Determine if triangle XYZ with coordinates X (2, 2), Y (6, 7), and Z (7, 3) is a right triangle. Use evidence to support your claim. If it is not a right triangle, what changes can be made to make it a right triangle?
@jim_thompson5910 can you help me
i do know that it is not a right triangle
@Hero can you help please
@amistre64 can you help me please
can someone please help me.
As you can see (7,3) is the common vertex so Find the slope of points (2,2) and (7,3) Then find the slope of points (6,7) and (7,3) If the slopes multiply to get -1, then the triangle is right.
so it is a right triangle
Only IF the slopes multiply to get -1. Have you found the slopes yet?
no doing that now
ok just did it and got .2 for (2,2) and (7,3) and -4 for (6,7) and (7,3) then when i multiplied them together i got -.8 so that means it is not a right triangle
-.8 is close to -1 but it needs to equal -1 in order for the triangle to be right. So no, the triangle is not right.
should i use what you said at first for the reasoning
@Hero Thanks for the Help
FULL EXPLANATION: There are different ways to find if it is a right triangle (like finding the slopes of the two shorter legs and seeing if they are opposite reciprocals). I am going to use the Pythagorean theorem to prove either if it is or if it is not a right triangle. For it to be a right triangle, the squares of the two legs have to be equal to the square of the hypotenuse. This is the Pythagorean theorem, a^2 + b^2 = c^2. Using the distance formula, I find that the two shorter legs have a length of a = 4.12 and b = 5.1 and the length of the largest side is c = 6.4. This is true, however, (4.12)^2 + (5.1)^2 = 42.98 and (6.4)^2 = 40.96. 42.98 and 40.96 are not equal to each other and therefore, triangle XYZ is not a right triangle. Even though it is not a right triangle, there are changes that can be made to make it a right triangle. You can shift point Z from (7, 3) to the left one unit and down another unit to make the new coordinate (6, 2). In this case, a^2 + b^2 would indeed equal c^2 and angle XZY would be fixed at a ninety degree angle.
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