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  

Answer:

\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.

Answer:

\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 .

Answer:

\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  

Answer:

\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  

Answer:

\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 .

Answer:

\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.

Answer:

\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 .

Answer:

\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 .

Answer:

\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]

Answer:

\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]

Answer:

\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 }   

Answer:

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]

Answer:

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 .

Answer:

\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).

Answer:

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 .

Answer:

\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 .

Answer:

\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 .

Answer:

\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]

Answer:

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  

Answer:

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  

Answer:

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 .

Answer:

\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]

Answer:

\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