Math homework help?

Question

Write an equation of the line containing the point (3,-2) and
a). perpendicular to the line 3x+4y=5
b). parallel to the line y=2x+5

Answer 1

PART A:

First arrange the line in slope-intercept form (y = mx + b):

3x + 4y = 5

Subtract 3x from both sides:
4y = -3x + 5

Divide both sides by 4:
y = (-3/4)x + 5/4

This line has a slope of -3/4. A perpendicular line will have a slope that is the *negative reciprocal*. Turn it upside down and change the sign. So the slope you want is 4/3.

Now given a point (3, -2) and a slope (4/3) you need a line. Use the "point-slope" form of the line:

y - y1 = m(x - x1)

Plug in your values from the point (x1, y1) and the slope (m) above:
y - (-2) = (4/3)(x - 3)

Simplify:
y + 2 = (4/3)(x - 3)

Now distribute the 4/3 through the parentheses on the right:
y + 2 = (4/3)x - (4/3)*3
y + 2 = (4/3)x - 4

Finally subtract 2 from both sides:
y = (4/3)x - 4 - 2
y = (4/3)x - 6

As a check, make sure the point (3, -2) is on the line. Let's plug in x = 3 and confirm we get -2...

y = (4/3)(3) - 6
y = 4 - 6
y = -2
Cool!

Therefore, the equation of the perpendicular line is:
y = (4/3)x - 6

PART B:

Same method, but you have a head-start. The line is already in slope-intercept form. The slope is therefore 2. A parallel through (3, -2) will have the same slope of 2.

Point (3, -2), Slope = 2.

Use the point-slope form and solve like PART A.

y - (-2) = 2(x - 3)
y + 2 = 2x - 6
y = 2x - 6 - 2
y = 2x - 8

Again, double-check with the point (3, -2) --> y = 2(3) - 8 = -2

Answer 2

a) perpendicular to the line 3x + 4y = 5
y = 4/3x + 5

b) parallel to the line y= 2x + 5
y = 2x + 5