Find the standard form of the equation?

Question

two problems
1)perpendicular to 4x+y=-7
2)parallel to 5x-2y=9 through (-1,0)

Answer 1

4x + y = - 7

4x + y - 4x = - 4x - 7

y = - 4x - 7

Slope is 4

Perpendiculal = - 1/4x

y = - 1/4x + b

- - - - - - - - - - - -s-

Answer 2

1) Perpendicular lines have opposite reciprocal slopes, so the first thing you'd have to do is find the slope. Put it in slope-intercept (y=mx+b, where m is the slope) form:
4x+y= -7
y= -4x-7
The slope of this line is -4. The perpendicular line will have a slope of +1/4. Substitute the 1/4 in for the -4, and put back into standard form:
y=(1/4)x-7
y-(1/4)x= - 7

2) Parallel lines have the same slopes, so put this into slope-intercept form again to determine the slope:
5x-2y=9
-2y= -5x+9
y=(5/2)x-(9/2)
The slope of this line is 5/2, so the slope of the parallel lines is also 5/2. Since you know the slope of the parallel line and also a point it passes through, you can put these into point-slope form (y-y1=m[x-x1]):
y-0=(5/2)(x+1)
y=(5/2)x+ (5/2)
y-(5/2)x=5/2

Answer 3

1)perpendicular to 4x+y=-7or y=-4x-7, its slope=-4
slope of the line perpendicular=1/4
equation of the perpendicularis:
y=(1/4)x + C.
2)parallel to 5x-2y=9
5x -2y +k=0 if it passes through(-1,0), then -5-0+k=0, so k=5
so the reqd. eEQ. is 5x-2y+5 =0