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

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

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

Answer:

Given line: 3x-4y+5=0

Therefore the line parallel to the given line is:

3x-4y+ \lambda =0    … … … … … i)

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

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

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

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

Answer:

Given line: x-3y+5=0

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

Therefore the line perpendicular to the given line is:

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

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

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

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

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Question 3: Find the equation of the perpendicular bisector of the line joining the points (1,3) and (3,1) .

Answer:

Let A( 1,3) and B(3, 1)

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

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

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

Hence the equation of the perpendicular bisector:

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

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

Answer:

Please refer to the adjoining figure.

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

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

Hence the equation of AD:

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

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

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

Hence the equation of BE:

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

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

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

Hence the equation of CF:

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

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

Answer:

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

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

Therefore the line perpendicular to the given line is:

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

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

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

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

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

Answer:

Given A ( 2, 1) and A'( 5, 2) .

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

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

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

Hence the equation of mirror:

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

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

Answer:

Given line: lx +my +n=0

Comparing lx +my +n=0 with ax+by+c = 0 we get a = l \ \ \ \& \ \ \ b = m

Therefore the line perpendicular to the given line is:

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

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

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

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

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

Answer:

Given line: 2x-3y=5

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

Therefore the line perpendicular to the given line is:

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

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

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

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

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

Answer:

Given line: 5 x -2y =8

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

Therefore the line perpendicular to the given line is:

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

Given points A(2,3) and B(4,5)

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

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

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

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

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

Answer:

Given line: 3 x-4y +11=0

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

Therefore the line perpendicular to the given line is:

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

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

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

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

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Question 11: Find the equation of the right bisector of the line segment joining the points (a,b) and (a_1,b_1) .

Answer:

Given A ( a, b) and B(a_1, b_1) .

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

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

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

Hence the right bisector:

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

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

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

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

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

Answer:

Given A ( 2,1 ) and B(a, b) .

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

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

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

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

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

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

Slope of mirror = - 1

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

\Rightarrow b-1 = a - 2

\Rightarrow 7-a - 1 = a - 2

\Rightarrow 2a = 8

\Rightarrow a = 4

Therefore b = 7 - 4 = 3

Hence the image of the point ( 2, 1) on mirror x + y -5 = 0 is ( 4, 3)

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

Answer:

Given A ( 2, 1) and A'( 5, 2) .

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

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

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

Hence the equation of mirror:

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

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

Answer:

Given line: 3 x -4y + 6 =0

Therefore the line parallel to the given line is:

3x-4y+ \lambda =0    … … … … … i)

Given A ( 2,3) and B(4, -1) .

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

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

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

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

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

Answer:

Given lines:

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

Line 3:    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:    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.

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

Answer:

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

The point of intersection with y axis means x = 0

Therefore for x = 0, y = 6

Hence the point of intersection is ( 0, 6)

Given line: 6x + 4y = 24

Comparing 6x + 4y = 24 with ax+by+c = 0 we get a = 6 \ \ \ \& \ \ \ b = 4

Therefore the line perpendicular to the given line is:

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

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

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

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

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

Answer:

Given line: y =mx + c      … … … … … i)

Therefore the slope of the line = -m

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

Therefore the equation of the required line:

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 ( -1, 2) .

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

Therefore from i)

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

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

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

Answer:

Given A ( 3,4) and B( -1, 2) .

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

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

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

Hence the equation of right bisector:

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

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

Answer:

Given line: 7x - 9y -19 = 0

Slope of this line = \frac{7}{9}

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

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

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

\Rightarrow - 14 = - 36 + 9 h

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

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

Answer:

Given line: x + 3y = 7      … … … … … i)

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

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

Let the image be A' ( a, b)

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

Equation of the perpendicular line:

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

M satisfies both line i) and ii), therefore

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

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

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

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

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

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

Solving iii) and iv) we get

a = -1   and b = -4

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

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

Answer:

Given line: 3x - 4y - 16 = 0      … … … … … i)

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

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

Equation of the perpendicular line:

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 \Big( \frac{68}{25} , \frac{-49}{25} \Big) which is also the foot of the perpendicular from ( -1, 3)

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Question 22: Find the projection of the point (1, 0) on the line joining the points (-1,2) and (5, 4) .

Answer:

Given A (-1,2) and B( 5,4) .

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

Therefore the slope of perpendicular to AB  = -3

Equation of mirror:

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:

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 \Big( \frac{1}{5} , \frac{12}{5} \Big)

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

Answer:

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

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

Therefore the line perpendicular to the given line is:

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

Therefore the equation of the required line:

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

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

Answer:

Given line: 2 x + 3 y = 12      … … … … … i)

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

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

Line ii) passes through (5, 5) , substituting we get

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

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

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

2021-01-29_13-15-42Point of intersection of line iii) and i) is E ( 3, 2)

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

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}

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

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

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

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

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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.

Answer:

Let x and y intercepts be ( 2a, 0) and (0, a)

Therefore the equation of line is:

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

Changing this into normal form

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

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

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

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

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

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

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

Answer:

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

2021-01-29_13-09-02Please refer to the adjoining figure.

Slope of AB = \frac{-1}{1} = -1

Therefore equation of AB :

y - ( -2) = -1 ( x - 1)

\Rightarrow y + 2 = - x + 1

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

Let the coordinate of B be ( x_1, y_1)

Now D is the mid point of AB , therefore

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

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

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

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

B  is also on line iii). Therefore, 

x_1 + y_1 + 1 = 0       … … … … … v)

Solving iv) and v) we get x_1 = -7  and y_1 = 6 

Therefore B ( x_1, y_1) = ( -7, 6) 

Similarly, we calculate C(x_2, y_2) 

Slope of AC = \frac{-1}{(-1/2)} = 2 

Therefore the slope of equation AC  :

y - ( -2) = 2 ( x - 1) 

\Rightarrow y + 2 = 2x - 2 

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

Let E  be the mid point of AC . Therefore,

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

E  lies on line ii). Therefore,

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

\Rightarrow 1 + x_2 - 4 + 2y_2 = 0 

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

Also C  lies on vi)

2x_2 - y_2 - 4 = 0       … … … … … viii)

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

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

Hence, equation of BC  :

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

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

\Rightarrow 14x + 23 y - 40 = 0