Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0

Asked by Abhisek | 1 year ago |  113

##### Solution :-

Let us consider the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)

So, let the slope of the line joining (-1, 3) and (a, b) be m1

m$$\dfrac{(b-3)}{(a+1)}$$

And let the slope of the line 3x – 4y – 16 = 0 be m2

y = $$\dfrac{3}{4}x$$ – 4

m2 = $$\dfrac{3}{4}$$

Since these two lines are perpendicular, m1 × m2 = -1

$$\dfrac{(b-3)}{(a+1)}$$ × ($$\dfrac{3}{4}$$) = -1

$$\dfrac{(3b-9)}{(4a+4)}$$ = -1

3b – 9 = -4a – 4

4a + 3b = 5 …….(1)

Point (a, b) lies on the line 3x – 4y = 16

3a – 4b = 16 ……..(2)

Solving equations (1) and (2), we get

a = $$\dfrac{68}{25}$$ and b = $$- \dfrac{49}{25}$$

The co-ordinates of the foot of perpendicular is ($$\dfrac{68}{25}$$, $$- \dfrac{49}{25}$$)

Answered by Pragya Singh | 1 year 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.