# What is the distance between the points of intersection of the graphs y=`x^2 `   and y= 6-x a.  √26                                                                      c. 2√37 b. 5√2                                                                       d. √170 I started by setting the equations equal to each other and solving for x, which gave me 2 and -3.  Then, I plugged these x-values back in and got the coordinate points (2,3) and (-3,-12).  Then, using the Pythagorean Theorem (distance formula), I tried to get the distance between the two points. √[(-3-2)`^2 ` + `(-12-3)^2` ] = √`c^2`  I ended up getting the square root of 250, or 5 on the square root of 10 equals c.  Of course, this isn't one of the answer choices.  I'd like to know where I went wrong (or if I did it totally wrong), and how I can correctly solve the equation. The first portion of your work is correct. The graphs of the two equations intersect at x=-3 and x=2. However, there is a discrepancy with the y-coordinates of the intersections. Let's go over it to clarify.

The given equations are:

`y=x^2`

`y= 6-x`

To solve for their intersections, set the two y's equal to each other.

`y = y`

`x^2=6-x`

`x^2+x-6=0`

`(x+3)(x-2)=0`

`x+3=0`                   `x-2=0`

`x=-3`                            `x=2`

Then, plug-in the x values to either of the original equations. Let's use the second equation y=6-x.

So, when  x=-3, the y value is:

`y=6-(-3)=9`

And, when x=2, the y value is:

`y= 6-2=4`

Hence, the points of intersection are (-3,9) and (2,4).

Next, to get the distance between these two points,  apply the formula:

`d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`d=sqrt((2-(-3))^2+(4-9)^2)`

`d=sqrt(5^2+(-5)^2)`

`d=sqrt50`

`d=5sqrt2`

Therefore, the distance between the points of intersection of the graphs of the two given equations is  `5sqrt2` units.

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