + y = 4

may be at a distance of 3 units from this point.

Let the line through $(-1, 2)$ make angle $\alpha$ with the $x$ -axis.

Parametric form of the line:
$x = -1 + r\cos\alpha, \quad y = 2 + r\sin\alpha$

where $r$ is the distance from $(-1,2)$ .

Substituting in $x + y = 4$ :
$(-1 + r\cos\alpha) + (2 + r\sin\alpha) = 4$
$1 + r(\cos\alpha + \sin\alpha) = 4$
$r(\cos\alpha + \sin\alpha) = 3$

Given $r = 3$ :
$3(\cos\alpha + \sin\alpha) = 3$
$\cos\alpha + \sin\alpha = 1$
$\sqrt{2}\sin\left(\alpha + 45°\right) = 1$
$\sin\left(\alpha + 45°\right) = \frac{1}{\sqrt{2}}$
$\alpha + 45° = 45° \text{ or } 135°$
$\alpha = 0° \text{ or } \alpha = 90°$

The line must be drawn either parallel to the $x$ -axis ( $\alpha = 0°$ ) or parallel to the $y$ -axis ( $\alpha = 90°$ ).

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16The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (-4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

Given: Hypotenuse has endpoints $A(1,3)$ and $B(-4,1)$ . The legs are parallel to the axes.

Since the legs are parallel to the coordinate axes, the right angle vertex $C$ must have coordinates $(1, 1)$ or $(-4, 3)$ .

Case 1: Right angle at $C(1, 1)$ :

Leg 1 (parallel to $y$ -axis through $A(1,3)$ and $C(1,1)$ ): $x = 1$
Leg 2 (parallel to $x$ -axis through $B(-4,1)$ and $C(1,1)$ ): $y = 1$

Case 2: Right angle at $C(-4, 3)$ :

Leg 1 (parallel to $x$ -axis through $A(1,3)$ and $C(-4,3)$ ): $y = 3$
Leg 2 (parallel to $y$ -axis through $B(-4,1)$ and $C(-4,3)$ ): $x = -4$

The equations of the legs are:
$\boxed{x = 1 \text{ and } y = 1}$
or
$\boxed{x = -4 \text{ and } y = 3}$

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17Find the image of the point (3, 8) with respect to the line

x + 3y = 7

assuming the line to be a plane mirror.

Let the image of $P(3,8)$ be $Q(h,k)$ .

Condition 1: The midpoint of $PQ$ lies on $x + 3y = 7$ :
$\frac{h+3}{2} + 3 \cdot \frac{k+8}{2} = 7$
$h + 3 + 3k + 24 = 14$
$h + 3k = -13 \quad \dots(1)$

Condition 2: $PQ$ is perpendicular to $x + 3y = 7$ (slope of line $= -\dfrac{1}{3}$ ):
$\frac{k-8}{h-3} \times \left(-\frac{1}{3}\right) = -1$
$\frac{k-8}{h-3} = 3$
$k - 8 = 3(h-3)$
$k - 8 = 3h - 9$
$3h - k = 1 \quad \dots(2)$

Solving (1) and (2):
From (2): $k = 3h - 1$ . Substituting in (1):
$h + 3(3h-1) = -13$
$h + 9h - 3 = -13$
$10h = -10 \implies h = -1$
$k = 3(-1) - 1 = -4$

The image of $(3,8)$ is $\boxed{(-1, -4)}$ .

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18If the lines

y = 3x + 1

and

2y = x + 3

are equally inclined to the line

y = mx + 4

, find the value of

m

Slopes: $m_1 = 3$ (line 1), $m_2 = \dfrac{1}{2}$ (line 2), $m_3 = m$ (required line).

Condition: Both lines are equally inclined to $y = mx + 4$ , so:
$\left|\frac{m_1 - m}{1 + m_1 m}\right| = \left|\frac{m_2 - m}{1 + m_2 m}\right|$
$\left|\frac{3-m}{1+3m}\right| = \left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right| = \left|\frac{1-2m}{2+m}\right|$

Case 1: $\dfrac{3-m}{1+3m} = \dfrac{1-2m}{2+m}$
$(3-m)(2+m) = (1-2m)(1+3m)$
$6 + 3m - 2m - m^2 = 1 + 3m - 2m - 6m^2$
$6 + m - m^2 = 1 + m - 6m^2$
$5m^2 + 5 = 0$
No real solution.

Case 2: $\dfrac{3-m}{1+3m} = -\dfrac{1-2m}{2+m}$
$(3-m)(2+m) = -(1-2m)(1+3m)$
$6 + 3m - 2m - m^2 = -(1 + 3m - 2m - 6m^2)$
$6 + m - m^2 = -1 - m + 6m^2$
$7m^2 - 2m - 7 = 0$
$m = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$

$\boxed{m = \frac{1+5\sqrt{2}}{7} \quad \text{or} \quad m = \frac{1-5\sqrt{2}}{7}}$

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19If sum of the perpendicular distances of a variable point

\mathbf{P}(x, y)

from the lines

x + y - 5 = 0

and

3x - 2y + 7 = 0

is always 10. Show that

\mathbf{P}

must move on a line.

Distance from $P(x,y)$ to $x + y - 5 = 0$ :
$d_1 = \frac{|x + y - 5|}{\sqrt{2}}$

Distance from $P(x,y)$ to $3x - 2y + 7 = 0$ :
$d_2 = \frac{|3x - 2y + 7|}{\sqrt{13}}$

Given: $d_1 + d_2 = 10$
$\frac{|x+y-5|}{\sqrt{2}} + \frac{|3x-2y+7|}{\sqrt{13}} = 10$

Considering the case where both expressions inside the absolute values are positive (the other cases lead to similar linear equations):
$\frac{x+y-5}{\sqrt{2}} + \frac{3x-2y+7}{\sqrt{13}} = 10$

Multiplying through by $\sqrt{2}\cdot\sqrt{13}$ :
$\sqrt{13}(x+y-5) + \sqrt{2}(3x-2y+7) = 10\sqrt{26}$
$(\sqrt{13}+3\sqrt{2})x + (\sqrt{13}-2\sqrt{2})y + (-5\sqrt{13}+7\sqrt{2}-10\sqrt{26}) = 0$

This is a linear equation in $x$ and $y$ , which represents a straight line.

In each of the four sign combinations of the absolute values, we get a linear equation. Hence, $P$ must move on a straight line. $\hspace{2cm}\blacksquare$

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20Find equation of the line which is equidistant from parallel lines

9x + 6y - 7 = 0

and

3x + 2y + 6 = 0

Rewrite both lines with the same coefficients. Multiply the second line by 3:
$9x + 6y + 18 = 0$

Now the two parallel lines are:
$9x + 6y - 7 = 0 \quad (C_1 = -7)$
$9x + 6y + 18 = 0 \quad (C_2 = 18)$

The equidistant line has $C = \dfrac{C_1 + C_2}{2} = \dfrac{-7+18}{2} = \dfrac{11}{2}$ :
$9x + 6y + \frac{11}{2} = 0$
$18x + 12y + 11 = 0$

$\boxed{18x + 12y + 11 = 0}$

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21A ray of light passing through the point (1, 2) reflects on the

x

-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Concept: The image of the point $(1,2)$ in the $x$ -axis is $(1,-2)$ . The reflected ray, when extended, passes through the image point.

Image of $(1,2)$ in $x$ -axis: $B(1,-2)$ .

The reflected ray passes through $B(1,-2)$ and $(5,3)$ .

Equation of line through $B(1,-2)$ and $(5,3)$ :
$\text{Slope} = \frac{3-(-2)}{5-1} = \frac{5}{4}$
$y + 2 = \frac{5}{4}(x-1)$
$4y + 8 = 5x - 5$
$5x - 4y - 13 = 0$

Point A is on the $x$ -axis, so $y = 0$ :
$5x - 13 = 0 \implies x = \frac{13}{5}$

$\boxed{A = \left(\frac{13}{5},\, 0\right)}$

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22Prove that the product of the lengths of the perpendiculars drawn from the points

\left(\sqrt{a^2 - b^2}, 0\right)

and

\left(-\sqrt{a^2 - b^2}, 0\right)

to the line

\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1

b^2

Rewrite the line: $bx\cos\theta + ay\sin\theta - ab = 0$

Here $A = b\cos\theta$ , $B = a\sin\theta$ , $C = -ab$ .

Distance from $\left(\sqrt{a^2-b^2},\, 0\right)$ :
$d_1 = \frac{|b\cos\theta \cdot \sqrt{a^2-b^2} + a\sin\theta \cdot 0 - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$
$= \frac{|b\sqrt{a^2-b^2}\cos\theta - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$

Distance from $\left(-\sqrt{a^2-b^2},\, 0\right)$ :
$d_2 = \frac{|-b\sqrt{a^2-b^2}\cos\theta - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$

Product:
$d_1 \cdot d_2 = \frac{|b\sqrt{a^2-b^2}\cos\theta - ab| \cdot |{-b\sqrt{a^2-b^2}\cos\theta - ab}|}{b^2\cos^2\theta + a^2\sin^2\theta}$

$= \frac{|-(b\sqrt{a^2-b^2}\cos\theta)^2 + (ab)^2|}{b^2\cos^2\theta + a^2\sin^2\theta}$

$= \frac{|a^2b^2 - b^2(a^2-b^2)\cos^2\theta|}{b^2\cos^2\theta + a^2\sin^2\theta}$

$= \frac{b^2|a^2 - (a^2-b^2)\cos^2\theta|}{b^2\cos^2\theta + a^2\sin^2\theta}$

$= \frac{b^2|a^2\sin^2\theta + b^2\cos^2\theta|}{a^2\sin^2\theta + b^2\cos^2\theta} = b^2 \hspace{2cm}\blacksquare$

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23A person standing at the junction (crossing) of two straight paths represented by the equations

2x - 3y + 4 = 0

and

3x + 4y - 5 = 0

wants to reach the path whose equation is

6x - 7y + 8 = 0

in the least time. Find equation of the path that he should follow.

Find the junction point (intersection of $2x-3y+4=0$ and $3x+4y-5=0$ ):

Multiply first by 4 and second by 3:
$8x - 12y + 16 = 0$
$9x + 12y - 15 = 0$

Adding: $17x + 1 = 0 \implies x = -\dfrac{1}{17}$

From first equation: $2\left(-\dfrac{1}{17}\right) - 3y + 4 = 0 \implies -3y = -4 + \dfrac{2}{17} = \dfrac{-68+2}{17} = \dfrac{-66}{17}$
$y = \frac{22}{17}$

Junction point: $\left(-\dfrac{1}{17},\, \dfrac{22}{17}\right)$

Least time means shortest distance, i.e., the person should walk along the perpendicular from the junction to the line $6x - 7y + 8 = 0$ .

Slope of $6x - 7y + 8 = 0$ : $m_1 = \dfrac{6}{7}$

Slope of perpendicular: $m = -\dfrac{7}{6}$

Equation of path through $\left(-\dfrac{1}{17},\, \dfrac{22}{17}\right)$ with slope $-\dfrac{7}{6}$ :
$y - \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right)$

Multiplying by $6 \times 17 = 102$ :
$6(17y - 22) = -7(17x + 1)$
$102y - 132 = -119x - 7$
$119x + 102y - 125 = 0$

$\boxed{119x + 102y - 125 = 0}$

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Frequently Asked Questions

What are the important topics in Straight Lines for CBSE Class 11 Mathematics?

Key topics in Straight Lines include Basics of Coordinate Geometry and Straight Lines, Slope and Inclination of a Line, Parallel and Perpendicular Lines; Angle Between Two Lines, Equations of a Line in Different Forms. Study these first, then practise questions on each for Class 11 exams.

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