we are given equation of line
[tex] y=\frac{2}{3} x+1 [/tex]
Equation of perpendicular line:
Firstly , we will find slope of line
[tex] y=\frac{2}{3} x+1 [/tex]
we can compare it with
y=mx+b
we get
[tex] m=\frac{2}{3} [/tex]
we know that
slope of perpendicular line will be -1/m
so, slope is
[tex] \frac{-3}{2} [/tex]
It passes through point (-4,5)
so, we can use point slope form of line
[tex] y-5=-\frac{3}{2}(x+4) [/tex]
we get
[tex] y=-\frac{3}{2}x-1 [/tex]..........Answer
Equation of parallel line:
Firstly , we will find slope of line
[tex] y=\frac{2}{3} x+1 [/tex]
we can compare it with
y=mx+b
we get
[tex] m=\frac{2}{3} [/tex]
we know that
slope of parallel line always equal
so, slope is
[tex] \frac{2}{3} [/tex]
It passes through point (-4,-5)
so, we can use point slope form of line
[tex] y+5=\frac{2}{3}(x+4) [/tex]
we get
[tex] y=\frac{2}{3}x-\frac{7}{3} [/tex]............Answer
