Question

point slope form and slope intercept form.
A line that passes through point A(5,8) and is perpendicular to the line passing through points B(-8,7) and C(-2,-1)

Answer 1

y = mx + b

Answer 2

Line given by points B[-8; 7] and C[-2; -1] has canonical form: 8x+6y+22 = 0

thus (8, 6) is normal vector of line BC and direct vector of line through A.
Thus on line lay point C = A + N = (5+8, 8+6 ) = (13, 14)

So line through A has two points: A(5,8) and C (13, 14); it's equation is:

y = 0.75x+4.25

or
6x-8y=-34

http://www.hackmath.net/en/calculator/line-slope?x0=5&y0=8&x1=13&y1=14&submit=Solve

Answer 3

slope of BC\[=\frac{ -1-7 }{ -2-(-8) }=\frac{ -8 }{ -2+8 }=\frac{ -8 }{ 6 }=-\frac{ 4 }{ 3 }\]
slope of line through A\[=-\frac{ 1 }{ \frac{ -4 }{ 3 } }=\frac{ 3 }{ 4 }\]
eq. of line through A(5,8) with slope 3/4 is
\[y-8=\frac{ 3 }{ 4 }\left( x-5 \right)\]
simplify it.