This question is Based on Equation of a Straight Line Chapter of M.L Aggarwal book for ICSE BOARD for class 10.

Here you have to find the equation of a line perpendicular from a point on a given line and Also find the co-ordinates of the foot of perpendicular.

This is the Question Number 23, Exercise 12.2 of M.L Aggarwal.

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# Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.

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Given line equation: 4x – 3y – 5 = 0

3y = 4x – 5

y = (4/3) x – 5

Slope of the line (m

_{1}) = 4/3Let the slope of the line perpendicular to the given line be m

_{2}Then, m

_{1}x m_{2}= -1(4/3) x m

_{2}= -1m

_{2}= -3/4Now, the equation of the line having slope m

_{2}and passing through the point (1, -2) will bey + 2 = (-3/4) (x – 1)

4y + 8 = -3x + 3

3x + 4y + 5 = 0

Next, for finding the co-ordinates of the foot of the perpendicular which is the point of intersection of the lines

4x – 3y – 5 = 0 …. (1) and

3x + 4y + 5 = 0 …. (2)

On multiplying (1) by 4 and (2) by 3, we get

16x – 12y – 20 = 0

9x + 12y + 15 = 0

Adding we get,

25x – 5 = 0

x = 5/25

x = 1/5

Putting the value of x in (1), we have

4(1/5) – 3y – 5 = 0

4/5 – 3y – 5 = 0

3y = 4/5 – 5 = (4 – 25)/5

3y = -21/5

y = -7/5

Thus, the co-ordinates are (1/5, -7/5)