Question

Find the length of the perpendicular
from the point (4,-7) to the line
joining the origin and the point of
intersection of the lines 2x-3y+14=0
and 5x+4y-7=0.​

Answer 1

The length of the perpendicular from the point (4, - 7) to the line joining the origin and the point of intersection of the lines 2x - 3y + 14 = 0 and 5x + 4y - 7 = 0 is 1 unit.

Step-by-step explanation:

Step 1. Finding the intersection of the given lines

The given lines are

2x - 3y + 14 = 0 ..... (1)

5x + 4y - 7 = 0 ..... (2)

Multiplying (1) by 5 and (2) by 2, we get

10x - 15y + 70 = 0

10x + 8y - 14 = 0

On subtraction, we get

- 15y - 8y + 70 + 14 = 0

or, - 23y + 84 = 0

or, y = 84/23

Putting y = 84/23 in (1), we have

2x - 3 (84/23) + 14 = 0

or, 2x - 252/23 + 14 = 0

or, 2x + 70/23 = 0

or, x = - 35/23

Thus the intersection of the lines (1) and (2) is (- 35/23, 84/23)

Step 2. Finding the line joining the origin (0, 0) and the point (- 35/23, 84/23)

The line joining the points (0, 0) and (- 35/23, 84/23) is

(y - 84/23)/(84/23 - 0) = (x + 35/23)/(- 35/23 - 0)

or, (23y - 84)/84 = (23x + 35)/(- 35)

or, (23y - 84)/12 = - (23x + 35)/5

or, 5 (23y - 84) = - 12 (23x + 35)

or, 23 (5y) - 420 = - 23 (12x) - 420

or, 5y = - 12x

or, 12x + 5y = 0

Step 3. Finding required distance of (4, - 7) from 12x + 5y = 0

Therefore the perpendicular distance of the point (4, - 7) from the line 12x + 5y = 0 is

= | (12 * 4) + 5 (- 7) |/√(12² + 5²) unit

= | 48 - 35 |/√(144 + 25) unit

= | 13 |/√169 unit

= 13/13 unit

= 1 unit