Straight Lines

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:
$$= 29 + \frac{63}{2} = \frac{58+63}{2} = \frac{121}{2} = 60.5 \text{ sq. units}$$

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:
$$(0, a),\quad (0, -a),\quad \text{and} \quad (\sqrt{3}\,a,\, 0) \text{ or } (-\sqrt{3}\,a,\, 0)$$

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., $y_1 = y_2$.
$$PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{(x_2-x_1)^2+0}$$
$$\boxed{PQ = |x_2 - x_1|}$$

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 $\left(\dfrac{15}{2},\, 0\right)$.

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 $-\dfrac{1}{2}$.

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. $\hspace{2cm}\blacksquare$

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 $-\sqrt{3}$.

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. $\hspace{2cm}\blacksquare$

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.

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 $2m = -1$.

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 $(h - x_1)$:
$$m(h - x_1) = k - y_1$$

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

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.
$$\boxed{x = 0}$$

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

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

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

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

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

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

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

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

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

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

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):
$$y = \frac{3}{4}x + 2$$
$$4y = 3x + 8$$
$$\boxed{3x - 4y + 8 = 0}$$

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 $5$:
$$y - 5 = 5(x - (-3))$$
$$y - 5 = 5x + 15$$
$$\boxed{5x - y + 20 = 0}$$

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

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

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:
$$\boxed{x + y = 5}$$

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 $\boxed{x + 2y - 6 = 0}$.

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 $(0, -2)$:
$$y = -\sqrt{3}\,x + (-2)$$
$$\boxed{\sqrt{3}\,x + y + 2 = 0}$$

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 $\dfrac{2}{9}$:
$$y - 9 = \frac{2}{9}(x - (-2))$$
$$9(y-9) = 2(x+2)$$
$$9y - 81 = 2x + 4$$
$$\boxed{2x - 9y + 85 = 0}$$

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

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.

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

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

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

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**. $\hspace{2cm}\blacksquare$

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 $c = 0$.

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.

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.

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 $\boxed{(-2, 0)}$.