Question 11: Find the equation of the line which is perpendicular to the line $\frac{x}{a}-\frac{y}{b}$ $=1$ at the point where this line meets $y-axis$.

Slope of line $\frac{x}{a}-\frac{y}{b}$ $=1$ is $\frac{b}{a}$

Therefore slope of line perpendicular to given line $= -\frac{a}{b}$

$y-intercept = b$

Therefore the equation of line passing through (0,b) and having slope of $-\frac{a}{b}$ is:

$y - b =$ $-\frac{a}{b} (x-0)$

$by-b^2 = -ax$

$ax+by=b^2$

$\\$

Question 12: $O (0, 0), A (3, 5) \ and \ B (-5, -3)$ are the vertices of a triangle $OAB$. Find:

(i) The equation of the median of triangle $OAB$ through vertex $O$

(ii) The equation of altitude of triangle $OAB$ through vertex $B$

Mid point of $AB =$ $(\frac{-5+3}{2}, \frac{-3+5}{2})$ $= (-1, 1)$

Therefore the equation of median of  $\triangle OAB$ through $O$ is

(i) $y - 0 = \frac{1-0}{-1-0} (x-0) \Rightarrow y = -x \ or \ y+x = 0$

(ii) Slope of $OA =$ $\frac{5-0}{3-0} = \frac{5}{3}$

Slope of line perpendicular to $OA = -\frac{3}{5}$

Therefore the equation of altitude of $\triangle OAB$ through $B$

$y - (-3) = -\frac{3}{5} (x-(-5))$

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

$5y+15 = -3x-15$

$5y+3x+30 = 0$

$\\$

Question 13: Determine whether the line through points $(-2, 3) \ and \ (4, 1)$ is perpendicular to the line $3x = y + 1$ . Does line $3x = y +1$ bisect the line segment joining the two given points?

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

Slope of $3x = y + 1$ is $3$

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

Therefore line passing through $(-2, 3)$ and $(4, 1)$ is perpendicular to $3x = y + 1$

Mid point of $(-2, 3)$ and $(4, 1) = (\frac{4-2}{2}, \frac{1+3}{2}) = (1,2)$

Substituting $(1, 2)$ in $3x=y+1$ we get that it satisfies the equation. Therefore $3x=y+1$ bisects the line joining $(-2, 3)$ and $(4, 1)$

$\\$

Question 14: Given a straight line $x cos \ 30^o+y sin\ 30^o=2$ . Determine the equation of the other line which is parallel to its and passes through $(4, 3)$ .

Given $x cos \ 30^o + y sin \ 30^o = 2$

Slope of this line $=$ $-\frac{ cos \ 30^o}{sin \ 30^o}$ $= -\sqrt{3}$

Equation of line with slope  $-\sqrt{3}$ and passing through $(4, 3)$ is

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

$y+\sqrt{3} x= 4 \sqrt{3}+3$

$\\$

Question 15: Find the value of $k$ such that the line $(k-2)x+(k+3)y-5=0$ is:

(i) Perpendicular to the line $2x - y + 7 = 0$

(ii) Parallel to it.

Given $(k-2)x+(k+3)y-5=0$

Slope of this line $= -\frac{k-2}{k+3}$

Slope of line $2x-y+7 = 0$ is $2$

Slope of line perpendicular to this line $= -\frac{1}{2}$

(i) If perpendicular

$-\frac{1}{2} = -\frac{k-2}{k+3}$

$k+3 = 2k-4$

$k= 7$

(ii) If parallel

$2 = -\frac{k-2}{k+3}$

$2k+6 = -k +2$

$3k = -4 \Rightarrow k = -\frac{4}{3}$

$\\$

Question 16: The vertices of a triangle $ABC$ are $A (0, 5), B (-1, -2) \ and \ C (11, 7)$ . Write down the equation of $BC$ . Find:

(i) The equation of the line through $A$ and perpendicular to $BC$.

(ii) The co-ordinates of the point $P$ , where the perpendicular through $A$ , as obtained in (i.), meets $BC$.

(i) Slope of $BC =$ $\frac{7-(-2)}{11-(-1)} = \frac{9}{4} = \frac{3}{4}$

Slope of line perpendicualr to $BC = -\frac{4}{3}$

Therefore equation of line passing through $A (0,5)$ with slope $-\frac{4}{3}$ is:

$y-5 = -\frac{4}{3} (x-0)$

$3y-15 = -4x$

$3y+4x=15$  … … … … (i)

(ii) Equation of $BC$

$y-7 = \frac{3}{4} (x-11)$

$4y - 28 = 3x - 33$

$4y-3x= -5$   … … … … (ii)

Solving (i) and (ii)  we get $x = 3$ and $latex y = 1 &s=0$.

Therefore $P$ is $(3, 1)$

$\\$

Question 17: From the given figure, find:

(i) The co-ordinates of $A, B, \ and \ C$ .

(ii) The equation of the line through $A$ and parallel to $BC$[2004]

Slope of $BC =$ $\frac{0-2}{3-(-1)} = \frac{-2}{4} = -\frac{1}{2}$

The equation of line parallel to $BC$ and passing through $A(2,3)$

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

$2y-6 = -x+2$

$2y+x=8$

$\\$

Question 18: $P (3, 4), Q (7, -2) \ and \ R (-2, -1)$ are the vertices of triangle $PQR$. Write down the equation of the median of the triangle through $R$[2005]

Mid point of $PQ =$ $(\frac{3+7}{2}, \frac{4-2}{2})$ $= (5,1)$

Therefore equation passing through $(5,1)$ and $P(-2,-1)$ is

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

$y-1 = \frac{2}{7} (x-5)$

$7y-7 = 2x-10$

$7y-2x+3=0$

$\\$

Question 19: $A ((8, -6), B (-4, 2) \ and \ C (0, -10)$ are vertices of a triangle $ABC$ . If $P$ is the mid-point of $AC$ , use co-ordinate geometry to show that $PQ$ is parallel to $BC$. Give a special name to quadrilateral $PBCQ$.

Coordinates of $P =$ $(\frac{-4+8}{2}, \frac{2-6}{2})$ $= (2, -2)$

Coordinates of $Q =$ $(\frac{8+0}{2}, \frac{-6-10}{2})$ $= (4,-8)$

Slope of $PQ =$ $\frac{-8-(-2)}{4-2} = \frac{-6}{2}$ $= -3$

Slope of $BC =$ $\frac{-10-2}{0-(-4)} = \frac{-12}{4}$ $= -3$

Therefore $PB \parallel BC$.

$PQBC$ is a trapezoid.

$\\$

Question 20: A line $AB$ meets the $x-axis$ at point $A$ and $y-axis$ at point $B$ . The point $P (-4, -2)$ divides the line segment $AB$ internally such that $AP : PB = 1 : 2$ . Find:

(i) The co-ordinates of $A \ and \ B$.

(ii) Equation of the line through $P$ and perpendicular to $AB$ .

$AP:PB = 1:2$

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

Therefore $-4 =$ $\frac{1 \times x+2 \times 0}{3}$ $\Rightarrow x = -6$

Similarly,  $-2 =$ $\frac{1 \times 0+2 \times y}{3}$ $\Rightarrow y= -3$

Therefore $B(-6, 0)$ and $A (0, -3)$

(ii)  Slope of  $AB =$ $\frac{-3-0}{0-(-6)} = \frac{-3}{6} = \frac{-1}{2}$

Slope of line perpendicular to $AB = 2$

Therefore the equation of line passing through $P(-4, -2)$ with slope $2$:

$y - (-2) = 2(x-(-4))$

$y+2 = 2(x+4)$

$y = 2x+6$

$\\$