# Answer to Question #21622 in Geometry for korena

Question #21622

The ENTRANCE TO MATT'S TENT MAKES AN ISOSCELES TRIANGLE. IF PLACED ON A COORDINATE GRID WITH THE BASE ON THE X-AXIS AND THE LEFT CORNER AT THE ORIGIN, THE RIGHT CORNER WOULD BE AT (6,0) AND THE VERTEX ANGLE WOULD BE AT (3,4)PROVE THAT IT IS AN ISOSCELES TRIANGLE.ALSO PLEASE EXPLAIN HOW YOU GOT IT STEP BY STEP.

Expert's answer

Vertices of triangle are (given): A(0, 0), B(6, 0) and C(3, 4).

Using the formula of distance between 2 points, calculate lengths of the sides:

AB = sqrt((xA - xB)^2 + (yA - yB)^2) = sqrt((0 - 6)^2 + (0 - 0)^2) = sqrt(36 + 0) = 6

BC = sqrt((xB - xC)^2 + (yB - yC)^2) = sqrt((6 - 3)^2 + (0 - 4)^2) = sqrt(9 + 16) = 5

AC = sqrt((xA - xC)^2 + (yA - yC)^2) = sqrt((0 - 3)^2 + (0 - 4)^2) = sqrt(9 + 16) = 5

We can see BC = AC => the triangle given is isosceles.

Using the formula of distance between 2 points, calculate lengths of the sides:

AB = sqrt((xA - xB)^2 + (yA - yB)^2) = sqrt((0 - 6)^2 + (0 - 0)^2) = sqrt(36 + 0) = 6

BC = sqrt((xB - xC)^2 + (yB - yC)^2) = sqrt((6 - 3)^2 + (0 - 4)^2) = sqrt(9 + 16) = 5

AC = sqrt((xA - xC)^2 + (yA - yC)^2) = sqrt((0 - 3)^2 + (0 - 4)^2) = sqrt(9 + 16) = 5

We can see BC = AC => the triangle given is isosceles.

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