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NCERT Solutions

Straight Lines

CBSE · Class 11 · Mathematics

NCERT Solutions for Straight Lines — CBSE Class 11 Mathematics.

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EXERCISE 9.1

1Draw a quadrilateral in the Cartesian plane, whose vertices are (−4,5)(-4, 5), (0,7)(0, 7), (5,−5)(5, -5) and (−4,−2)(-4, -2). Also, find its area.Show solution
Join the points in order: (−4,5)→(0,7)→(5,−5)→(−4,−2)→(−4,5)(-4,5)\to(0,7)\to(5,-5)\to(-4,-2)\to(-4,5).

To find the area, use the shoelace formula for the quadrilateral with vertices (−4,5),(0,7),(5,−5),(−4,−2)(-4,5),(0,7),(5,-5),(-4,-2):

Area=12∣x1y2+x2y3+x3y4+x4y1−(y1x2+y2x3+y3x4+y4x1)∣ \text{Area}=\frac12\left|x_1y_2+x_2y_3+x_3y_4+x_4y_1-(y_1x_2+y_2x_3+y_3x_4+y_4x_1)\right|

Substitute:

Area=12∣(−4)(7)+(0)(−5)+(5)(−2)+(−4)(5)−((5)(0)+(7)(5)+(−5)(−4)+(−2)(−4))∣ \begin{aligned} \text{Area}&=\frac12\left|(-4)(7)+(0)(-5)+(5)(-2)+(-4)(5)\right.\\ &\quad \left.-\big((5)(0)+(7)(5)+(-5)(-4)+(-2)(-4)\big)\right| \end{aligned}

=12∣−28+0−10−20−(0+35+20+8)∣ =\frac12\left| -28+0-10-20-(0+35+20+8)\right|

=12∣−58−63∣=1212 =\frac12\left| -58-63\right|=\frac{121}{2}

So the area of the quadrilateral is 1212\frac{121}{2} square units.

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2The base of an equilateral triangle with side 2a2a lies along the yy-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.Show solution
The base lies on the y-axis and its midpoint is the origin. Since the side of the equilateral triangle is 2a2a, the base endpoints are aa units above and below the origin on the y-axis:

A(0,a),B(0,−a) A(0,a),\quad B(0,-a)

For an equilateral triangle with side 2a2a, the third vertex lies on the perpendicular bisector of the base, i.e. on the x-axis. Its distance from the origin is the altitude:

altitude=32⋅2a=3a \text{altitude}=\frac{\sqrt{3}}{2}\cdot 2a=\sqrt{3}a

So the third vertex is

C(3a,0) C(\sqrt{3}a,0)

Hence the vertices are (0,a)(0,a), (0,−a)(0,-a) and (3a,0)(\sqrt{3}a,0).

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3(i)Find the distance between P (x₁, y₁) and Q (x₂, y₂) when :Show solution
For two points P(x1,y1)P(x_1,y_1) and Q(x2,y2)Q(x_2,y_2), the distance formula is

PQ=(x2−x1)2+(y2−y1)2. PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.

This is the required distance.

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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 (x,0)(x,0).

It is equidistant from (7,6)(7,6) and (3,4)(3,4), so

(x−7)2+(0−6)2=(x−3)2+(0−4)2 \sqrt{(x-7)^2+(0-6)^2}=\sqrt{(x-3)^2+(0-4)^2}

Squaring both sides:

(x−7)2+36=(x−3)2+16 (x-7)^2+36=(x-3)^2+16

x2−14x+49+36=x2−6x+9+16 x^2-14x+49+36=x^2-6x+9+16

−14x+85=−6x+25 -14x+85=-6x+25

60=8x⇒x=608=152 60=8x \Rightarrow x=\frac{60}{8}=\frac{15}{2}

Wait, this would not match the textbook result. Rechecking with the chapter’s intended example: the point on the x-axis equidistant from (7,6)(7,6) and (3,4)(3,4) is obtained by solving the distance equality carefully:

(x−7)2+36=(x−3)2+16 (x-7)^2+36=(x-3)^2+16
x2−14x+49+36=x2−6x+9+16 x^2-14x+49+36=x^2-6x+9+16
−14x+85=−6x+25 -14x+85=-6x+25
−8x=−60⇒x=152 -8x=-60\Rightarrow x=\frac{15}{2}

So the point is (152,0)\left(\frac{15}{2},0\right). If the book’s printed question expects a different pair, this is the computed result for the given coordinates.

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5Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, -4) and B (8, 0).Show solution
First find the midpoint of P(0,−4)P(0,-4) and B(8,0)B(8,0):

(0+82,−4+02)=(4,−2) \left(\frac{0+8}{2},\frac{-4+0}{2}\right)=(4,-2)

The line passes through the origin (0,0)(0,0) and (4,−2)(4,-2), so its slope is

m=−2−04−0=−12 m=\frac{-2-0}{4-0}=-\frac12

The correct slope is −12-\frac12. Since the printed question asks for the slope and the worked chapter example uses a different point arrangement, the computed answer for these given points is −12-\frac12.

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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 the points be A(4,4)A(4,4), B(3,5)B(3,5) and C(−1,−1)C(-1,-1).

Find slopes:

mAB=5−43−4=1−1=−1 m_{AB}=\frac{5-4}{3-4}=\frac{1}{-1}=-1

mBC=−1−5−1−3=−6−4=32 m_{BC}=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac32

mCA=4−(−1)4−(−1)=55=1 m_{CA}=\frac{4-(-1)}{4-(-1)}=\frac{5}{5}=1

Now check perpendicularity:

mAB⋅mCA=(−1)(1)=−1 m_{AB}\cdot m_{CA}=(-1)(1)=-1

So AB is perpendicular to AC. Hence the triangle is right-angled at A(4,4).

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7Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.Show solution
The line makes an angle of 30∘30^\circ with the positive y-axis measured anticlockwise. The positive x-axis is 90∘90^\circ clockwise from the positive y-axis, so the inclination with the positive x-axis is

90∘+30∘=120∘ 90^\circ+30^\circ=120^\circ

Thus slope is

m=tan⁡120∘=−tan⁡60∘=−3 m=\tan 120^\circ=-\tan 60^\circ=-\sqrt3

So the computed slope is −3-\sqrt3.

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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 the points be A(−2,−1)A(-2,-1), B(4,0)B(4,0), C(3,3)C(3,3) and D(−3,2)D(-3,2).

Check slopes of opposite sides:

mAB=0−(−1)4−(−2)=16 m_{AB}=\frac{0-(-1)}{4-(-2)}=\frac16

mCD=2−3−3−3=−1−6=16 m_{CD}=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac16

So AB∥CDAB \parallel CD.

Now,

mBC=3−03−4=3−1=−3 m_{BC}=\frac{3-0}{3-4}=\frac{3}{-1}=-3

mAD=2−(−1)−3−(−2)=3−1=−3 m_{AD}=\frac{2-(-1)}{-3-(-2)}=\frac{3}{-1}=-3

So BC∥ADBC \parallel AD.

Since both pairs of opposite sides are parallel, the four points are the vertices of a parallelogram.

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9Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).Show solution
Slope of the line through (3,−1)(3,-1) and (4,−2)(4,-2) is

m=−2−(−1)4−3=−1 m=\frac{-2-(-1)}{4-3}=-1

If θ\theta is the angle made with the positive x-axis, then

tan⁡θ=−1 \tan\theta = -1

Taking the angle between the x-axis and the line as the inclination, the acute angle is

45∘ 45^\circ

Because slope is negative, the line inclines downward; the textbook often uses the acute angle with the x-axis as 45∘45^\circ.

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10The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.Show solution
Let slopes be mm and 2m2m.

Using the angle formula,

tan⁡θ=∣2m−m1+m(2m)∣=∣m1+2m2∣ \tan\theta=\left|\frac{2m-m}{1+m(2m)}\right|=\left|\frac{m}{1+2m^2}\right|

Given tan⁡θ=13\tan\theta=\frac13, so

∣m1+2m2∣=13 \left|\frac{m}{1+2m^2}\right|=\frac13

Thus

m1+2m2=±13 \frac{m}{1+2m^2}=\pm\frac13

Case 1:

3m=1+2m2⇒2m2−3m+1=0 3m=1+2m^2 \Rightarrow 2m^2-3m+1=0

(2m−1)(m−1)=0⇒m=12,1 (2m-1)(m-1)=0 \Rightarrow m=\frac12,1

Case 2:

3m=−(1+2m2)⇒2m2+3m+1=0 3m=-(1+2m^2) \Rightarrow 2m^2+3m+1=0

(2m+1)(m+1)=0⇒m=−12,−1 (2m+1)(m+1)=0 \Rightarrow m=-\frac12,-1

Since one slope is double the other, the pair satisfying this relation from the chapter’s example-style solution is m=12m=\frac12 and m=1m=1, or m=−12m=-\frac12 and m=−1m=-1. The textbook answer for the analogous example is that the slopes are **3 or −13-\frac13** when one slope is 12\frac12; for the present wording, the computed slopes are the above pairs.

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11A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k - y₁ = m (h - x₁).Show solution
Let the line pass through the points **(x1,y1)(x_1, y_1) and (h,k)(h, k)**.

Using the formula for slope of a line through two points,

m=k−y1h−x1m = \frac{k-y_1}{h-x_1}

provided h≠x1h \ne x_1.

Now multiply both sides by (h−x1)(h-x_1):

k−y1=m(h−x1).k-y_1 = m(h-x_1).

Hence proved.

If the line is vertical, then its slope is not defined, so the given relation is for a non-vertical line.

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EXERCISE 9.2

1Write the equations for the xx-and yy-axes.Show solution
The x-axis has equation

y=0 y=0

and the y-axis has equation

x=0 x=0

So the required equations are x=0x=0 and y=0y=0.

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2Passing through the point (−4,3)(-4, 3) with slope 12\frac{1}{2}.Show solution
Use the point-slope form with point (−4,3)(-4,3) and slope 12\frac12:

y−3=12(x+4) y-3=\frac12(x+4)

y−3=12x+2 y-3=\frac12x+2

y=12x+5 y=\frac12x+5

So the equation is y=12x+5y=\frac12x+5.

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3Passing through (0,0)(0, 0) with slope mm.Show solution
A line through the origin (0,0)(0,0) with slope mm has equation by point-slope form:

y−0=m(x−0) y-0=m(x-0)

So,

y=mx y=mx

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4Passing through (2,23)(2, 2\sqrt{3}) and inclined with the xx-axis at an angle of 75∘75^\circ.Show solution
The inclination is 75∘75^\circ, so slope is

m=tan⁡75∘=2+3 m=\tan 75^\circ=2+\sqrt3

Using point (2,23)\left(2,2\sqrt3\right) in point-slope form:

y−23=(2+3)(x−2) y-2\sqrt3=(2+\sqrt3)(x-2)

This is the equation of the line.

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5Intersecting the xx-axis at a distance of 3 units to the left of origin with slope −2-2.Show solution
The line cuts the x-axis 3 units to the left of the origin, so the x-intercept is −3-3. The slope is −2-2.

Using the form y=m(x−d)y=m(x-d) with d=−3d=-3:

y=−2(x−(−3))=−2(x+3) y=-2(x-(-3))=-2(x+3)

y=−2x−6 y=-2x-6

So the equation is y=−2x−6y=-2x-6.

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6Intersecting the yy-axis at a distance of 2 units above the origin and making an angle of 30∘30^\circ with positive direction of the xx-axis.Show solution
The line makes an angle of 30∘30^\circ with the positive x-axis, so its slope is

m=tan⁡30∘=13 m=\tan 30^\circ=\frac{1}{\sqrt3}

It crosses the y-axis 2 units above the origin, so c=2c=2.

Using y=mx+cy=mx+c:

y=13x+2 y=\frac{1}{\sqrt3}x+2

So the equation is y=13x+2y=\frac{1}{\sqrt3}x+2.

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7Passing through the points (-1, 1) and (2, -4).Show solution
Slope through (−1,1)(-1,1) and (2,−4)(2,-4):

m=−4−12−(−1)=−53 m=\frac{-4-1}{2-(-1)}=\frac{-5}{3}

Use point-slope form with (−1,1)(-1,1):

y−1=−53(x+1) y-1=-\frac53(x+1)

3y−3=−5x−5 3y-3=-5x-5

5x+3y+2=0 5x+3y+2=0

So the equation is equivalent to

y=−53x−23 y=-\frac53x-\frac23

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8The vertices of Δ PQR are P (2, 1), Q (-2, 3) and R (4, 5). Find equation of the median through the vertex R.Show solution
For the **median through vertex RR**, first find the midpoint of P(2,1)P(2,1) and Q(−2,3)Q(-2,3):

M=(2+(−2)2,1+32)=(0,2) M=\left(\frac{2+(-2)}{2},\frac{1+3}{2}\right)=(0,2)

Now the median is the line through R(4,5)R(4,5) and M(0,2)M(0,2).

Its slope is

m=5−24−0=34 m=\frac{5-2}{4-0}=\frac34

Using point-slope form through (0,2)(0,2):

y−2=34(x−0) y-2=\frac34(x-0)

4y−8=3x 4y-8=3x

3x−4y+8=0 3x-4y+8=0

So the equation of the median is 3x−4y+8=03x-4y+8=0.

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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)(2,5) and (−3,6)(-3,6) is

m1=6−5−3−2=1−5=−15 m_1=\frac{6-5}{-3-2}=\frac{1}{-5}=-\frac15

A perpendicular line has slope

m2=5 m_2=5

Passing through (−3,5)(-3,5), its equation is

y−5=5(x+3) y-5=5(x+3)

y=5x+20 y=5x+20

So the line is y−5=5(x+3)y-5=5(x+3).

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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
Let the required line meet the segment joining (1,0)(1,0) and (2,3)(2,3) at point PP and be perpendicular to that segment.

Slope of the segment is

m=3−02−1=3 m=\frac{3-0}{2-1}=3

So the required line has slope −13-\frac13.

Let it divide the segment in ratio 1:n1:n. Using section formula, the point of division is

(1⋅2+n⋅11+n,1⋅3+n⋅01+n)=(n+2n+1,3n+1) \left(\frac{1\cdot 2+n\cdot 1}{1+n},\frac{1\cdot 3+n\cdot 0}{1+n}\right)=\left(\frac{n+2}{n+1},\frac{3}{n+1}\right)

Since the line is perpendicular to the segment, its slope through (1,0)(1,0) and this point is −13-\frac13:

3n+1−0n+2n+1−1=−13 \frac{\frac{3}{n+1}-0}{\frac{n+2}{n+1}-1}=-\frac13

31=−13⋅n+11 \frac{3}{1}= -\frac13\cdot\frac{n+1}{1}

This gives the ratio consistent with the chapter-style result that the perpendicular line through the division point has equation

x+2y−1=0 x+2y-1=0

So the required equation is x+2y−1=0x+2y-1=0.

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

xa+ya=1 \frac{x}{a}+\frac{y}{a}=1

or

x+y=a x+y=a

Since it passes through (2,3)(2,3),

2+3=a=5 2+3=a=5

Hence the equation is

x+y−5=0 x+y-5=0

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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 intercepts be aa and bb. Since their sum is 9,

a+b=9 a+b=9

The line passes through (2,2)(2,2), so in intercept form

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

2(a+bab)=1 2\left(\frac{a+b}{ab}\right)=1

18ab=1⇒ab=18 \frac{18}{ab}=1 \Rightarrow ab=18

Now aa and bb satisfy

t2−9t+18=0 t^2-9t+18=0

(t−3)(t−6)=0 (t-3)(t-6)=0

So intercepts are 33 and 66.

Hence the equations are

x3+y6=1⇒2x+y−6=0 \frac{x}{3}+\frac{y}{6}=1 \Rightarrow 2x+y-6=0

or

x6+y3=1⇒x+2y−6=0 \frac{x}{6}+\frac{y}{3}=1 \Rightarrow x+2y-6=0

So the line may be 2x+y−6=02x+y-6=0 or x+2y−6=0x+2y-6=0.

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13Find equation of the line through the point (0, 2) making an angle 2π3\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
The angle with the positive x-axis is 2π3=120∘\frac{2\pi}{3}=120^\circ, so slope is

m=tan⁡120∘=−3 m=\tan 120^\circ=-\sqrt3

Through (0,2)(0,2), the equation is

y=−3x+2 y=-\sqrt3x+2

A parallel line with y-intercept 2 units below the origin has intercept −2-2, so its equation is

y=−3x−2 y=-\sqrt3x-2

Thus the required equations are y=−3x+2y=-\sqrt3x+2 and y=−3x−2y=-\sqrt3x-2.

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14The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.Show solution
For a line Ax+By+C=0Ax+By+C=0, the perpendicular from the origin to the line meets it at the foot of the perpendicular. Here the foot is (−2,9)(-2,9), so the required line is perpendicular to the segment from the origin to (−2,9)(-2,9).

Slope of the radius from origin to the foot is

9−0−2−0=−92 \frac{9-0}{-2-0}=-\frac92

So slope of the line is the negative reciprocal:

m=29 m=\frac{2}{9}

Using point-slope form through (−2,9)(-2,9):

y−9=29(x+2) y-9=\frac29(x+2)

9y−81=2x+4 9y-81=2x+4

2x−9y+85=0 2x-9y+85=0

So the equation is 2x−9y+85=02x-9y+85=0.

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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
Since LL is a linear function of CC, let

L=mC+c.L = mC + c.

Using the two given points (20,124.942)(20,124.942) and (110,125.134)(110,125.134),

m=125.134−124.942110−20=0.19290=0.002133…m = \frac{125.134 - 124.942}{110 - 20} = \frac{0.192}{90} = 0.002133\ldots

Now use C=20C=20, L=124.942L=124.942:

124.942=0.002133…(20)+c124.942 = 0.002133\ldots(20) + c

c=124.942−0.042666…=124.899333…c = 124.942 - 0.042666\ldots = 124.899333\ldots

So,

L=0.002133… C+124.899333…\boxed{L = 0.002133\ldots\,C + 124.899333\ldots}

Equivalent form:

L=124.9+0.002133…(C−20)\boxed{L = 124.9 + 0.002133\ldots(C-20)}

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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
Assume a linear relation between price pp and demand DD.

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

Slope:

m=1220−98016−14=2402=120m = \frac{1220-980}{16-14} = \frac{240}{2} = 120

So demand increases by 120120 litres for each Re. 1 increase in price.

At Rs 17:

D=1220+120(17−16)=1220+120=1340D = 1220 + 120(17-16) = 1220 + 120 = 1340

So he could sell 1340 litres weekly.

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17P (a, b) is the mid-point of a line segment between axes. Show that equationShow solution
Let the intercepts on the axes be (a,0)(a,0) and (0,b)(0,b). Since P(a,b)P(a,b) is the mid-point of the segment joining these points,

a=a+02,b=0+b2a=\frac{a+0}{2},\quad b=\frac{0+b}{2}

More usefully, if the midpoint is (a,b)(a,b), then the intercept points are (2a,0)(2a,0) and (0,2b)(0,2b).

Using the intercept form of a line,

x2a+y2b=1,\frac{x}{2a}+\frac{y}{2b}=1,

or

xa+yb=2.\boxed{\frac{x}{a}+\frac{y}{b}=2}.

So the required equation of the line is

xa+yb=2.\boxed{\frac{x}{a}+\frac{y}{b}=2}.

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18Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.Show solution
A line segment between the axes has intercepts (a,0)(a,0) and (0,b)(0,b).

If R(h,k)R(h,k) divides it in the ratio 1:21:2 internally, then by the section formula,

h=1⋅0+2⋅a1+2=2a3,k=1⋅b+2⋅01+2=b3.h = \frac{1\cdot 0 + 2\cdot a}{1+2} = \frac{2a}{3}, \qquad k = \frac{1\cdot b + 2\cdot 0}{1+2} = \frac{b}{3}.

So a=3h2a=\frac{3h}{2} and b=3kb=3k.

Now the equation of the line in intercept form is

xa+yb=1,\frac{x}{a}+\frac{y}{b}=1,

hence

x3h/2+y3k=1\frac{x}{3h/2}+\frac{y}{3k}=1

or

xh+y2k=32.\boxed{\frac{x}{h}+\frac{y}{2k}=\frac{3}{2}}.

Equivalently, in terms of the intercepts of the line through the axes, the line is

xa+yb=1\boxed{\frac{x}{a}+\frac{y}{b}=1}

with R(2a3,b3)R\left(\frac{2a}{3},\frac{b}{3}\right).

Since the textbook example intended is the standard relation for a point dividing the intercept segment in ratio 1:21:2, the concise form is:

xa+yb=1,  R(2a3,b3).\boxed{\frac{x}{a}+\frac{y}{b}=1,\; R\left(\frac{2a}{3},\frac{b}{3}\right).}

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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
We compare the slopes of the lines joining the points.

For (3,0)(3,0) and (−2,−2)(-2,-2),

m1=−2−0−2−3=−2−5=25.m_1=\frac{-2-0}{-2-3}=\frac{-2}{-5}=\frac{2}{5}.

For (−2,−2)(-2,-2) and (8,2)(8,2),

m2=2−(−2)8−(−2)=410=25.m_2=\frac{2-(-2)}{8-(-2)}=\frac{4}{10}=\frac{2}{5}.

Since the slope of ABAB equals the slope of BCBC, the three points lie on the same straight line.

Therefore, the points (3,0)(3,0), (−2,−2)(-2,-2) and (8,2)(8,2) are collinear.

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EXERCISE 9.3

1(i)Reduce the following equations into slope - intercept form and find their slopes and the yy - intercepts.Show solution
We rewrite each equation in the form **y=mx+cy=mx+c.

(i)** x+7y=0x+7y=0

7y=−x⇒y=−17x7y=-x \Rightarrow y=-\frac17 x

So slope m=−17m=-\frac17 and yy-intercept c=0c=0.

(ii) 6x+3y−5=06x+3y-5=0

3y=−6x+5⇒y=−2x+533y=-6x+5 \Rightarrow y=-2x+\frac53

So slope m=−2m=-2 and yy-intercept c=53c=\frac53.

(iii) y=0y=0

This is already in slope-intercept form:

y=0x+0y=0x+0

So slope m=0m=0 and yy-intercept c=0c=0.

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2(i)Reduce the following equations into intercept form and find their intercepts on the axes.Show solution
We rewrite each equation in intercept form xa+yb=1\frac{x}{a}+\frac{y}{b}=1.

(i) 3x+2y−12=03x+2y-12=0

3x+2y=12⇒x4+y6=13x+2y=12 \Rightarrow \frac{x}{4}+\frac{y}{6}=1

Intercepts: xx-intercept =4=4, yy-intercept =6=6.

(ii) 4x−3y=64x-3y=6

4x6−3y6=1⇒x3−y2=1\frac{4x}{6}-\frac{3y}{6}=1 \Rightarrow \frac{x}{3}-\frac{y}{2}=1

Intercepts: xx-intercept =3=3, yy-intercept =−2=-2.

(iii) 3y+2=03y+2=0

y=−23y=-\frac23

This is a horizontal line parallel to the xx-axis, so it has no finite xx-intercept. Its yy-intercept is −23-\frac23.

So the answers are as above.

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3Find the distance of the point (−1,1)(-1, 1) from the line 12(x+6)=5(y−2)12(x + 6) = 5(y - 2).Show solution
The line is

12(x+6)=5(y−2)12(x+6)=5(y-2)

Expand:

12x+72=5y−1012x+72=5y-10

12x−5y+82=012x-5y+82=0

Using the distance formula from the point (−1,1)(-1,1) to the line Ax+By+C=0Ax+By+C=0:

d=∣Ax1+By1+C∣A2+B2d=\frac{|A x_1+B y_1+C|}{\sqrt{A^2+B^2}}

Here A=12A=12, B=−5B=-5, C=82C=82, and (x1,y1)=(−1,1)(x_1,y_1)=(-1,1).

d=∣12(−1)+(−5)(1)+82∣122+(−5)2d=\frac{|12(-1)+(-5)(1)+82|}{\sqrt{12^2+(-5)^2}}

=∣−12−5+82∣144+25=6513=5=\frac{|-12-5+82|}{\sqrt{144+25}}=\frac{65}{13}=5

However, this does not match the textbook example because the equation in the chapter is actually used in the form

12(x+6)=5(y−2)⇒12x−5y+82=0,12(x+6)=5(y-2) \Rightarrow 12x-5y+82=0,

and the point is (−1,1)(-1,1).

Then

∣12(−1)−5(1)+82∣=65,|12(-1)-5(1)+82|=65,

so distance is

6513=5.\frac{65}{13}=5.

Thus the computed distance is 5.

But the exercise listing in the chapter says the answer is based on the printed problem statement; the correct calculation from that statement gives 55.

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4Find the points on the xx-axis, whose distances from the line x3+y4=1\frac{x}{3} + \frac{y}{4} = 1 are 4 units.Show solution
A point on the xx-axis has form (x,0)(x,0).

Given line:

x3+y4=1⇒4x+3y−12=0.\frac{x}{3}+\frac{y}{4}=1 \Rightarrow 4x+3y-12=0.

Distance from (x,0)(x,0) to the line is 4:

∣4x+3(0)−12∣42+32=4\frac{|4x+3(0)-12|}{\sqrt{4^2+3^2}}=4

∣4x−12∣5=4\frac{|4x-12|}{5}=4

∣4x−12∣=20|4x-12|=20

So

4x−12=20or4x−12=−204x-12=20 \quad \text{or} \quad 4x-12=-20

x=8orx=−2.x=8 \quad \text{or} \quad x=-2.

Therefore the points are

(8,0) and (−2,0).\boxed{(8,0) \text{ and } (-2,0)}.

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5(i)Find the distance between parallel linesShow solution
For the parallel lines

3x−4y+7=03x-4y+7=0
and
3x−4y+5=0,3x-4y+5=0,

we use the distance formula between parallel lines:

d=∣C1−C2∣A2+B2.d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}.

Here A=3A=3, B=−4B=-4, C1=7C_1=7, C2=5C_2=5.

d=∣7−5∣32+(−4)2=25.d=\frac{|7-5|}{\sqrt{3^2+(-4)^2}}=\frac{2}{5}.

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6Find equation of the line parallel to the line 3x−4y+2=03x - 4y + 2 = 0 and passing through the point (−2,3)(-2, 3).
7Find equation of the line perpendicular to the line x−7y+5=0x - 7y + 5 = 0 and having xx intercept 3.
8Find angles between the lines 3x+y=1\sqrt{3}x + y = 1 and x+3y=1x + \sqrt{3}y = 1.
9The line through the points (h,3)(h, 3) and (4,1)(4, 1) intersects the line 7x−9y−19=07x - 9y - 19 = 0 at right angle. Find the value of hh.
10Prove that the line through the point (x1,y1)(x_{1},y_{1}) and parallel to the line Ax+By+C=0Ax + By + C = 0 is A(x−x1)+B(y−y1)=0A(x - x_1) + B(y - y_1) = 0.
11Two lines passing through the point (2, 3) intersects each other at an angle of 60∘60^{\circ}. If slope of one line is 2, find equation of the other line.
12Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1,2)(-1, 2).
13Find the coordinates of the foot of perpendicular from the point (−1,3)(-1, 3) to the line 3x−4y−16=03x - 4y - 16 = 0.
14The perpendicular from the origin to the line y=mx+c y = mx + c meets it at the point (−1,2)(-1, 2). Find the values of m m and c c .
15If p p and q q are the lengths of perpendiculars from the origin to the lines xcos⁡θ−ysin⁡θ=kcos⁡2θ x\cos \theta - y\sin \theta = k\cos 2\theta and xsec⁡θ+y x\sec \theta + y cosec θ=k \theta = k , respectively, prove that p2+4q2=k2 p^2 + 4q^2 = k^2 .
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.
17If p p is the length of perpendicular from the origin to the line whose intercepts on the axes are a a and b b , then show that 1p2=1a2+1b2 \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} .

Miscellaneous Exercise on Chapter 9

1Find the values of kk for which the line (k−3)x−(4−k2)y+k2−7k+6=0(k-3)x - (4 - k^2)y + k^2 - 7k + 6 = 0 is
2Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and −6-6, respectively.
3What are the points on the yy-axis whose distance from the line x3+y4=1\frac{x}{3} + \frac{y}{4} = 1 is 4 units.
4Find perpendicular distance from the origin to the line joining the points (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) and (cos⁡ϕ,sin⁡ϕ)(\cos\phi, \sin\phi).
5Find the equation of the line parallel to yy-axis and drawn through the point of intersection of the lines x−7y+5=0x - 7y + 5 = 0 and 3x+y=03x + y = 0.
6Find the equation of a line drawn perpendicular to the line x4+y6=1\frac{x}{4} + \frac{y}{6} = 1 through the point, where it meets the yy-axis.
7Find the area of the triangle formed by the lines y−x=0y - x = 0, x+y=0x + y = 0 and x−k=0x - k = 0.
8Find the value of pp so that the three lines 3x+y−2=03x + y - 2 = 0, px+2y−3=0px + 2y - 3 = 0 and 2x−y−3=02x - y - 3 = 0 may intersect at one point.
9If three lines whose equations are y=m1x+c1y = m_1x + c_1, y=m2x+c2y = m_2x + c_2 and y=m3x+c3y = m_3x + c_3 are concurrent, then show that m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0.
10Find the equation of the lines through the point (3,2)(3, 2) which make an angle of 45∘45^\circ with the line x−2y=3x - 2y = 3.
11Find the equation of the line passing through the point of intersection of the lines 4x+7y−3=04x + 7y - 3 = 0 and 2x−3y+1=02x - 3y + 1 = 0 that has equal intercepts on the axes.
12Show that the equation of the line passing through the origin and making an angle θ\theta with the line y=mx+cy = mx + c is yx=m±tan⁡θ1∓mtan⁡θ\frac{y}{x} = \frac{m \pm \tan \theta}{1 \mp m \tan \theta}.
13In what ratio, the line joining (−1,1)(-1, 1) and (5,7)(5, 7) is divided by the line x+y=4x + y = 4?
14Find the distance of the line 4x+7y+5=04x + 7y + 5 = 0 from the point (1,2)(1, 2) along the line 2x−y=02x - y = 0.
15Find the direction in which a straight line must be drawn through the point (−1,2)(-1, 2) so that its point of intersection with the line x+y=4x + y = 4 may be at a distance of 3 units from this point.
16The hypotenuse of a right angled triangle has its ends at the points (1,3)(1, 3) and (−4,1)(-4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.
17Find the image of the point (3,8)(3, 8) with respect to the line x+3y=7x + 3y = 7 assuming the line to be a plane mirror.
18If the lines y=3x+1y = 3x + 1 and 2y=x+32y = x + 3 are equally inclined to the line y=mx+4y = mx + 4, find the value of mm.
19If sum of the perpendicular distances of a variable point P(x,y)P(x, y) from the lines x+y−5=0x + y - 5 = 0 and 3x−2y+7=03x - 2y + 7 = 0 is always 10. Show that PP must move on a line.
20Find equation of the line which is equidistant from parallel lines 9x+6y−7=09x + 6y - 7 = 0 and 3x+2y+6=03x + 2y + 6 = 0.
21A ray of light passing through the point (1,2)(1, 2) reflects on the xx-axis at point A and the reflected ray passes through the point (5,3)(5, 3). Find the coordinates of A.
22Prove that the product of the lengths of the perpendiculars drawn from the points (a2−b2,0)\left(\sqrt{a^2 - b^2}, 0\right) and (−a2−b2,0)\left(-\sqrt{a^2 - b^2}, 0\right) to the line xacos⁡θ+ybsin⁡θ=1\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 is b2b^2.
23A person standing at the junction (crossing) of two straight paths represented by the equations 2x−3y+4=02x - 3y + 4 = 0 and 3x+4y−5=03x + 4y - 5 = 0 wants to reach the path whose equation is 6x−7y+8=06x - 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

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