HELP!!! RST has vertices R(-3, - 1), S(0,3), and t(3,0). What type of triangle is RST?

Mathematics

1 answer:

VladimirAG [237]2 years ago

5 0

<h3>Answer: C) scalene</h3>

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Explanation:

Use the distance formula to calculate the distance from R to S. This is identical to the length of segment RS.

$R = (x_1,y_1) = (-3,-1) \text{ and } S = (x_2, y_2) = (0,3)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-3-0)^2 + (-1-3)^2}\\\\d = \sqrt{(-3)^2 + (-4)^2}\\\\d = \sqrt{9 + 16}\\\\d = \sqrt{25}\\\\d = 5\\\\$

Segment RS is exactly 5 units long.

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Repeat similar steps to find the length of segment ST

$S = (x_1,y_1) = (0,3) \text{ and } T = (x_2, y_2) = (3,0)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(0-3)^2 + (3-0)^2}\\\\d = \sqrt{(-3)^2 + (3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d \approx 4.2426\\\\$

ST is roughly 4.2426 units long.

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Lastly, let's calculate the length of segment TR.

$T = (x_1,y_1) = (3,0) \text{ and } R = (x_2, y_2) = (-3,-1)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-(-3))^2 + (0-(-1))^2}\\\\d = \sqrt{(3+3)^2 + (0+1)^2}\\\\d = \sqrt{(6)^2 + (1)^2}\\\\d = \sqrt{36 + 1}\\\\d = \sqrt{37}\\\\d \approx 6.0828\\\\$