Write the equation of the line that is perpendicular to the line 8y-16=5x through the point (5,-5)

Question

A y=-8/5x-3

B. Y=8/5x+3

C. Y=-8/5x+3

D. Y=8/5x+3

Answer 1

I answered your question about the parallel line equation. Follow the same steps except on step 2. The line perpendicular to a line has a slope that is the negative reciprocal. Example. If the slope of a line is 3/4 then the slope of the line perpendicular to it has a slope of -4/3.

Answer 2

1) Turn this equation into slope intercept form y = mx + b. (m = slope and b = y-intercept.)

2) We need to take 8y - 16 = 5x and turn it into slope intercept form (solve for y).

3) I get 8y = 5x + 16. Then divide both sides of the equation by 8 to get y by itself. You'll get:

y = 5/8x + 2

4) The slope is the number in front of your x variable, which is 5/8.

5) Lines that are perpendicular are opposite reciprocals of each other (Same thing as their products being -1)

6) Remember, the slope of the original equation is 5/8, so I need to figure out what I can multiply 5/8 by to get to -1.

7) Instead of me figuring that out, I can alternatively take -1 and divide by 5/8. When I do that, I get -8/5. What that means is that the slope of the line perpendicular to 8y - 16 = 5x is -8/5.

8) I'm now going to take my NEW slope -8/5 and put it in slope intercept form, which looks like: y = -8/5 x + b

9) Now, take (5, -5) and plug each in for x and y. x = 5 and y = -5

10) You should get this: -5 = -8/5 (5) + b. Simplify to get -5 = -8 + b

11) Add 8 to both sides to get b = 3.

12) Put your new slope and your new y-intercept (b) together to get y = -8/5 x + 3. Your answer is C. Hope this helps!