# What is the equation for the line that passes through the point (1, -5), and that is perpendicular to the line y=-2x+5?

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We want the line passing through the point (1,-5) that is perpendicular to the line y=-2x+5.

If two lines are perpendicular, then their slopes are opposite reciprocals of each other. Thus the slope of the line we are seeking is `1/2` .

We now a point on the line and its slope. We use the point-slope form:

`y-y_1=m(x-x_1)` where `(x_1,y_1)` is the given point and m is the slope.

Thus the equation of the line is `y-(-5)=1/2(x-1)` or `y+5=1/2(x-1)` .-------------------------------------------------------------------

**In slope-intercept form the equation of the line is `y=1/2x-11/2` **

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(The lines do not appear perpendicular due to the scale.)

**Sources:**

y = -2x + 5

To be perpendicular the slope will have to be **1/2 **

Now plug the points (1, -5) into the equation

**-5 = 1/2 (1) + b **multiply 1/2 with 1

By multiplying, you should get

**-5 = 1/2 + b **now subtract 1/2 on both sides

By subtracting you should get

**11/2 = b**

So your equation would look like in slope intercept form :

**y = 1/2x + 11/2**

You use the equation Ax + By + C =0

X – 2y = (1) - 2(-5)

Then it becomes:

X – 2y = 1 + 10

X – 2y = 11

X – 2y -11 = 0

Sorry correction on the last couple steps when you multiply by 2 you should get -10=1+b so you then subtract one over to give yiu -11 as your b

You know that y=mx+b is the basic equation where b is the y intercept and m is the slope. Whenever you have a perpendicular line to something you take the given slope and do the opposite recipricol so int his case for the slope you have -2 which is the same as -2/1 and its opposite recipricol would be +1/2.

Inorder to fill in an equation you need a slope and a y intercept. To find the y intercept we need a point on the graph so the new equation so far is

Y=1/2x+b and were trying to find b.

The point you are given is (1,-5)in the form of (x, y)as in x=1and y=-5. So you simply plug those into the equation and solve for b

-5=1/2 (1)+b

-5=1/2+b

Multiply by 2

-10=b

So your equation is then y=1/2-10 for the perpendicular line.

the general equation of a line is Ax + By + C =0. So you should make y=-2x+ 5 into general equation..

so it will be 2x + y - 5 =0.

there is a theorem states that Ax + By + C =0 is perpendicular to ±Bx ± Ay + C =0..

So we will make 2x + y in the given line into x – 2y.

The solution will be:

X – 2y = (1) - 2(-5) the (1) & (-5) are your given points, in

the right side of the Equal sign, we

substitute the x to 1 and y to -5 from the left

side

X – 2y = 1 + 10

X – 2y = 11

X – 2y -11 = 0 is the equation of the line.