`(1, -3), 4x - 3y = -7` Find the distance between the point and the line.

Given (1, -3) `4x-3y=-7`

`4x-3y=-7` 

`4x-3y+7=0`

Let A=4, B=-3, C=7, x1=1, 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)(1)+(-3)(-3)+7|/sqrt(4^2+(-3)^2)`

`d=|4+9+7|/sqrt(16+9)=20/5=4`

The distance between the point and the line is 4 units.