Answer:
\(-1.5x-3.5y=-31.5\)
Step-by-step explanation:
we know that
If two lines are perpendicular
then
the product of their slopes is equal to minus one
so
\(m1*m2=-1\)
In this problem line AB and line PQ are perpendicular
Step 1
Find the slope of the line AB
The equation of the line AB is
\(-7x+3y=-21.5\)
isolate the variable y
\(3y=7x-21.5\) ------> \(y=(7/3)x-21.5/3\)
The slope of the line AB is equal to
\(m1=7/3\)
Step 2
Find the slope of the line PQ
remember that
\(m1*m2=-1\)
we have
\(m1=7/3\) ----> slope line AB
so
substitute and solve for m2
\((7/3)*m2=-1\)
\(m2=-3/7\)
Step 3
Find the equation of the line PQ
The equation of the line into point-slope form is equal to
\(y-y1=m(x-x1)\)
we have
\(m=-3/7\)
\(P(7,6)\)
substitute
\(y-6=(-3/7)(x-7)\)
\(y=(-3/7)x+3+6\)
\(y=(-3/7)x+9\) -----> multiply by \(7\) both sides
\(7y=-3x+63\)
\(7y+3x=63\) -----> divide by \(2\) both sides
\(1.5x+3.5y=31.5\) -----> multiply by \(-1\) both sides
\(-1.5x-3.5y=-31.5\)
