First, we need to find the slope of the line which passes through #(-2, 4)# and #(-7, 2)#. The slope can be found by using the formula: #m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))#
Where #m# is the slope and (#color(blue)(x_1, y_1)#) and (#color(red)(x_2, y_2)#) are the two points on the line.
Substituting the values from the points in the problem gives:
#m = (color(red)(2) - color(blue)(4))/(color(red)(-7) - color(blue)(-2)) = (color(red)(2) - color(blue)(4))/(color(red)(-7) + color(blue)(2)) = (-2)/-5 = 2/5#
A perpendicular slope is the negative inverse of the original slope. Let's call the perpendicular slope #m_p#.
The we can say: #m_p = -1/m#
Or, for this problem:
#m_p = -1/(2/5) = -5/2#
We can now use the point-slope formula to find the equation of the line passing through #(-1, 3)# with a slope of #-5/2#. The point-slope form of a linear equation is: #(y - color(blue)(y_1)) = color(red)(m)(x - color(blue)(x_1))#
Where #(color(blue)(x_1), color(blue)(y_1))# is a point on the line and #color(red)(m)# is the slope.
Substituting the slope we calculated and the values from the point in the problem gives:
#(y - color(blue)(3)) = color(red)(-5/2)(x - color(blue)(-1))#
#(y - color(blue)(3)) = color(red)(-5/2)(x + color(blue)(1))#
If we want this slope-intercept form we can solve for #y# giving:
#y - color(blue)(3) = (color(red)(-5/2) xx x) + (color(red)(-5/2) xx color(blue)(1))#
#y - color(blue)(3) = -5/2x - 5/2#
#y - color(blue)(3) + 3 = -5/2x - 5/2 + 3#
#y - 0 = -5/2x - 5/2 + (2/2 xx 3)#
#y = -5/2x - 5/2 + 6/2#
#y = -5/2x + 1/2#
