Question 1: Find the slope and $y-intercept$ of the line:

i) $y = 4$

ii) $ax - by = 0$

iii) $3x - 4y = 5$

i)     Given equation is $y = 4$

$\Rightarrow y = (0) x+4$

$\Rightarrow Slope (m) = 0$ and $y-intercept = 4$

ii)    Given equation is $ax - by = 0$

$\Rightarrow by = (a) x$

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

$\Rightarrow Slope (m) =$ $\frac{a}{b}$ and $y-intercept = 0$

iii)  Given equation is $3x - 4y = 5$

$\Rightarrow 4y=3x-5$

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

$\Rightarrow Slope (m) =$ $\frac{3}{4}$ and $y-intercept = -\frac{5}{4}$

$\\$

Question 2: The equation of a line is $x - y = 4$ . Find its slope and $y-intercept$ . Also, find its inclination.

Given equation is $x - y = 4$

$\Rightarrow y=x-4$

$\Rightarrow Slope (m) =1$ and $y-intercept = -4$

$m= tan \ \theta \Rightarrow tan \ \theta = 1 \Rightarrow \theta = 45^o$

$\\$

Question 3:

i) Is the line $3x + 4y + 7 = 0$ perpendicular to the line $28x - 21y + 50 = 0$ ?

ii) Is the line $x - 3y = 4$ perpendicular to the line $3x - y = 7$ ?

iii) Is the line $3x + 2y = 5$ parallel to the line $x + 2y = 1$ ?

iv) Determine $x$ so that the slope of the line through $(1, 4) \ and \ (x, 2) \ is \ 2$ .

i)    Given equation is $3x + 4y + 7 = 0$

$\Rightarrow 4y=-3x-7$

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

$\Rightarrow Slope (m_1) = -\frac{3}{4}$

Given equation is $28x - 21y + 50 = 0$

$\Rightarrow 21y=28x+50$

$\Rightarrow y = (\frac{4}{3}) x+\frac{50}{21}$

$\Rightarrow Slope (m_2) = (\frac{4}{3})$

Since $m_1.m_2 = (-\frac{3}{4}).(\frac{4}{3}) = -1$, the two lines are perpendicular to each other.

ii)     Given equation is $x - 3y = 4$

$\Rightarrow 3y = x-4$

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

$\Rightarrow Slope (m_1) = \frac{1}{3}$

Given equation is $3x - y = 7$

$\Rightarrow y=3x-7$

$\Rightarrow Slope (m_2) = 3$

Since $m_1.m_2 = ( \frac{1}{3}).(3) = 1 \neq -1$, the two lines are NOT perpendicular to each other.

iii)    Given equation is $3x + 2y = 5$

$\Rightarrow 2y=-3x+5$

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

$\Rightarrow Slope (m_1) = -\frac{3}{2}$

Given equation is $x+2y=1$

$\Rightarrow 2y=-x+1$

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

$\Rightarrow Slope (m_2) = -\frac{1}{2}$

Since $m_1.m_2 = ( -\frac{3}{2}).( -\frac{1}{2}) = \frac{3}{4} \neq -1$, the two lines are NOT perpendicular to each other.

iv)  Given $(1, 4) \ and \ (x, 2) \ is \ 2$

Slope :  $\frac{2-4}{x-1} =2$

$\Rightarrow -2=2(x-1)$

$\Rightarrow -2=2x-2$

$\Rightarrow x=0$

$\\$

Question 4: Find the slope of the line which is parallel to:

i) $x + 2y +3 = 0$

ii) $\frac{x}{2}-\frac{y}{3}-1=0$

i)     Given equation is $x + 2y +3 = 0$

$\Rightarrow 2y=-x-3$

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

$\Rightarrow Slope (m) = -\frac{1}{2}$

Therefore the slope of line parallel to the given line is $= -\frac{1}{2}$

ii)    Given equation is $\frac{x}{2}-\frac{y}{3}$ $-1=0$

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

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

$\Rightarrow Slope (m) = \frac{3}{2}$

Therefore the slope of line parallel to the given line is $= \frac{3}{2}$

$\\$

Question 5: Find the slope of the line which is perpendicular to:

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

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

i)     Given equation is $x -\frac{y}{2}+3=0$

$\Rightarrow \frac{y}{2} =x+3$

$\Rightarrow y = 2x+6$

$\Rightarrow Slope (m_1) = 2$

Let  the slope of line perpendicular to the given line is: $m_2$

Therefore $m_1.m_2 = -1 \Rightarrow m_2 = \frac{-1}{2} \Rightarrow m_2=\frac{-1}{2}$

ii)   Given equation is $\frac{x}{3}-2y=4$

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

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

$\Rightarrow Slope (m_1) = \frac{1}{6}$

Let  the slope of line perpendicular to the given line is: $m_2$

Therefore $m_1.m_2 = -1 \Rightarrow m_2 = \frac{-1}{\frac{1}{6}} \Rightarrow m_2=6$

$\\$

Question 6:

i) Lines $2x-by+5=0 \ and \ ax+3y=2$ are parallel to each other. Find the relation connecting $a \ and \ b$ .

ii) Lines $mx+3y+7=0 \ and \ 5x-ny-3=0$ are perpendicular to each other. Find the relation connecting $m \ and \ n$ .

i)    Given equation is $2x-by+5=0$

$\Rightarrow by=2x+5$

$\Rightarrow y = \frac{2}{b}x+\frac{5}{b}$

$\Rightarrow Slope (m_1) = \frac{2}{b}$

Given equation is $ax+3y=2$

$\Rightarrow 3y=-ax+2$

$\Rightarrow y = \frac{-a}{3}x+\frac{2}{3}$

$\Rightarrow Slope (m_2) = \frac{-a}{3}$

Since they are parallel, $m_1 = m_2$

$\Rightarrow \frac{2}{b}=\frac{-a}{3}$

$\Rightarrow ab=-6$

ii)    Given equation is $mx+3y+7=0$

$\Rightarrow 3y=-mx-7$

Slope $\Rightarrow y = \frac{-m}{3}x-\frac{7}{3}$

$\Rightarrow (m_1) = -\frac{m}{b}$

Given equation is $5x-ny-3=0$

$\Rightarrow ny=5x-3$

$\Rightarrow y = \frac{5}{n}x-\frac{3}{n}$

$\Rightarrow Slope (m_2) = \frac{5}{n}$

Since they are perpendicular, $m_1 . m_2 =-1$

$\Rightarrow \frac{-m}{3}.\frac{5}{n}=-1$

$\Rightarrow mn=\frac{3}{5}$

$\\$

Question 7: Find the value of $p$ if the lines, whose equations are $2x - y + 5 = 0 \ and \ px + 3y = 4$ are perpendicular to each other.

Given equation is $2x - y + 5 = 0$

$\Rightarrow y=2x+5$

$\Rightarrow Slope (m_1) = 2$

Given equation is $px + 3y = 4$

$\Rightarrow 3y=-px+4$

$\Rightarrow y = \frac{-p}{3}x+\frac{4}{3}$

$\Rightarrow Slope (m_2) = \frac{-p}{3}$

Since they are perpendicular, $m_1 . m_2 =-1$

$\Rightarrow 2. \frac{-p}{3}=-1$

$\Rightarrow p=\frac{3}{2}$

$\\$

Question 8: The equation of a line $AB$ is $2x - 2y + 3 = 0$ .

i) Find the slope of the line $AB$ .

ii) Calculate the angle that the line $AB$ makes with the positive direction of the $x-axis$ .

Given equation is $2x - 2y + 3 = 0$

$\Rightarrow 2y=2x+3$

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

$\Rightarrow Slope (m) =1$

$m= tan \ \theta \Rightarrow tan \ \theta = 1 \Rightarrow \theta = 45^o$

$\\$

Question 9: The lines represented by $4x + 3y = 9 \ and \ px - 6y + 3=0$ are parallel. Find the value of $p$ .

Given equation is $4x + 3y = 9$

$\Rightarrow 3y=-4x+9$

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

$\Rightarrow Slope (m_1) = \frac{-4}{3}$

Given equation is $px - 6y + 3=0$

$\Rightarrow 6y=px+3$

$\Rightarrow y = \frac{p}{6}x+\frac{1}{2}$

$\Rightarrow Slope (m_2) = \frac{p}{6}$

Since they are parallel, $m_1 = m_2$

$\Rightarrow \frac{-4}{3}=\frac{p}{6}$

$\Rightarrow p=-8$

$\\$

Question 10: If the lines $y = 3x + 7 \ and \ 2y + px = 3$  are perpendicular to each other, find the value of $p$. [2006]

Given equation is $y = 3x + 7$

$\Rightarrow Slope (m_1) = 3$

Given equation is $2y + px = 3$

$\Rightarrow 2y=-px+3$

$\Rightarrow y = \frac{-p}{2}x+\frac{3}{2}$

$\Rightarrow Slope (m_2) = \frac{-p}{2}$

Since they are perpendicular, $m_1 . m_2 =-1$

$\Rightarrow 3.\frac{-p}{2}=-1$

$\Rightarrow p=\frac{2}{3}$

$\\$

Question 11: The line through $A (-2, 3) \ and \ B (4, b)$ is perpendicular to the line $2x - 4y = 5$ . Find the value of $b$ [2012]

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

Given equation is $2x - 4y = 5$

$\Rightarrow 4y=2x-5$

$\Rightarrow y = \frac{1}{2}x-\frac{5}{4}$

$\Rightarrow Slope (m_2) = \frac{1}{2}$

Since they are perpendicular, $m_1 . m_2 =-1$

$\Rightarrow \frac{b-3}{6}. \frac{1}{2}=-1$

$\Rightarrow p=-9$

$\\$

Question 12: Find the equation of the line passing through $(-5, 7)$ and parallel to:

i) $x-axis$

ii) $y-axis$

i)     Line parallel to $x-axis$ has a slope of $0$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

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

$\Rightarrow y-7=0(x-5)$

$\Rightarrow y = 7$

ii)    Line parallel to $y-axis$ has a slope of $\infty$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

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

$\Rightarrow y-7= \infty (x-(-5))$

$\Rightarrow x=-5$

$\\$

Question 13:  i) Find the equation of the line passing through $(5, -3)$ and parallel to $x - 3y = 4$ .

ii) Find the equation of the line parallel to the line $3x + 2y = 8$ and passing through the point $(0, 1)$ [2007]

i)    Given Point $(x_1, y_1)=(5,-3)$

Given equation is $x - 3y = 4$

$\Rightarrow 3y=x-4$

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

$\Rightarrow Slope (m) = \frac{1}{3}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

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

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

$\Rightarrow 3y+9=x-5$

$\Rightarrow x-3y-14=0$

ii)   Given Point $(x_1, y_1)=(0,1)$

Given equation is $3x+2y=8$

$\Rightarrow 2y=-3x+8$

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

$\Rightarrow Slope (m) = \frac{-3}{2}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

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

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

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

$\Rightarrow 2y+3x=2$

$\\$

Question 14: Find the equation of the line passing through $(-2, 1)$ and perpendicular to $4x + 5y = 6$ .

Given Point $(x_1, y_1)=(-2,1)$

Given equation is $4x + 5y = 6$

$\Rightarrow 5y=-4x+6$

$\Rightarrow y = \frac{-4}{5}x+\frac{6}{5}$

$\Rightarrow Slope (m) = \frac{-4}{5}$

Therefore slope of the new line $= m = \frac{5}{4}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

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

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

$\Rightarrow 4y-4=5x+10$

$\Rightarrow 4y=5x+14$

$\\$

Question 15: Find the equation of the perpendicular bisector of the line segment obtained on joining the points $(6, -3) \ and \ (0, 3)$.

Let P be the bisector of the points $(6, -3) \ and \ (0, 3)$

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

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

Slope of the line joining the two given points $= \frac{3-(-3)}{0-6}=\frac{6}{-6}=-1$

Therefore the slope of the line perpendicular  to the line joining $(6, -3) \ and \ (0, 3) = \frac{-1}{-1} = 1$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is

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

$\Rightarrow y-0= 1 (x-3)$

$\Rightarrow y=x-3$

$\\$

Question 16:  $B(-5,6) \ and \ D(1,4)$ are the vertices of rhombus $ABCD$ . Find the equations of the diagonals $BD \ and \ AC$ .

Slope of $BD = \frac{6-4}{-5-1}=\frac{2}{-6}=-\frac{1}{3}$

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

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$

$\Rightarrow y-6= -\frac{1}{3} (x-(-5))$

$\Rightarrow 3y-18=-x-5$

$\Rightarrow 3y+x=13$

Midpoint of BD $(x, y) = (\frac{-5+1}{2}, \frac{6+4}{2}) = (-2,5)$

Equation of AC $y-y_1=m(x-x_1)$

$\Rightarrow y-5= 3 (x-(-2))$

$\Rightarrow y-5=3x+6$

$\Rightarrow y=3x+11$

$\\$

Question 17: $A = (7, -2) \ and \ C = (-1, -6)$ are the vertices of a square $ABCD$. Find the equations of the diagonals $AC \ and \ BD$.

Slope of $AC = \frac{-6-(-2)}{-1-(7)}=\frac{-4}{-8}=\frac{1}{2}$

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

Equation of AC :

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

$\Rightarrow 2y+4=x-7$

$\Rightarrow 2y=x-11$

Midpoint of BD $(x, y) = (\frac{-1+7}{2}, \frac{-6-2}{2}) = (3,-4)$

Equation of BD $y-y_1=m(x-x_1)$

$\Rightarrow y-(-4)= -2 (x-3)$

$\Rightarrow y+4=-2x+6$

$\Rightarrow y+2x=2$

$\\$

Question 18:  $A (1, -5), B (2, 2) \ and \ C (-2, 4)$ are the vertices of triangle $ABC$ . Find the equation of:

i) The median of the triangle through $A$ .

ii) The altitude of the triangle through $B$ .

iii) The line through $C$ and parallel to $AB$ .

i)    Midpoint of  $BC = D(x, y) =$ $(\frac{2+2}{2}, \frac{4+2}{2})$ $= (0,3)$

Slope of $AD$ $= \frac{3-(-5)}{0-1}=\frac{8}{-1}$ $=-8$

Equation of BD $y-y_1=m(x-x_1)$

$\Rightarrow y-3= -8 (x-0)$

$\Rightarrow y+8x=3$

ii)    Slope of $AC$ $= \frac{4-(-5)}{-2-1}=\frac{9}{-3}$ $=-3$

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

Equation of line $y-y_1=m(x-x_1)$

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

$\Rightarrow 3y-6=x-2$

$\Rightarrow 3y=x+4$

iii)   Slope of $AB$ $= \frac{2-(-5)}{2-1}=\frac{7}{1}$ $=7$

Slope of line parallel to this $= 7$

Equation of line $y-y_1=m(x-x_1)$

$\Rightarrow y-4= 7(x-(-2))$

$\Rightarrow y=7x+18$

$\\$

Question 19:  i) Write down the equation of the line $AB$ , through $(3, 2)$ and perpendicular to the line $2y = 3x + 5$ .

ii) $AB$ meets the $x-axis \ at \ A$ and the $y-axis$ at $B$ . write down the co-ordinates of $A \ and \ B$. Calculate the area of triangle $OAB$ , where $O$ is origin. [1995]

i)     Given Point $(x_1, y_1)=(3,2)$

Given equation is $2y=3x+5$

$\Rightarrow y = \frac{3}{2}x+5$

$\Rightarrow Slope (m) = \frac{3}{2}$

Therefore slope of the new line $= m = \frac{-1}{\frac{3}{2}}$ $= -\frac{2}{3}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$

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

$\Rightarrow 3y-6=-2x+6$

$\Rightarrow 3y+2x=12$

ii)   Equation of $AB$ is $\Rightarrow 3y+2x=12$

When $y = 0, x = 6$. Therefore $A (6,0)$

When $c = 0, y =4$. Therefore $B(0,4)$

Area of the triangle $= \frac{1}{2} \times 6 \times 4 = 12$ sq. units.

$\\$

Question 20: The line $4x - 3y + 12 = 0$ meets $x-axis \ at \ A$ . write the co-ordinates of $A$ . Determine the equation of the line through $A$ and perpendicular to $4x - 3y + 12 = 0$

Given equation $4x - 3y + 12 = 0$

When $y = 0, x = -3$

Therefore $A(-3,0)$

Given equation is $4x - 3y + 12 = 0$

$\Rightarrow 3y=4x+12$

$\Rightarrow y = \frac{4}{3}x+4$

$\Rightarrow Slope (m) = \frac{4}{3}$

Therefore slope of the line perpendicular to this line $= m = \frac{-1}{\frac{4}{3}} = -\frac{3}{4}$

Equation of a line with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$

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

$\Rightarrow 4y=-3(x+3)$

$\Rightarrow 4y+3x+9=0$

$\\$

Question 21: The point $P$ is the foot of perpendicular from $A (-5, 7)$ to the line whose equation $2x-3y+18=0$ Determine:

i) The equation of the line $AP$

ii) The co-ordinates of $P$

i)    Given equation is $2x-3y+18=0$

$\Rightarrow 3y=2x+18$

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

$\Rightarrow Slope (m) = \frac{2}{3}$

Therefore slope of the line perpendicular to this line $= m = \frac{-1}{\frac{2}{3}} = -\frac{3}{2}$

Equation of AP with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$

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

$\Rightarrow 2y-14=-3x-15$

$\Rightarrow 2y+3x+1=0$

ii)   The coordinate of P is the intersection of the two lines:

$2x-3y+18=0$ and $2y+3x+1=0$.

Solving the two equations, we get $x = -3 \ and \ y = 4$. Therefor $P(-3,4)$

$\\$

Question 22: The points $A, B \ and \ C$ are $(4, 0), (2, 2) \ and \ (0, 6)$ respectively. Find the equations of $AB \ and \ BC$ . If $AB$ cuts the $y-axis$ at $P \ and \ BC$ cuts the $x-axis$ at $Q$ , find the co-ordinates of $P \ and \ Q$ .

Given points $A(4,0), B(2,2) \ and \ C(0,6)$

Slope of $AB$ $= \frac{2-0}{2-4}=\frac{2}{-2}$ $=-1$

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

Equation of AB with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$

$\Rightarrow y-2= -1(x-2)$

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

$\Rightarrow y+x=4$

Equation of AB with slope $m$ and passing through $(x_1, y_1)$ is $y-y_1=m(x-x_1)$

$\Rightarrow y-6= -2(x-0)$

$\Rightarrow y-6=-2x$

$\Rightarrow y+2x=6$

Intercept of $AB$ on $y-axis \ (i.e. x=0) = P(0,4)$

Intercept of $BC$ on $x-axis \ (i.e. y=0) = Q(3, 10)$

$\\$

Question 23: Find the value of a for the points $A (a, 3), B (2, 1) \ and \ C (5, a)$ are collinear. Hence, find the equation of the line. [2014]

Given points $A (a, 3), B (2, 1) \ and \ C (5, a)$

Slope of $AB$ $= \frac{1-3}{2-a}=\frac{-2}{2-a}$

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

Because $A, B, \ and \ C$ are collinear:

$\frac{-2}{2-a}=\frac{a-1}{3}$

$-6=(2-a)(a-1)$

$-6=2a-2-a^2+a$

$-6=3a-a^2-2$

$a^2-3a-4=0$

$(a-4)(a+1)=0 \Rightarrow a=4, \ or \ -1$

$\\$