What is the distance between the points of intersection of the graphs y=`x^2 ` and y= 6-xa. √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.
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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:
To solve for their intersections, set the two y's equal to each other.
`y = y`
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:
And, when x=2, the y value is:
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)`
Therefore, the distance between the points of intersection of the graphs of the two given equations is `5sqrt2` units.
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