Help with maths! PLEASE!!!?

parallel - use the same 'm'
perpendicular - use '-m'
a) y=mx+b, m = -1/4, for (x,y) =(-2,5) , the new b is 5=-1/4*(-2)+b or b = 4.5
that gives y = 4.5-1/4x
b) similar to above, for (x,y)=(-4,0), the new b is y= x/2+2 => y=mx+b => 0=(1/2)*-4+b or b=2
that gives y = 2+1/2x
3)y=3x+4 m=3 so new m = -3
y=-3x+b for (x,y)=(3,2) , the new b is 2 = -3*3+b or b = 11
that gives y= -3x+11

a) passes through (-2,5) and is parallel to the line y=2-1/4x
m = -1/4, point (-2, 5)
y - y1 = m(x - x1)
y - 5 = -1/4(x + 2)
y - 5 = (-1/4)x - 1/2
y = (-1/4)x + 9/2 in slope-intercept form

b) passes through (-4, 0) and is parallel to the line 2y=x+4
m = 1/2 , point (-4, 0)
y - 0 = 1/2(x + 4)
y = (1/2)x + 2 in slope-intercept form

2)The line, L1 passes through (3,2) and is perpendicular to L2 with equation y=3x+4 find the equation of L1
Perpendicular lines have inverse negative slopes:
L1 m = -1/3 and point (3, 2)
y - 2 = -1/3(x - 3)
y - 2 = (-1/3)x + 1
y = (-1/3)x + 3 in slope-intercept form

3) Find the equation of the line that passes through (2,2) and is perpendicular to the line y=2/5x-3
m = -5/2 , point (2, 2)
y - 2 = -5/2(x - 2)
y - 2 = (-5/2)x + 5
y = (-5/2)x + 7

- .--