Question 1: Point $P$ divides the line segment joining the points $A (8, 0) \ and \ B (16, -8)$ in the ratio $3 : 5$ . Find the co-ordinates of point $P$ . Also, find the equation of the line through $P$ and parallel to $3x + 5y = 7$.

Given $P$ divides $A (8, 0) \ and \ B (16, -8)$ in the ratio $3 : 5$

Ratio: $m_1:m_2 = 3 : 5$

Let the coordinates of the point $P \ be \ (x, y)$. Therefore

$x =$ $\frac{3 \times 16+5 \times 8}{3+5} = \frac{88}{8}$ $= 11$

$y =$ $\frac{3 \times (-8)+5 \times 0}{3+5} = \frac{-24}{8}$ $= -3$

Therefore $P = (11, -3)$

Equation of the line given : $3x+5y=7 \Rightarrow slope = m = \frac{-3}{5}$

Therefore the required equation is

$y - y_1 = m(x-x_1)$

$y -(-3) = \frac{-3}{5}(x-11)$ or $5y +3x=18$

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Question 2: The line segment joining the points $A (3, -4) \ and \ B (-2, 1)$ is divided in the ratio $1 : 3$ at point $P$ in it. Find the co-ordinates of $P$ . Also, find the equation of the line through $P$ and perpendicular to the line $5x - 3y = 4$ .

Given $P$ divides $A (3,-4) \ and \ B (-2,1)$ in the ratio $1:3$

Ratio: $m_1:m_2 = 1:3$

Let the coordinates of the point $P \ be \ (x, y)$. Therefore

$x = \frac{1 \times (-2)+3 \times 3}{1+3} = \frac{7}{4}$

$y = \frac{1 \times 1+3 \times (-4)}{1+3} = \frac{-11}{4}$

Therefore $P = (\frac{7}{4} , \frac{-11}{4})$

Equation of the line given : $5x - 3y = 4 \Rightarrow slope = m_1 = \frac{5}{3}$

Therefore the slope of the line perpendicular to the above line $= \frac{-3}{5}$

Therefore the required equation is

$y - y_1 = m(x-x_1)$

$y -\frac{-11}{4} =\frac{-3}{5}(x-\frac{7}{4})$  or  $10y +6x+17 = 0$

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Question 3: A line $5x + 3y + 15 = 0$ meets $y-axis$ at point $P$ . Find the co-ordinates of point $P$ . Find the equation of a line through $P$ and perpendicular to $x - 3y + 4 = 0$.

At $y-axis, \ x = 0 \Rightarrow y = -5$

Therefore the coordinate of $P = (0, -5)$

$x - 3y + 4 = 0$ $\Rightarrow slope = \frac{1}{3}$

Therefore the slope of a like perpendicular to this line $= -3$

Hence the line passing through $(0, -5)$ with a slope of $-3$ is

$y-(-5) = -3(x-0) \Rightarrow y+3x+5 = 0$

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Question 4: Find the value of $k$ for which the lines $kx - 5y + 4 = 0$ and $5x - 2y + 5 = 0$ are perpendicular to each other. [2003]

Slope of  $kx - 5y + 4 = 0$  $\Rightarrow m_1 = \frac{k}{5}$

Slope of $5x - 2y + 5 = 0$  $\Rightarrow m_2 = \frac{5}{2}$

Since the two lines are perpendicular, $m_1. m_2 = -1$

$\Rightarrow \frac{k}{5} . \frac{5}{2}$ $= -1$

$\Rightarrow k = -2$

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Question 5: A straight line passes through the points $P (-1, 4) \ and \ Q (5, -2)$. It intersects the co-ordinate axes at points $A \ and \ B$ . $M$ is the mid-point of the line segment $AB$ . Find:

• The equation of line
• The co-ordinates of $A \ and \ B$
• The co-ordinates of $M$ [2003]

$P (-1, 4) \ and \ Q (5, -2)$

The equation of the line:

$y - 4 = \frac{-2-4}{5-(-1)} (x+1)$

$\Rightarrow y-4 = -1 (x+1)$

$\Rightarrow y+x=3$

The $x-intercept A = (3,0)$ and the $y-intercept B = (0,3)$

The coordinate of $M = (\frac{3+0}{2}, \frac{0+3}{2}) = (\frac{3}{2}, \frac{3}{2})$

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Question 6: $(1, 5) \ and \ (-3, -1)$ are the co-ordinates of vertices $A \ and \ C$ respectively of rhombus $ABCD$ . Find the equations of the diagonals $AC \ and \ BD$.

Midpoint $M = (\frac{-3+1}{2}, \frac{-1+5}{2}) = (-1, 2)$

Slope of $AC = \frac{-1-5}{-3-1} = \frac{3}{2}$

Equation of  $AC$:

$y - 5 = \frac{3}{2} (x-1) \Rightarrow 3x-2y+7=0$

Slope of $BD = -\frac{2}{3}$

Therefore equation of $BD$:

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

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Question 7: Show that $A (3, 2), B (6, -2) \ and \ C (2, -5)$ can be vertices of a square.

• Find the co-ordinates of its fourth vertex $D$ , if $ABCD$ is a square
• Without using the co-ordinates of vertex $D$ , find the equation of side $AD$ of the square and the equation of diagonal $BD$ .

Mid point of $AC =$ $(\frac{3+2}{2}, \frac{-5+2}{2}) = (\frac{5}{2}, \frac{-3}{2})$

Let the coordinate of $D$ be $(x, y)$

In a square, the diagonals bisect each other. Therefore

$\frac{x+6}{2}= \frac{5}{2} \Rightarrow x = -1$

$\frac{y-2}{2}= \frac{-3}{2} \Rightarrow y = -1$

Hence $D$ is $(-1, -1)$

Slope  of $AB = \frac{-2-2}{6-3} = \frac{-4}{3}$

Since $AB \perp AD$, slope of $AD = \frac{3}{4}$

Hence the equation of $AD$:

$y - 2 = \frac{3}{4} (x-3) \Rightarrow 4y-3x+1 = 0$

Slope of $BD = \frac{-2+1}{6+1} = -\frac{1}{7}$

Hence the equation of $BD$:

$y +2 = -\frac{1}{7}(x-6) \Rightarrow 7y+x+8=0$

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Question 8: A line through origin meets the line $x = 3y + 2$ at right angles at point $X$ . find the co-ordinates of point $X$ .

Given $x = 3y + 2$ … … … … (i)

Slope of line $x = 3y + 2$ is $\frac{1}{3}$

Slope of perpendicular $= -3$

The equation of a line passing through $(0,0)$ and having slope $3$ is

$y - 0 = -3 (x-0) \Rightarrow y = -3x$  … … … … (i)

Solving equations (i) and (ii)

$x= 3 (-3x) + 2 \Rightarrow x = \frac{1}{5}$

Hence $y = -\frac{3}{5}$

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Question 9: A straight line passes through the point $(3, 2)$ and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Let y-intercept be $(0, y)$ and x-intercept be $(x, 0)$

Given $(3,2)$ is the mid point of $(0,y)$ and $(x,0)$. Therefore:

$y-intercept = (0, 4)$

$x-intercept = (6, 0)$

Slope of line $= \frac{0-4}{6-0} = -\frac{2}{3}$

Equation of line:

$y-0 = -\frac{2}{3} (x-6) \Rightarrow 3y+2x=12$

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Question 10: Find the equation of the line passing through the point of intersection of $7x + 6y = 71 \ and \ 5x - 8y = -23$ ; and perpendicular to the line $4x -2y = 1$.

Solve equations

$7x + 6y = 71$   … … … … (i)

$5x - 8y = -23$   … … … … (ii)

Multiply (i) by 4 and (ii) by 3 and then add the equations, we get

$43x = 215 \Rightarrow x = 5$

Substituting $x = 5$ in (i) we get

$7(5)+6y=71 \Rightarrow y = 6$

Therefore the intercept is $(5, 6)$

Sloe of line $4x -2y = 1$ is $2$

Therefore the slope of perpendicular $= -\frac{1}{2}$

Hence the equation of the perpendicular:

$y - 6 = -\frac{1}{2} (x-5)$

$2y -12 = -x+5$

$2y+x=17$

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