What is the distance between the points of intersection of the graphs y=`x^2 ` and y= 6-x a. √26 ...
What is the distance between the points of intersection of the graphs y=`x^2 ` and y= 6-x
a. √26 c. 2√37b. 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.
1 Answer
lemjay | High School Teacher | (Level 3) Senior Educator
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.
