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# Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5, – 2) and (2, 2).

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An Important Question of class 10 Based on Equation of a Straight Line Chapter of M.L Aggarwal for ICSE BOARD.
Here given that line that is perpendicular to a line and passes through the mid-point of the line segment joining the points given in this question.
This is the Question Number 20, Exercise 12.2 of M.L Aggarwal.

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1. Given line: 3x + 2y – 8 = 0

2y = -3x + 8

y = (-3/2) x + 4

Here, slope (m1) = -3/2

Now, the co-ordinates of the mid-point of the line segment joining the points (5, -2) and (2, 2) will be

((5 + 2)/7, (-2 + 2)/7) = (7/2, 0)

Let’s consider the slope of the line perpendicular to the given line be m2

Then,

m1 x m2 = -1

(-3/2) x m2 = -1

m2 = 2/3

So, the equation of the line with slope m2 and passing through (7/2, 0) will be

y – 0 = (2/3) (x – 7/2)

3y = 2x – 7

2x – 3y – 7 = 0

Thus, the required line equation is 2x – 3y – 7 = 0.

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