Question 1: Find the slope and $\displaystyle \text{ y-intercept }$ of the line:

$\displaystyle \text{i) } y = 4 \hspace{1.0cm} \text{ii) } ax - by = 0 \hspace{1.0cm} \text{iii) } 3x - 4y = 5$

$\displaystyle \text{i) } \text{Given equation is } y = 4$

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

$\displaystyle \Rightarrow \text{ Slope } (m) = 0 \text{ and } \text{ y-intercept } = 4$

$\displaystyle \text{ii) } \text{Given equation is } ax - by = 0$

$\displaystyle \Rightarrow by = (a) x$

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

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

$\displaystyle \text{iii) } \text{Given equation is } 3x - 4y = 5$

$\displaystyle \Rightarrow 4y=3x-5$

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

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

$\displaystyle \\$

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

$\displaystyle \text{Given equation is } x - y = 4$

$\displaystyle \Rightarrow y=x-4$

$\displaystyle \Rightarrow \text{ Slope } (m) =1 \text{ and } \text{ y-intercept } = -4$

$\displaystyle m= \tan \theta \Rightarrow \tan \theta = 1 \Rightarrow \theta = 45^{\circ}$

$\displaystyle \\$

Question 3:

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

$\displaystyle \text{ii) } \text{Is the line } x - 3y = 4 \text{ perpendicular to the line } 3x - y = 7 ?$

$\displaystyle \text{iii) } \text{Is the line } 3x + 2y = 5 \text{ parallel to the line } x + 2y = 1 ?$

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

$\displaystyle \text{i) } \text{Given equation is } 3x + 4y + 7 = 0$

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

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

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

$\displaystyle \text{Given equation is } 28x - 21y + 50 = 0$

$\displaystyle \Rightarrow 21y=28x+50$

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

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

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

$\displaystyle \text{ii) } \text{Given equation is } x - 3y = 4$

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

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

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

$\displaystyle \text{Given equation is } 3x - y = 7$

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

$\displaystyle \Rightarrow \text{ Slope } (m_2) = 3$

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

$\displaystyle \text{iii) } \text{Given equation is } 3x + 2y = 5$

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

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

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

$\displaystyle \text{Given equation is } x+2y=1$

$\displaystyle \Rightarrow 2y=-x+1$

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

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

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

$\displaystyle \text{iv) Given } (1, 4)\text{ and } (x, 2) \text{ is } 2$

$\displaystyle \text{Slope : } \frac{2-4}{x-1} =2$

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

$\displaystyle \Rightarrow -2=2x-2$

$\displaystyle \Rightarrow x=0$

$\displaystyle \\$

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

$\displaystyle \text{i) } x + 2y +3 = 0 \hspace{1.0cm} \text{ii) } \frac{x}{2} - \frac{y}{3} -1=0$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{i) } x - \frac{y}{2} +3=0 \hspace{1.0cm} \text{ii) } \frac{x}{3} -2y=4$

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

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

$\displaystyle \Rightarrow y = 2x+6$

$\displaystyle \Rightarrow \text{ Slope } (m_1) = 2$

$\displaystyle \text{Let the slope of line perpendicular to the given line is: } m_2$

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

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

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

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

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

$\displaystyle \text{Let the slope of line perpendicular to the given line is: } m_2$

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

$\displaystyle \\$

Question 6:

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

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

$\displaystyle \text{i) } \text{Given equation is } 2x-by+5=0$

$\displaystyle \Rightarrow by=2x+5$

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

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

$\displaystyle \text{Given equation is } ax+3y=2$

$\displaystyle \Rightarrow 3y=-ax+2$

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

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

$\displaystyle \text{Since they are parallel, } m_1 = m_2$

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

$\displaystyle \Rightarrow ab=-6$

$\displaystyle \text{ii) } \text{Given equation is } mx+3y+7=0$

$\displaystyle \Rightarrow 3y=-mx-7$

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

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

$\displaystyle \text{Given equation is } 5x-ny-3=0$

$\displaystyle \Rightarrow ny=5x-3$

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

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

$\displaystyle \text{Since they are perpendicular, } m_1 . m_2 =-1$

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

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

$\displaystyle \\$

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

$\displaystyle \text{Given equation is } 2x - y + 5 = 0$

$\displaystyle \Rightarrow y=2x+5$

$\displaystyle \Rightarrow \text{ Slope } (m_1) = 2$

$\displaystyle \text{Given equation is } px + 3y = 4$

$\displaystyle \Rightarrow 3y=-px+4$

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

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

$\displaystyle \text{Since they are perpendicular, } m_1 . m_2 =-1$

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \text{Given equation is } 2x - 2y + 3 = 0$

$\displaystyle \Rightarrow 2y=2x+3$

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

$\displaystyle \Rightarrow \text{ Slope } (m) =1$

$\displaystyle m= \tan \theta \Rightarrow \tan \theta = 1 \Rightarrow \theta = 45^{\circ}$

$\displaystyle \\$

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

$\displaystyle \text{Given equation is } 4x + 3y = 9$

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

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

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

$\displaystyle \text{Given equation is } px - 6y + 3=0$

$\displaystyle \Rightarrow 6y=px+3$

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

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

$\displaystyle \text{Since they are parallel, } m_1 = m_2$

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

$\displaystyle \Rightarrow p=-8$

$\displaystyle \\$

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

$\displaystyle \text{Given equation is } y = 3x + 7$

$\displaystyle \Rightarrow \text{ Slope } (m_1) = 3$

$\displaystyle \text{Given equation is } 2y + px = 3$

$\displaystyle \Rightarrow 2y=-px+3$

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

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

$\displaystyle \text{Since they are perpendicular, } m_1 . m_2 =-1$

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

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

$\displaystyle \\$

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

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

$\displaystyle \text{Given equation is } 2x - 4y = 5$

$\displaystyle \Rightarrow 4y=2x-5$

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

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

$\displaystyle \text{Since they are perpendicular, } m_1 . m_2 =-1$

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

$\displaystyle \Rightarrow p=-9$

$\displaystyle \\$

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

$\displaystyle \text{i) x-axis } \hspace{1.0cm} \text{ii) y-axis }$

i) Line parallel to $\displaystyle x-axis$ has a $\displaystyle \text{Slope of } 0$

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1)$ is

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

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

$\displaystyle \Rightarrow y = 7$

ii) Line parallel to $\displaystyle y-axis$ has a $\displaystyle \text{Slope of } \infty$

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1)$ is

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

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

$\displaystyle \Rightarrow x=-5$

$\displaystyle \\$

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

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

i) Given Point $\displaystyle (x_1, y_1)=(5,-3)$

$\displaystyle \text{Given equation is } x - 3y = 4$

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

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

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

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1)$ is

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

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

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

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

ii) Given Point $\displaystyle (x_1, y_1)=(0,1)$

$\displaystyle \text{Given equation is } 3x+2y=8$

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

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

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

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1)$ is

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

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

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

$\displaystyle \Rightarrow 2y+3x=2$

$\displaystyle \\$

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

$\displaystyle \text{Given Point } (x_1, y_1)=(-2,1)$

$\displaystyle \text{Given equation is } 4x + 5y = 6$

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

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

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

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

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1)$ is

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

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

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

$\displaystyle \Rightarrow 4y=5x+14$

$\displaystyle \\$

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

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

Let the coordinates of $\displaystyle P \text{ be } (x, y)$

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

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

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

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1)$ is

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

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

$\displaystyle \Rightarrow y=x-3$

$\displaystyle \\$

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

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

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

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1) \text{ is } y-y_1=m(x-x_1)$

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

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

$\displaystyle \Rightarrow 3y+x=13$

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

$\displaystyle \text{Equation of AC } y-y_1=m(x-x_1)$

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

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

$\displaystyle \Rightarrow y=3x+11$

$\displaystyle \\$

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

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

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

Equation of AC :

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

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

$\displaystyle \Rightarrow 2y=x-11$

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

$\displaystyle \text{ Equation of BD } y-y_1=m(x-x_1)$

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow y+8x=3$

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

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

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

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

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

$\displaystyle \Rightarrow 3y=x+4$

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

Slope of line parallel to this $\displaystyle = 7$

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

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

$\displaystyle \Rightarrow y=7x+18$

$\displaystyle \\$

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

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

i) Given Point $\displaystyle (x_1, y_1)=(3,2)$

$\displaystyle \text{Given equation is } 2y=3x+5$

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

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

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

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1) \text{ is } y-y_1=m(x-x_1)$

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

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

$\displaystyle \Rightarrow 3y+2x=12$

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

$\displaystyle \text{When } y = 0, x = 6 . \text{Therefore } A (6,0)$

$\displaystyle \text{When } c = 0, y =4 . \text{Therefore } B(0,4)$

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

$\displaystyle \\$

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

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

When $\displaystyle y = 0, x = -3$

$\displaystyle \text{Therefore } A(-3,0)$

$\displaystyle \text{Given equation is } 4x - 3y + 12 = 0$

$\displaystyle \Rightarrow 3y=4x+12$

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

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

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

Equation of a line with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1) \text{ is } y-y_1=m(x-x_1)$

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

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

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

$\displaystyle \\$

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

i) The equation of the line $\displaystyle AP$ ii) The co-ordinates of $\displaystyle P$

i) $\displaystyle \text{Given equation is } 2x-3y+18=0$

$\displaystyle \Rightarrow 3y=2x+18$

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

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

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

Equation of AP with $\displaystyle \text{Slope } m$ and passing through $\displaystyle (x_1, y_1) \text{ is } y-y_1=m(x-x_1)$

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

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

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

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

$\displaystyle 2x-3y+18=0 \text{ and } 2y+3x+1=0 .$

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

$\displaystyle \\$

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

$\displaystyle \text{Given points } A(4,0), B(2,2) \text{ and } C(0,6)$

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

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

$\displaystyle \text{Equation of AB with Slope } m \text{ and passing through } \\ \\ (x_1, y_1) \text{ is } y-y_1=m(x-x_1)$

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

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

$\displaystyle \Rightarrow y+x=4$

$\displaystyle \text{Equation of AB with Slope } m \text{ and passing through } \\ \\ (x_1, y_1) \text{ is } y-y_1=m(x-x_1)$

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

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

$\displaystyle \Rightarrow y+2x=6$

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \text{Because } A, B, \text{ and } C \text{are collinear:}$

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

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

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

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

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

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