line m contains the points -3,4 and 1,0. write the equation of a line that would be perpendicular to this one and pass through the point -2,6 How do you solve that

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:[tex]y = mx + b[/tex]Where:m: It's the slopeb: It is the cut-off point with the y axisAccording to the statement we have two points through which the line passes:[tex](x_ {1}, y_ {1}): (- 3,4)\\(x_ {2}, y_ {2}) :( 1,0)[/tex]We found the slope:[tex]m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0-4} {1 - (- 3)} = \frac {-4} { 1 + 3} = \frac {-4} {4} = - 1[/tex]By definition, if two lines are perpendicular then the product of their slopes is -1.Thus, a perpendicular line will have a slope: [tex]m_ {2} = \frac {-1} {- 1} = 1[/tex]Thus, the equation will be of the form:[tex]y = x + b[/tex]We substitute the given point and find "b":[tex]6 = -2 + b\\6 + 2 = b\\b = 8[/tex]Finally, the equation is:[tex]y = x + 8[/tex]Answer:[tex]y = x + 8[/tex]