Question 21: A line intersects $x-axis$ at point $(-2, 0)$ and cuts off an intercept of $3$ units from the positive side of $y-axis$ . Find the equation of the line. [1992]

$x-intercept = (-2, 0)$

$y-intercept = (0, 3)$

Equation of line

$y-3 = \frac{3-0}{0-(-2)} (x-0)$

$y-3 = \frac{3}{2} x$

$2y -6 = 3x$

$2y = 3x+6$

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Question 22: Find the equation of a line passing through the point $(2, 3)$ and having the $x-intercept$ of $4$ units. [2002]

$x-intercept = (4, 0)$

Equation of line passing through $(2,3)$ and $(4, 0)$

$y - 0 = \frac{0-3}{4-2} (x-2)$

$y = -\frac{3}{2} (x-4)$

$2y = -3x+12$

$2y + 3x = 12$

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Question 23: The given figure (not drawn to scale) shows two straight lines $AB \ and \ CD$ . If equation of the line $AB is y =x+1$ and equation of $CD$ is $y = \sqrt{3}x-1$ . Write down the inclination of lines $AB \ and \ CD$ ; also, find the angle between $AB \ and \ CD$ [1989]

$AB: y = x+1$

$CD: y = \sqrt{3}x - 1$

Slope of $AB = 1$

$tan \alpha_1 = 1 \Rightarrow \alpha_1 = 45^o$

Slope of $CD = \sqrt{3}$

$tan \alpha_2 = 1 \Rightarrow \alpha_2 = 60^o$

Therefore $45^o+(180^o-60^o) + \theta = 180^o$

$\Rightarrow \theta = 15^o$

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Question 24: Write down the equation of the line whose gradian is $\frac{2}{3}$ and which passes through $P$ , where $P$ divides the line segment joining $A (-2, 6) \ and \ B (3, -4)$ in the ratio $2 : 3$ [2001]

Given  $P$ divides the line segment joining $A (-2, 6) \ and \ B (3, -4)$ in the ratio $2 : 3$

Let the coordinates of $P = (x, y)$

Therefore

$x = \frac{2 \times 3 + 3 \times (-2)}{2+3} = \frac{6-6}{5} = 0$

$y = \frac{2 \times (-4) + 3 \times 6}{2+3} = \frac{-8+18}{5} = 2$

$\therefore P(0,2)$

Equation of a line passing through $P(0,2)$ with slope $\frac{3}{2}$

$y-2 = \frac{3}{2} (x-0)$

$2y - 4 = 3x \Rightarrow 2y = 3x+4$

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Question 25: The ordinate of a point lying on the line joining points $(6, 4) \ and \ (7, -5) \ is \ -23$. Find the co-ordinates of that point.

Let the ordinate of a point lying on the line joining points $(6, 4) \ and \ (7, -5) \ is \ -23$ be $(x, -23)$

Equation of line passing through $(6,4)$ and $(7, -5)$

$y - 4 = \frac{-5-4}{7-6} (x-6)$

$y-4 = -9(x-6)$

$y+9x=58$

Therefore if $y = -23$, then

$-23+9x=58 \Rightarrow x = 9$

Therefore the point is $(9, -23)$

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Question 26: Point $A \ and \ B$ have co-ordinates $(7, 3) \ and \ (1, 9)$ respectively. Find:

(i) The slope of $AB$

(ii) The equation of perpendicular bisector of the line segment $AB$

(iii) The value of $p \ of \ (-2, p)$ lies on it [2008]

(i) Slope of $AB = \frac{9-(-3)}{1-7} = \frac{12}{-6} = -2$

(ii) Mid point of $AB = (\frac{7+1}{2}, \frac{-3+9}{2}) = (4,3)$

Therefore equation of line passing through $(4,3)$ and slope $\frac{1}{2}$ is

$y - 3 = \frac{1}{2} (x-4)$

$2y -6 = x-4$

$2y= x+2$

(iii) $p \ of \ (-2, p)$ lies on it

Therefore $2(p) = (-2)+2 \Rightarrow p = 0$

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Question 27: $A \ and \ B$ are two points on the $x-axis \ and \ y-axis$ respectively. $P (2, -3)$ is the mid-point of $AB$ . Find the

(i) Co-ordinates of $A \ and \ B$

(ii) Slope of line $AB$

(iii) Equation of line $AB$ [2010]

Let $A(x,0)$ and $B(0, y)$

$P(2, -3)$ is the mid point

(i) Therefore  $2 = \frac{0+x}{2} \Rightarrow x = 4$

$-3 = \frac{y+0}{2} \Rightarrow y = -6$

Hence $A(4,0)$ and $B(0, -6)$

(ii) Slope of $AB = \frac{-6-0}{0-4} = \frac{-6}{-4} = \frac{3}{2}$

(iii) Equation of $AB$

$y = 0 = \frac{3}{2} (x-4)$

$2y = 3x-12$

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Question 28: The equation of a line is $3x + 4y - 7 = 0$ . Find:

(i) Slope of the line.

(ii) The equation of a line perpendicular to the given line and passing through the intersection of the lines $x -y + 2 = 0$ and $3x + y- 10 = 0$ [2010]

(i) Slope $= -\frac{3}{4}$

(ii) Slope of perpendicular $= \frac{4}{3}$

For point of intersection solve $x -y + 2 = 0$ and $3x + y- 10 = 0$

$y = 4$ and $x = 2$

Therefore intersection $= (2, 4)$

Therefore equation of line

$y-4 = \frac{4}{3} (x-2)$

$3y-12 = 4x-8$

$3y = 4x+4$

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Question 29: $ABCD$ is a parallelogram where $A (x, y), B (5, 8), C (4, 7) \ and \ D (2, -4)$ . Find:

(i) Co-ordinates of $A$

(ii) Equation of diagonal $BD$ [2011]

(i) Mid point of $BD = (\frac{5+2}{2}, \frac{-4+8}{2})= (\frac{}{2}, 2)$

Therefore we have $A(x, 0), O(\frac{}{2}, 2)$ and $C(4, 7)$

$O$ is the mid point of $AC$ as well  (diagonals of a parallelogram bisect each other)

Hence $\frac{x+4}{2} = \frac{7}{2} \Rightarrow x = 3$

and $\frac{y+7}{2} = 2 \Rightarrow y = -3$

Hence $A ( 3, -3)$

(ii) Equation of $BD$

$y - 8 = \frac{-4-8}{2-5} (x-5)$

$y-8 = 4(x-5)$

$y - 8 = 4x - 20$

$y + 12 = 4x$

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Question 30: Given equation of line $L_1$ is $y = 4$ .

(i) Write the slope of line $L_2$ is $L_2$ is the bisector of angle $O$

(ii) Write the co-ordinates of point $P$ .

(iii) Find the equation of $L_2$

$L_1 : y = 4$

(i) $\angle \alpha = 45^o$

Therefore slope $m = tan 45^o = 1$

(ii) Therefore $P (4, 4)$

(iii) Equation of line $L_2$

$y - 0 = 1 (x-0) \Rightarrow y = x$

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