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Question

# At what point do the lines of equations 3x + 2y + 5 = 0 and 2x − 3y − 1 = 0 intersects each other? Find one more point on each of these lines. Prove that these lines are perpendicular.

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Solution

## Intersection point of two lines is given by the coordinates obtained on solving the two linear equations. Given equations of lines are 3x + 2y + 5 = 0 and 2x − 3y − 1 = 0. On solving, we get x = −1, y = −1 Therefore, the intersection point of these lines is (−1, −1). Let the point (x1, y1) lie on the line, 3x + 2y + 5 = 0. ⇒ 3x1 + 2y1 + 5 = 0 Any point on this line can be found out by assuming x to have any value, say 1 and putting it in the above equation. ⇒ 3(1) + 2y1 + 5 = 0 ⇒ 2y1 = −8 ⇒ y1 = −4 Thus, the point (1, −4) lie on the line 3x + 2y + 5 = 0. Similarly, let the point (x2, y2) lie on the line 2x − 3y − 1 = 0. ⇒ 2x1 − 3y1 − 1 = 0 Any point on this line can be found out by assuming x to have any value, say 2 and putting it in the above equation. ⇒ 2(2) − 3y1 − 1 = 0 ⇒ −3y1 = −3 ⇒ y1 = 1 Thus, the point (2, 1) lie on the line 2x − 3y − 1 = 0. Now, let the slope of the line 3x + 2y + 5 = 0 be p and the slope of 2x − 3y − 1 = 0 be q. We have just now found the points lying on these lines. Points (−1, −1) and (1, −4) lie on the line 3x + 2y + 5 = 0. ∴ p = Similarly, points (−1, −1) and (2, 1) lie on the line 2x − 3y − 1 = 0. ∴ q = For two lines to be perpendicular, the product of their slopes should be equal to −1, i.e., pq = −1 pq = × = −1 Thus, the given lines are perpendicular.

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