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 answer for algebra 1 need help

The equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6) in point slope form is y = x + 8Solution:Given that line m contains points (-3, 4) and (1, 0)We are asked to find the equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6)Let us first find slope of the line "m"Given two points are (-3, 4) and (1, 0)[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex][tex]\left(x_{1}, y_{1}\right)=(-3,4) \text { and }\left(x_{2}, y_{2}\right)=(1,0)[/tex][tex]m=\frac{0-4}{1-(-3)}=-1[/tex]Thus slope of line m is -1We know that product of slope of given line and perpendicular line are always -1So, we get[tex]\begin{array}{l}{\text { slope of line } m \times \text { slope of perpendicular line }=-1} \\\\ {-1 \times \text { slope of perpendicular line }=-1} \\\\ {\text { slope of perpendicular line }=1}\end{array}[/tex]So we have got the slope of perpendicular line is 1 and it passes through (-2, 6)Let us use the point slope form to find the required equationThe point slope form is given as:[tex]y - y_1 = m(x - x_1)[/tex][tex](x_1, y_1) = (-2, 6)[/tex] and m = 1y - 6 = 1(x - (-2))y - 6 = x + 2y = x + 8Thus equation of required line in point slope form is y = x + 8