ICSE Board Question Based on Equation of a Straight Line Chapter of M.L Aggarwal for class10
In this question find the equation of the line through a point and perpendicular to the line joining the points given in this question.
This is the Question Number 31, Exercise 12.2 of M.L Aggarwal.
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Find the equation of the line through (0, – 3) and perpendicular to the line joining the points (– 3, 2) and (9, 1).
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The slope of the line joining the points (-3, 2) and (9, 1) is
m1 = (1 – 2)/ (9 + 3) = -1/12
Now, let the slope of the line perpendicular to the above line be m2
Then, m1 x m2 = -1
(-1/12) x m2 = -1
m2 = 12
So, the equation of the line passing through (0, -3) and having slope of m2 will be
y – (-3) = 12 (x – 0)
y + 3 = 12x
12x – y = 3
Thus, the required line equation is 12x – y = 3.