`(2, -3), 4x - 5y = -2` Find the distance between the point and the line.

Given (2, -3) `4x-5y=-2`

`4x-5y=-2`

`4x-5y+2=0`

A=4, B=-5, C=2, x1=2, and y1=-3

Find the distance between the point and the line using the formula

`d=|Ax_1+By_1+C|/sqrt(A^2+B^2)`

`d=|(4)(2)+(-5)(-3)+2|/sqrt(4^2+(-5)^2)`

`d=|8+15+2|/sqrt(16+25)=25/sqrt(41)=3.9`

The distance between the point and the line is 3.9 units.