Note: If $\displaystyle ax+by+c = 0$ is a line, then the equation perpendicular to this line is given by $\displaystyle bx-ay+\lambda=0$ where $\displaystyle \lambda = \text{constant}$

If $\displaystyle ax+by+c = 0$ is a line, then the equation parallel to this line is given by $\displaystyle ax+by+\lambda= 0$ where $\displaystyle \lambda = \text{constant}$.

Question 1: Find the equation of a line passing through the point $\displaystyle (2, 3)$ and parallel to the line $\displaystyle 3x-4y+5=0$.

$\displaystyle \text{Given line: } 3x-4y+5=0$

Therefore the line parallel to the given line is:

$\displaystyle 3x-4y+ \lambda =0$ … … … … … i)

This line passes through $\displaystyle ( 2, 3)$. Therefore substituting this in i) we get

$\displaystyle 3(2) - 4 ( 3) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = 12-6=6$

Therefore the equation of the required line is $\displaystyle 3x-4y+6=0$

$\displaystyle \\$

Question 2: Find the equation of a line passing through $\displaystyle (3, 2)$ and perpendicular to the line $\displaystyle x-3y+5=0$

$\displaystyle \text{Given line: } x-3y+5=0$

$\displaystyle \text{Comparing } x-3y+5=0 \text{ with } ax+by+c = 0 \text{ we get } a = 1 \ \ \ \& \ \ \ b = -3$

Therefore the line perpendicular to the given line is:

$\displaystyle (-3)x-(1) y + \lambda = 0 \hspace{0.5cm} \Rightarrow 3x + y - \lambda = 0$ … … … … … i)

This line passes through $\displaystyle ( 3, -2)$. Therefore substituting this in i) we get

$\displaystyle 3(3) + (-2) - \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = 7$

Therefore the equation of the required line is $\displaystyle 3x + y - 7 = 0$

$\displaystyle \\$

Question 3: Find the equation of the perpendicular bisector of the line joining the points $\displaystyle (1,3) \text{ and } (3,1)$.

$\displaystyle \text{Let } A( 1,3) \text{ and } B(3, 1)$

$\displaystyle \text{Therefore the midpoint of } AB \text{ is } C = \Big( \frac{1+3}{2} , \frac{3+1}{2} \Big) = (2,2)$

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

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

Hence the equation of the perpendicular bisector:

$\displaystyle y - 2 = 1 ( x - 2) \hspace{0.5cm} \Rightarrow x - y = 0$

$\displaystyle \\$

Question 4: Find the equation of the altitude of a $\displaystyle \triangle ABC$ whose vertices are $\displaystyle A(1,4), B(-3,2) \text{ and } C (-5,-3)$.

$\displaystyle \text{Slope of } BC = \frac{-3-2}{-5-(-3)} = \frac{-5}{-2} = \frac{5}{2}$

Therefore the slope of AD $\displaystyle = \frac{-1}{(5/2)} = \frac{-2}{5}$

$\displaystyle y - 4 = \frac{-2}{5} ( x - 1) \hspace{0.5cm} \Rightarrow 5y-20= - 2x + 2 \hspace{0.5cm} \Rightarrow 2x + 5y - 22 = 0$

$\displaystyle \text{Slope of } AC = \frac{4-(-3)}{1-(-5)} = \frac{7}{6}$

Therefore the slope of BE $\displaystyle = \frac{-1}{(7/6)} = \frac{-6}{7}$

Hence the equation of BE:

$\displaystyle y - 2 = \frac{-6}{7} [ x - (-3)] \hspace{0.5cm} \Rightarrow 7y-14= -6x-18 \hspace{0.5cm} \Rightarrow 6x+7y+4 = 0$

$\displaystyle \text{Slope of } AB = \frac{4-2}{1-(-3)} = \frac{2}{4} = \frac{1}{2}$

Therefore the slope of CF $\displaystyle = \frac{-1}{(1/2)} = -2$

Hence the equation of CF:

$\displaystyle y - (-3) = -2 [ x - (-5)] \hspace{0.5cm} \Rightarrow y+3=-2x-10 \hspace{0.5cm} \Rightarrow 2x+y+13 = 0$

$\displaystyle \\$

Question 5: Find the equation of a line which is perpendicular to the line $\displaystyle \sqrt{3}x -y + 5 = 0$ and which cuts off an intercept of $\displaystyle 4$ units with the negative direction of y-axis.

$\displaystyle \text{Given line: } \sqrt{3}x -y + 5 = 0$

$\displaystyle \text{Comparing } \sqrt{3}x -y + 5 = 0 \text{ with } ax+by+c = 0 \text{ we get } a = \sqrt{3} \ \ \ \& \ \ \ b = -1$

Therefore the line perpendicular to the given line is:

$\displaystyle (-1)x-(\sqrt{3}) y + \lambda = 0$ … … … … … i)

This line passes through $\displaystyle ( 0,-4)$. Therefore substituting this in i) we get

$\displaystyle (-1)(0) - \sqrt{3} ( -4) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = -4\sqrt{3}$

Therefore the equation of the required line is $\displaystyle x+ \sqrt{3}y+ 4\sqrt{3} = 0$

$\displaystyle \\$

Question 6: If the image of the point $\displaystyle (2, 1)$ with respect to a line mirror is $\displaystyle (5,2)$ find the equation of the mirror.

$\displaystyle \text{Given } A ( 2, 1) \text{ and } A'( 5, 2)$.

$\displaystyle \text{Therefore the midpoint of } AA' \text{ is } M = \Big( \frac{2+5}{2} , \frac{1+2}{2} \Big) = \Big( \frac{7}{2} , \frac{3}{2} \Big)$

$\displaystyle \text{Slope of } AA' = \frac{2-1}{5-2} = \frac{1}{3}$

$\displaystyle \text{Therefore the slope of mirror } = \frac{-1}{(1/3)} = -3$

Hence the equation of mirror:

$\displaystyle y - ( \frac{3}{2} ) = -3 \Big[ x - \frac{7}{2} \Big] \hspace{0.5cm} \Rightarrow 2y-3=-3(2x-7) \hspace{0.5cm} \Rightarrow 3x+y-12 = 0$

$\displaystyle \\$

Question 7: Find the equation of the straight line through the point $\displaystyle (\alpha, \beta)$ and perpendicular to the line $\displaystyle lx +my +n=0$.

$\displaystyle \text{Given line: } lx +my +n=0$

$\displaystyle \text{Comparing } lx +my +n=0 \text{ with } ax+by+c = 0 \text{ we get } a = l \ \ \ \& \ \ \ b = m$

Therefore the line perpendicular to the given line is:

$\displaystyle mx-l y + \lambda = 0$ … … … … … i)

This line passes through $\displaystyle ( \alpha, \beta )$. Therefore substituting this in i) we get

$\displaystyle m \alpha - l \beta + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = l \beta - m \alpha$

Therefore the equation of the required line is $\displaystyle mx - ly + l \beta - m \alpha = 0 \hspace{0.5cm} \Rightarrow m(x-\alpha) = l ( y - \beta)$

$\displaystyle \\$

Question 8: Find the equation of the straight line perpendicular to $\displaystyle 2x-3y=5$ and cutting off an intercept $\displaystyle L$ on the positive direction of the x-axis.

$\displaystyle \text{Given line: } 2x-3y=5$

$\displaystyle \text{Comparing } 2x-3y=5 \text{ with } ax+by+c = 0 \text{ we get } a = 2 \ \ \ \& \ \ \ b = -3$

Therefore the line perpendicular to the given line is:

$\displaystyle (-3)x-(2) y + \lambda = 0$ … … … … … i)

This line passes through $\displaystyle ( 1,0)$. Therefore substituting this in i) we get

$\displaystyle (-3)(1) -(2) (0) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = 3$

Therefore the equation of the required line is $\displaystyle -3x-2y+3=0 \hspace{0.5cm} \Rightarrow 3x+2y-3=0$

$\displaystyle \\$

Question 9: Find the equation of the straight line perpendicular to $\displaystyle 5 x -2y =8$ and which passes through the mid-point of the line segment joining $\displaystyle (2,3) \text{ and } (4,5)$.

$\displaystyle \text{Given line: } 5 x -2y =8$

$\displaystyle \text{Comparing } 5 x -2y =8 \text{ with } ax+by+c = 0 \text{ we get } a = 5 \ \ \ \& \ \ \ b = -2$

Therefore the line perpendicular to the given line is:

$\displaystyle (-2)x-(5) y + \lambda = 0$ … … … … … i)

Given points $\displaystyle A(2,3) \text{ and } B(4,5)$

$\displaystyle \text{Therefore the midpoint of } AB \text{ is } C = \Big( \frac{2+4}{2} , \frac{3+5}{2} \Big) = ( 3, 4 )$

This line passes through $\displaystyle ( 3,4)$. Therefore substituting this in i) we get

$\displaystyle (-2)(3)-(5) (4) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = 26$

Therefore the equation of the required line is $\displaystyle -2x-5y+26=0 \hspace{0.5cm} \Rightarrow 2x+5y-26=0$

$\displaystyle \\$

Question 10: Find the equation of the straight line line which has y-intercept equal to $\displaystyle 4/3$ and is perpendicular to $\displaystyle 3 x-4y +11=0$.

$\displaystyle \text{Given line: } 3 x-4y +11=0$

$\displaystyle \text{Comparing } 3 x-4y +11=0 \text{ with } ax+by+c = 0 \text{ we get } a = 3 \ \ \ \& \ \ \ b = -4$

Therefore the line perpendicular to the given line is:

$\displaystyle (-4)x-(3) y + \lambda = 0$ … … … … … i)

This line passes through $\displaystyle ( 0, \frac{4}{3} )$. Therefore substituting this in i) we get

$\displaystyle (-4)(0)-(3) \Big( \frac{4}{3} \Big) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = 4$

Therefore the equation of the required line is $\displaystyle -4x-3y+4=0 \hspace{0.5cm} \Rightarrow 4x+3y-4=0$

$\displaystyle \\$

Question 11: Find the equation of the right bisector of the line segment joining the points $\displaystyle (a,b) \text{ and } (a_1,b_1)$.

$\displaystyle \text{Given } A ( a, b) \text{ and } B(a_1, b_1)$.

$\displaystyle \text{Therefore the midpoint of } AB \text{ is } C = \Big( \frac{a+a_1}{2} , \frac{b+b_1}{2} \Big)$

$\displaystyle \text{Slope of } AB = \frac{b_1-b}{a_1-a}$

Therefore the slope of right bisector $\displaystyle = - \frac{a_1-a}{b_1-b}$

Hence the right bisector:

$\displaystyle y - \frac{b+b_1}{2} = - \frac{a_1-a}{b_1-b} \Big[ x - \frac{a+a_1}{2} \Big]$

$\displaystyle \Rightarrow 2y(b_1-b) - ( b+b_1) (b_1-b) = -2x ( a_1-a)+ ( a_1-a)(a+a_1)$

$\displaystyle \Rightarrow 2x ( a_1-a) + 2y(b_1-b) + ( b^2 - {b_1}^2) + ( a^2 - {a_1}^2)= 0$

$\displaystyle \Rightarrow 2x ( a_1-a) + 2y(b_1-b) + ( a^2 + b^2) - ( {a_1}^2+{b_1}^2) = 0$

$\displaystyle \\$

Question 12: Find the image of the point $\displaystyle (2, 1)$ with respect to the line mirror $\displaystyle x + y -5 = 0$.

$\displaystyle \text{Given } A ( 2,1 ) \text{ and } B(a, b)$.

$\displaystyle \text{Therefore the midpoint of } AB \text{ is } C = \Big( \frac{2+a}{2} , \frac{1+b}{2} \Big)$

Mid point C lies on $\displaystyle x + y -5 = 0$, therefore

$\displaystyle \frac{2+a}{2} + \frac{1+b}{2} - 5 = 0$

$\displaystyle \Rightarrow 2 + a + 1 + b - 10 = 0$

$\displaystyle \Rightarrow a + b = 7$ … … … … … i)

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

Slope of mirror $\displaystyle = - 1$

$\displaystyle \text{Therefore } \frac{b-1}{a-2} \times (-1) = -1$

$\displaystyle \Rightarrow b-1 = a - 2$

$\displaystyle \Rightarrow 7-a - 1 = a - 2$

$\displaystyle \Rightarrow 2a = 8$

$\displaystyle \Rightarrow a = 4$

$\displaystyle \text{Therefore } b = 7 - 4 = 3$

Hence the image of the point $\displaystyle ( 2, 1)$ on mirror $\displaystyle x + y -5 = 0 \text{ is } ( 4, 3)$

$\displaystyle \\$

Question 13: If the image of the point $\displaystyle (2, 1)$ with respect to the line mirror be $\displaystyle (5, 2)$ find the equation of the mirror.

$\displaystyle \text{Given } A ( 2, 1) \text{ and } A'( 5, 2)$.

$\displaystyle \text{Therefore the midpoint of } AA' \text{ is } M = \Big( \frac{2+5}{2} , \frac{1+2}{2} \Big) = \Big( \frac{7}{2} , \frac{3}{2} \Big)$

$\displaystyle \text{Slope of } AA' = \frac{2-1}{5-2} = \frac{1}{3}$

$\displaystyle \text{Therefore the slope of mirror } = \frac{-1}{(1/3)} = -3$

Hence the equation of mirror:

$\displaystyle y - \Big( \frac{3}{2} \Big) = -3 \Big[ x - \frac{7}{2} \Big] \hspace{0.5cm} \Rightarrow 2y-3=-3(2x-7) \hspace{0.5cm} \Rightarrow 3x+y-12 = 0$

$\displaystyle \\$

Question 14: Find the equation to the straight line parallel to $\displaystyle 3 x -4y + 6 =0$ and passing through the middle point of the join of points $\displaystyle (2, 3) \text{ and } ( 4, - 1)$.

$\displaystyle \text{Given line: } 3 x -4y + 6 =0$

Therefore the line parallel to the given line is:

$\displaystyle 3x-4y+ \lambda =0$ … … … … … i)

$\displaystyle \text{Given } A ( 2,3) \text{ and } B(4, -1)$.

$\displaystyle \text{Therefore the midpoint of } AB \text{ is } C = \Big( \frac{2+4}{2} , \frac{3-1}{2} \Big) = ( 3, 1)$

This line passes through $\displaystyle ( 3,1 )$. Therefore substituting this in i) we get

$\displaystyle 3(3) - 4 ( 1) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = -5$

Therefore the equation of the required line is $\displaystyle 3x-4y-5=0$

$\displaystyle \\$

Question 15: Prove that the lines $\displaystyle 2x-3y+1=0, \ \ x+y=3, \ \ 2x-3y=2 \text{ and } x+y=4$ form a parallelogram

Given lines:

Line 1: $\displaystyle 2x-3y+1=0 \hspace{0.5cm} \Rightarrow y = \frac{2}{3} x+ \frac{1}{3} {\hspace{0.5cm} \Rightarrow \text{Slope} = } \frac{2}{3}$

Line 2: $\displaystyle x+y=3 \hspace{0.5cm} \Rightarrow y = -x+ 2 \hspace{0.5cm} \Rightarrow \text{Slope} = -1$

Line 3: $\displaystyle 2x-3y=2 \hspace{0.5cm} \Rightarrow y = \frac{2}{3} x- \frac{2}{3} {\hspace{0.5cm} \Rightarrow \text{Slope} =} \frac{2}{3}$

Line 4: $\displaystyle x+y=4 \hspace{0.5cm} \Rightarrow y = -x+ 4 \hspace{0.5cm} \Rightarrow \text{Slope} = -1$

Therefore slope of Line 1 and Line 3 is equal. Therefore the lines are parallel to each other.

Similarly, slope of Line 2 and Line 4 is equal. Therefore the lines are parallel to each other.

Hence the four lines will form a parallogram.

$\displaystyle \\$

Question 16: Find the equation of a line drawn perpendicular to the line $\displaystyle \frac{x}{4} + \frac{y}{6} =1$ through the point where it meets the y-axis.

$\displaystyle \text{Given line: } \frac{x}{4} + \frac{y}{6} =1 \Rightarrow 6x + 4y = 24$

The point of intersection with y axis means $\displaystyle x = 0$

Therefore for $\displaystyle x = 0, y = 6$

Hence the point of intersection is $\displaystyle ( 0, 6)$

$\displaystyle \text{Given line: } 6x + 4y = 24$

$\displaystyle \text{Comparing } 6x + 4y = 24 \text{ with } ax+by+c = 0 \text{ we get } a = 6 \ \ \ \& \ \ \ b = 4$

Therefore the line perpendicular to the given line is:

$\displaystyle (4)x-(6) y + \lambda = 0 \hspace{0.5cm} \Rightarrow 4x -6y + \lambda = 0$ … … … … … i)

This line passes through $\displaystyle ( 0,6)$. Therefore substituting this in i) we get

$\displaystyle 4(0) - 6(6) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = 36$

Therefore the equation of the required line is $\displaystyle 4x -6 y + 36 = 0 \Rightarrow 2x - 3y + 18 = 0$

$\displaystyle \\$

Question 17: The perpendicular from the origin to the line $\displaystyle y =mx + c$ meets it at the point $\displaystyle (-1, 2$). Find the values of $\displaystyle m \text{ and } c$.

$\displaystyle \text{Given line: } y =mx + c$ … … … … … i)

Therefore the slope of the line $\displaystyle = -m$

Hence the slope of the line perpendicular to the given line $\displaystyle = - \frac{1}{m}$

Therefore the equation of the required line:

$\displaystyle y - 0 = - \frac{1}{m} ( x - 0) \hspace{0.5cm} \Rightarrow my + x = 0$ … … … … … ii)

The point of intersection of line i) and ii) is given as $\displaystyle ( -1, 2)$.

Substituting $\displaystyle x =1$ in equation ii) we get $\displaystyle m = \frac{1}{2}$

Therefore from i)

$\displaystyle c = 2 + \frac{1}{2} = \frac{5}{2}$

Hence, $\displaystyle m = \frac{1}{2} \text{ and } c = \frac{5}{2}$

Question 18: Find the equation of the right bisector of the given line: segment, joining the points $\displaystyle (3, 4) \text{ and } ( -1, 2)$.

$\displaystyle \text{Given } A ( 3,4) \text{ and } B( -1, 2)$.

$\displaystyle \text{Therefore the midpoint of } AB \text{ is } C = \Big( \frac{3-1}{2} , \frac{4+2}{2} \Big) = ( 1,3 )$

$\displaystyle \text{Slope of } AB = \frac{2-4}{-1-3} = \frac{-2}{-4} = \frac{1}{2}$

Therefore the slope of right bisector $\displaystyle = \frac{-1}{(1/2)} = 2$

Hence the equation of right bisector:

$\displaystyle y - 3 = -2 (x-1) \hspace{0.5cm} \Rightarrow 2x+y-5=0$

$\displaystyle \\$

Question 19: The line through $\displaystyle (h, 3) \text{ and } (4, 1)$ intersects the line $\displaystyle 7x - 9y -19 = 0$ at right angle. Find the value of $\displaystyle h$.

$\displaystyle \text{Given line: } 7x - 9y -19 = 0$

$\displaystyle \text{Slope of this line } = \frac{7}{9}$

$\displaystyle \text{Therefore slope of a line perpendicular to this line } = \frac{-1}{(7/9)} = \frac{-9}{7}$

$\displaystyle \text{Slope of } A(h, 3) \text{ and } B( 4, 1) = \frac{1-3}{4-h} = \frac{-2}{4-h}$

$\displaystyle \therefore \frac{-9}{7} = \frac{-2}{4-h}$

$\displaystyle \Rightarrow - 14 = - 36 + 9 h$

$\displaystyle \Rightarrow h = \frac{22}{9}$

$\displaystyle \\$

Question 20: Find the image of the point $\displaystyle (3, 8)$ with respect to the line $\displaystyle x + 3y = 7$ assuming the line to be a plane mirror.

$\displaystyle \text{Given line: } x + 3y = 7$ … … … … … i)

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

Therefore slope of a line perpendicular to mirror $\displaystyle = \frac{-1}{(-1/3)} = 3$

Let the image be $\displaystyle A' ( a, b)$

$\displaystyle \text{Therefore the midpoint of } AA" \text{ is } M = \Big( \frac{3+a}{2} , \frac{8+b}{2} \Big)$

Equation of the perpendicular line:

$\displaystyle y - 8 = 3 ( x - 3) \hspace{0.5cm} \Rightarrow 3x - y - 1 = 0$ … … … … … ii)

$\displaystyle M$ satisfies both line i) and ii), therefore

$\displaystyle \frac{3+a}{2} +3 \Big( \frac{8+b}{2} \Big) = 7$

$\displaystyle \Rightarrow 3 +a + 24 + 3b = 14$

$\displaystyle \Rightarrow a + 3b + 13 = 0$ … … … … … iii)

Also, $\displaystyle 3 \Big( \frac{3+a}{2} \Big) - \frac{8+b}{2} -1 = 0$

$\displaystyle \Rightarrow 9 + 3a - 8 - b - 2 = 0$

$\displaystyle \Rightarrow 3a-b-1=0$ … … … … … iv)

Solving iii) and iv) we get

$\displaystyle a = -1 \text{ and } b = -4$

Therefore the image of the point $\displaystyle ( 3, 8)$ with respect to line $\displaystyle x + 3y = 7 \text{ is } ( -1, -4)$.

$\displaystyle \\$

Question 21: Find the coordinates of the foot of the perpendicular from the point $\displaystyle (-1,3)$ to the line $\displaystyle 3x - 4y - 16 = 0$.

$\displaystyle \text{Given line: } 3x - 4y - 16 = 0$ … … … … … i)

Slope of the line $\displaystyle = - \frac{3}{4}$

Therefore slope of a line perpendicular $\displaystyle = \frac{-1}{(3/4)} = - \frac{4}{3}$

Equation of the perpendicular line:

$\displaystyle y - 3 = - \frac{4}{3} [ x - (-1)] \hspace{0.5cm} \Rightarrow 3y-9=-4x-4 \hspace{0.5cm} \Rightarrow 4x+3y-5=0$ … … … ii)

Solving i) and ii) we get the point of intersection which is $\displaystyle \Big( \frac{68}{25} , \frac{-49}{25} \Big)$ which is also the foot of the perpendicular from $\displaystyle ( -1, 3)$

$\displaystyle \\$

Question 22: Find the projection of the point $\displaystyle (1, 0)$ on the line joining the points $\displaystyle (-1,2) \text{ and } (5, 4)$.

$\displaystyle \text{Given } A (-1,2) \text{ and } B( 5,4)$.

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

Therefore the slope of perpendicular to $\displaystyle AB = -3$

Equation of mirror:

$\displaystyle y - 4 = \frac{1}{3} (x-5) \hspace{0.5cm} \Rightarrow 3y-12=x-5 \hspace{0.5cm} \Rightarrow x-3y+7=0$ … … … … … i)

Equation of perpendicular:

$\displaystyle y - 0 = - 3 ( x - 1) \hspace{0.5cm} \Rightarrow 3x+y - 3 = 0$ … … … … … ii)

Solving i) and ii) we get the point of intersection which is $\displaystyle \Big( \frac{1}{5} , \frac{12}{5} \Big)$

$\displaystyle \\$

Question 23: Find the equation of a line perpendicular to the line $\displaystyle \sqrt{3} x - y + 5 = 0$ and at a distance of $\displaystyle 3$, units from the origin.

$\displaystyle \text{Given line: } \sqrt{3} x - y + 5 = 0$

$\displaystyle \text{Comparing } \sqrt{3} x - y + 5 = 0 \text{ with } ax+by+c = 0 \text{ we get } a = \sqrt{3} \ \ \ \& \ \ \ b = -1$

Therefore the line perpendicular to the given line is:

$\displaystyle (-1)x-(\sqrt{3}) y + \lambda = 0 \hspace{0.5cm} \Rightarrow x + \sqrt{3} y - \lambda = 0$ … … … … … i)

Given that line i) is 3 units

Therefore , $\displaystyle \frac{-\lambda}{\sqrt{1+3}} = 3 \hspace{0.5cm} \Rightarrow \lambda = \pm 6$

Therefore the equation of the required line:

$\displaystyle x + \sqrt{3} y \pm 6 = 0$

$\displaystyle \\$

Question 24: The line $\displaystyle 2 x + 3 y = 12$ meets the x-axis at $\displaystyle A$ and y-axis at $\displaystyle B$. The line through $\displaystyle (5, 5)$ perpendicular to $\displaystyle AB$ meets the x-axis and the line $\displaystyle AB$ at $\displaystyle C \text{ and } E$ respectively. If $\displaystyle O$ is the origin of coordinates, find the area of figure $\displaystyle OCEB$.

$\displaystyle \text{Given line: } 2 x + 3 y = 12$ … … … … … i)

The x and y intercept are $\displaystyle A(6,0) \ \ \& \ \ B(0, 4)$ respectively.

Line perpendicular to i) is $\displaystyle 3x-2y+ \lambda = 0$ … … … … … ii)

Line ii) passes through $\displaystyle (5, 5)$, substituting we get

$\displaystyle 3(5) - 2(5) + \lambda = 0 \hspace{0.5cm} \Rightarrow \lambda = - 5$

Therefore equation of the perpendicular is $\displaystyle 3x-2y-5=0$ … … … … … iii)

Line iii) meets x-axis at $\displaystyle C \Big( \frac{5}{3} ,0 \Big)$

Point of intersection of line iii) and i) is $\displaystyle E ( 3, 2)$

Area of $\displaystyle OCEB =$ Area $\displaystyle \triangle BCO +$ Area $\displaystyle \triangle BCE$

$\displaystyle CE = \sqrt{ ( 3 - \frac{5}{3})^2 + ( 2-0)^2 } = \sqrt{ (\frac{4}{3})^2 + 4 } = \frac{\sqrt{52}}{3} = \frac{2\sqrt{13}}{3}$

$\displaystyle BE = \sqrt{ (3-0)^2 + ( 2-4)^2 } = \sqrt{9+14} = \sqrt{13}$

Therefore Area of $\displaystyle \triangle BCE = \frac{1}{2} \times \frac{2\sqrt{13}}{3} \times \sqrt{13} = \frac{13}{3}$

Area of $\displaystyle \triangle BCO = \frac{1}{2} \times \frac{5}{3} \times 4 = \frac{10}{3}$

Therefore the area of $\displaystyle OCEB = \frac{13}{3} + \frac{10}{3} = \frac{26}{3}$ sq. units.

$\displaystyle \\$

Question 25: Find the equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin.

$\displaystyle \text{Let } x \text{ and } y$ intercepts be $\displaystyle ( 2a, 0) \text{ and } (0, a)$

Therefore the equation of line is:

$\displaystyle \frac{x}{2a} + \frac{y}{a} = 1 \hspace{0.5cm} \Rightarrow x + 2y = 2a$

Changing this into normal form

$\displaystyle \frac{x}{\sqrt{1^2+2^2}} + \frac{2y}{\sqrt{1^2+2^2}} = \frac{2a}{\sqrt{1^2+2^2}}$

$\displaystyle \frac{x}{\sqrt{5}} + \frac{2y}{\sqrt{5}} = \frac{2a}{\sqrt{5}}$

$\displaystyle \text{Therefore } p = \frac{2a}{\sqrt{5}}$ ( this is the length of perpendicular from origin)

$\displaystyle \text{Given } p = 1 \hspace{0.5cm} \Rightarrow \frac{2a}{\sqrt{5}} = 1 \hspace{0.5cm} \Rightarrow a = \pm \frac{\sqrt{5}}{2}$

$\displaystyle \pm$ sign is because the line could be on both sides of origin.

Therefore the equation of the required line is $\displaystyle x + 2y \pm \sqrt{5} = 0$.

$\displaystyle \\$

Question 26: The equations of perpendicular bisectors of the sides $\displaystyle AB \text{ and } AC$ of a $\displaystyle \triangle ABC$ are $\displaystyle x-y+5=0 \text{ and } x +2y = 0$ respectively. If the point $\displaystyle A \text{ is } (1,-2)$ find the equation of the line.

Given lines: $\displaystyle x-y+5=0$ … … … … … i) $\displaystyle x +2y = 0$ … … … … … ii)

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

Therefore equation of $\displaystyle AB$:

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

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

$\displaystyle \Rightarrow x + y + 1 =0$ … … … … … iii)

Let the coordinate of $\displaystyle B$ be $\displaystyle ( x_1, y_1)$

Now $\displaystyle D$ is the mid point of $\displaystyle AB$, therefore

$\displaystyle D = \Big( \frac{1+x_1}{2} , \frac{-2+y_1}{2} \Big). \ \ \ D$ lies on equation i)

$\displaystyle \text{Therefore } \frac{1+x_1}{2} - \Big( \frac{-2+y_1}{2} \Big) + 5 = 0$

$\displaystyle \Rightarrow 1 + x_1 + 2 - y_1 + 10 = 0$

$\displaystyle \Rightarrow x_1 - y_1 + 13 = 0$ … … … … … iv)

$\displaystyle B$ is also on line iii). Therefore,

$\displaystyle x_1 + y_1 + 1 = 0$ … … … … … v)

Solving iv) and v) we get $\displaystyle x_1 = -7 \text{ and } y_1 = 6$

$\displaystyle \text{Therefore } B ( x_1, y_1) = ( -7, 6)$

Similarly, we calculate $\displaystyle C(x_2, y_2)$

$\displaystyle \text{Slope of } AC = \frac{-1}{(-1/2)} = 2$

Therefore the slope of equation $\displaystyle AC$:

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

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

$\displaystyle \Rightarrow 2x - y - 4 = 0$ … … … … … vi)

$\displaystyle \text{Let } E$ be the mid point of $\displaystyle AC$. Therefore,

$\displaystyle E = \Big( \frac{1+x_2}{2} , \frac{-2+y_2}{2} \Big)$

$\displaystyle E$ lies on line ii). Therefore,

$\displaystyle \frac{1+x_2}{2} + 2 \Big( \frac{-2+y_2}{2} \Big) = 0$

$\displaystyle \Rightarrow 1 + x_2 - 4 + 2y_2 = 0$

$\displaystyle \Rightarrow x_2 + 2 y_2 - 3 = 0$ … … … … … vii)

Also $\displaystyle C$ lies on vi)

$\displaystyle 2x_2 - y_2 - 4 = 0$ … … … … … viii)

Solving vii) and viii) we get $\displaystyle x_2 = \frac{11}{5}$ , and $\displaystyle y_2 = \frac{2}{5}$

$\displaystyle \text{Therefore } C (x_2, y_2) = \Big( \frac{11}{5} , \frac{2}{5} \Big)$

Hence, equation of $\displaystyle BC$:

$\displaystyle y - 6 = \Bigg( \frac{\frac{2}{5} - 6}{\frac{11}{5}+7} \Bigg) ( x + 7)$

$\displaystyle \Rightarrow y - 6 = \frac{-28}{46} ( x+7)$

$\displaystyle \Rightarrow 14x + 23 y - 40 = 0$