The equation of the line segment through A and B is given as -7x + 3y = -21.5
In standard form, 3y = 7x - 21.5 y = (7/3)x - 7.1667 Line AB has a slope of 7/3.
Let the equation of line segment PQ be y = mx + b
Because line segments AB and PQ are perpendicular, therefore (7/3)*m = -1 m = -3/7
The equation of PQ is y = -(3/7)x + b To find b, note that the line passes through the point (7,6). Therefore 6 = -(3/7)*7 + b 6 = -3 + b b = 9 The equation of PQ is y = - (3/7)x + 9 or 7y = -3x + 63 3x + 7y = 63