Math problem

Question

Find r so that the line through (2/3, 5) and (-3/5, r) is perpendicular to x+2y=1.

 

For some reason, I don't even know where to start on this problem. If you could explain how to solve it, that would be greatly appreciated. Thanks!

Answer 1

Hi Nick,

 

Find r so that the line through (2/3, 5) and (-3/5, r) is perpendicular to x+2y=1.

For some reason, I don't even know where to start on this problem. If you could explain how to solve it, that would be greatly appreciated. Thanks!

 

 

First start by putting x+2y=1 into slope-intercept form, y=mx+b

 

x+2y=1

2y=-x+1

y=-1/2x+1/2

 

The slope of this line is -1/2.

A line perpendicular to this line has a slope which is the negative reciprocal.

So the slope would be 2.

 

Right now, we know that y=2x+b

 

We need to calculate the y-intercept, b.

So substitute in the given point (2/3, 5) and (-3/5, r) 

 

y=2x+b

5=2(2/3)+b

5=4/3+b

11/3=b

 

Substitute b into equation

y=2x+11/3

 

One last step to find the r-value for (-3/5, r)

substitute in -3/5 for x

 

y=2(-3/5)+11/3

y=-6/5+11/3

y=37/15    in this case y =r

 

ANSWER:  r=37/15