• elleblythe
If the vertices of an equilateral triangle are (-4, -3) and (4, 1), how do we find the remaining vertex? Please show complete solution and answer
  • Stacey Warren - Expert brainly.com
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  • katieb
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  • wolf1728
Well if nothing else, I figured I might as well make a graphic. The 3rd vertex would be found in quadrant II or quadrant IV
1 Attachment
  • ybarrap
What your essentially being asked is to find the center of a third circle(s) with center (x,y) and radius equal to the distance between <-4,-3> and <4,1> and whose points pass through these two points: |dw:1378001128405:dw| So, we need to use the equation of a circle to create an equation for each point. For the point <-4,-3> we have: \((x+4)^2+(y+3)^2=r^2\), where r is the distance between the two given points: \(r=\sqrt{-4-4)^2+(-3-1)^2}=\sqrt{48}\). The equation of a circle for point <4,1> is \((x-4)^2+(y-1)^2=r^2\). Using these two equations, we have can solve for x and y, the center of the third circle, which is the intersection between the two circles formed by the two given vertices as center, each with radius r. We'll have two solutions, therefore, there are two places where you can put a vertex to make equilateral triangles: http://www.wolframalpha.com/input/?i=%28x%2B4%29^2%2B%28y%2B3%29^2%3D16*9%2C%28x-4%29^2%2B%28y-1%29^2%3D16*9 That's it! Let me know if you have any questions.
  • ybarrap
The two solutions are: \( <-2 \sqrt{31/5}, 4 \sqrt{31/5}-1>\) and \(<2 \sqrt{31/5}, -1-4 \sqrt{31/5} >\) Note: \(d_1\) in my sketch above is \(r\), the radius of all circles.

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