Find equations of the lines passing through (−2, 3) and having the following characteristics.

a.y=13/16x+37/8

b.y=3/2x+6

c.y=3/4x+9/2

d.x=-2

Step-by-step explanation:

All lines must pass through (-2,3)

a. The slope must be 13/16;

y=13/16x+b, SInce it must pass through (-2,3), plug in -2 as the x-value and 3 as the y-value

3=13/16(-2)+b, solve for b.

24/8=-13/8+b

37/8=b

y=13/16x+37/8

b, It must be parallel to the line 3x-2y=2; first, find the slope by converting to y=mx+b

3x-2y=2

-2y=-3x+2

y=3/2x-1, Parallel lines have the same slope so a line parallel to 3x-2y=2 and passing through (-2,3) will have a slope of 3/2.

Plug in (-2,3) to find the b-value again.

(3)=3/2(-2)+b

3=-3+b

6=b

y=3/2x+6

c. It must have a line perpendicular to 4x+3y=6.  Again like in b, find the slope.

y=-4/3+2, the slope is -4/3.

Now, perpendicular lines have the opposite inverse slopes which means that you add a negative sign and flip the numerator and denominator

-(-3/4), this is just 3/4

Ok, the slope of the line is 3/4, plug in (-2,3) to find the b-value again.

(3)=3/4(-2)+b

6/2=-3/2+b

9/2=b

y=3/4x+9/2

d. The line is paralell to the y-axis.

This just means that the line is vertical. The line has no slope.

The vertical line that passes through the x-value -2 is x=-2

x=-2

I did this stuff a few years ago.. I think it is 60 If I remember right

The formula I learned is A²+B²=C²