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
Answer: 3x + 7y = 63
-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
Answer: 3x + 7y = 63