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# (i) The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A. (ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0.

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An Important Question of M.L Aggarwal book of class 10 Based on Equation of a Straight Line Chapter for ICSE BOARD.
Here Given that:
(i) A line meets the x-axis at A. Write down the co-ordinates of A.
(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0.
This is the Question Number 18, Exercise 12.2 of M.L Aggarwal.

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1. Given line: 4x – 3y + 12 = 0

(i) When this line meets the x-axis, its y co-ordinate becomes 0.

So, putting y = 0 in the given equation, we get

4x – 3(0) + 12 = 0

4x + 12 = 0

x = -12/4

x = -3

Hence, the line meets the x-axis at A (-3, 0).

(ii) Now, the slope of the line is given by

4x – 3y + 12 = 0

3y = 4x + 12

y = (4/3) x + 4

⇒ m1 = 4/3

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

Then, m1 x m2 = -1

4/3 x m2 = -1

m2 = -3/4

Thus, the equation of the line perpendicular to the given line passing through A will be

y – 0 = -3/4 (x + 3)

4y = -3(x + 3)

3x + 4y + 9 = 0

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