Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Asked by Pragya Singh | 11 months ago |  77

##### Solution :-

Points are (2, 5) and (-3, 6).

We know that slope, m = $$\dfrac{y_2-y_1}{x_2-x_1}$$

$$\dfrac{(6 –5)}{(-3-2)}$$

= $$\dfrac{1}{-5}$$

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Then, m = ($$-\dfrac{1}{m}$$)

$$\dfrac{-1}{\dfrac{-1}{5}}$$

= 5

We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0)

Then, y – 5 = 5(x – (-3))

y – 5 = 5x + 15

5x + 15 – y + 5 = 0

5x – y + 20 = 0

The equation of the line is 5x – y + 20 = 0

Answered by Abhisek | 11 months ago

### Related Questions

#### Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

#### Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

#### Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.