The line perpendicular to 5y+2x=10 and passes through the point (3,6)

Question

Write an equation of the line satisfying given conditions. Write the answer in slope intercept or standard form. Express numbers as integers or simplified fractions. 

Answer 1

First, put your linear equation in the standard form of y=mx +b, where m is the slope and b is the y-intercept:

5y +2x=10. We want to get y by itself, so subtract 2x from both sides:

5y = -2x + 10

Divide everything by 5 to get y by itself, and you have
5y/5 = -2x/5 +10/5, or y = (-2/5)x + 2

So m = -2/5. 

We need the slope of our second line (let's call that y₂.) to be the negative reciprocal (negative and upside down) of our original slope.

The slope of y₂ = 5/2. We can write the general equation:

y₂ = (5/2)x +b. This is a different line, and it will have a different y-intercept. We can find this by plugging the point they give us in for the x and y values:

y₂ = (5/2)x +b 
with (3,6): 6=(5/2)(3)+b

6=(15/2)+b. You can solve for b with a calculator, or you can use fractions with like denominators:

12/2 = 15/2+b. Subtract 15/2 from both sides:

-3/2=b.

Finally, your perpendicular line has the equation of y=(5/2)x-3/2.