At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

Mathematics

Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

SOLVED

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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.