1) Given equation: [tex]y=\frac{-3}{2} x + 6[/tex].
Slope of the given equation = [tex]-\frac{3}{2}[/tex].
Parallel lines have equal slopes.
Therefore, slope of parallel line is also [tex]-\frac{3}{2}[/tex].
Given point (2,-1).
Applying point slope form y-y1 = m(x-x1)
y - (-1) =[tex]-\frac{3}{2}[/tex](x-2)
y+1 = [tex]-\frac{3}{2}x+3[/tex]
Subtracting 1 from both sides, we get
[tex]y= -\frac{3}{2}x+2[/tex].
2) Given equation x=-3.
The given line is a vertical line.
Therefore, parallel line would also be a vertical line.
For the given point (4,2) we would have equation for vertical line x = 4.
Therefore, required equation of line is x = 4.
3) Given equation: [tex]y=\frac{1}{2} x -1[/tex].
Slope of the given line is : [tex]\frac{1}{2}[/tex].
Slope of perpendicular line is negative reciprocal.
Therefore, slope of perpendicular line = -2.
Given point (-2,3).
Applying point-slope form, we get
y-3 = -2(x-(-2))
y -3 = -2 (x+2)
y -3 = -2x -4 .
Adding 3 on both sides, we get
y = -2x -1.
Therefore, equation of perpendicular line is y = -2x -1.
