Question

PROBLEMS ON DISTANCES
The distance of the point (2,3) from the
2x-3y+9=0 measured along a line x-y+1=0 is

​

Answer 1

The given point is (2,3).

The given lines are

2x - 3y + 9 = 0                                                        ... (1)

T- y+1=0                                                                 ...(2)

Multiply equation 2 by 3.

3x - 3y +3 = 0                                                        ... (3)

Subtract equation (3) from (1).

2x - 3y +9 - (3x - 3y + 3) = 0

2r - 3y +9 - 3x + 3y -3 = 0

- x+6=0

x = 6

Put this value in (1).

2(6) - 3y + 9 =0

21 - 3y = 0

y = 7

Therefore intersection point of both lines is (6,7).

Distance formula:

D = $\sqrt(x2-x1)^2+(y2-y1)^2$

The distance between (2,3) and (6,7) is

D = $\sqrt(6-2)^2+(7-3)^2 = 4\sqrt2$

Therefore the distance of the point (2 3) from the line 2x-3y+9=0 measured

along the line x-y+1=0 is $4\sqrt2$ units.

I HOPE IT WILL HELP YOU.

THANK YOU.