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.

Answer 1

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.