Question

Find the distance of the point (2 3) from the line 2x-3y+9=0 measured along the line x-y+1=0

Answer 1

Answer:

The distance of the point (2 3) from the line 2x-3y+9=0 measured along the line x-y+1=0 is $4\sqrt{2}$ units.

Step-by-step explanation:

The given point is (2,3).

The given lines are

$2x-3y+9=0$     .... (1)

$x-y+1=0$           ....(2)

Multiply equation 2 by 3.

$3x-3y+3=0$        ... (3)

Subtract equation (3) from (1).

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

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

$-x+6=0$

$x=6$

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:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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

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

Therefore the distance of the point (2 3) from the line 2x-3y+9=0 measured along the line x-y+1=0 is $4\sqrt{2}$ units.

Answer 2

Answer:

Yo bro here is your answer

Step-by-step explanation:

All steps in the attachement