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A right triangle in the first quadrant is bounded by lines y equals 0 y equals x and y equals -x plus 5 Find its area?
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∙ 14y ago
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Lorenza Eichmann
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∙ 14y agoy = 0
y = x
y = -x + 5
Let's find the intersection point of the lines with equation y = x and y = -x + 5.
y = x
y = -x + 5 (add both equations)
2y = 5(divide by 2 both sides)
y = 5/2 = 2.5 which yields that x = 2.5.
So the intersection point (2.5, 2.5) tells us that the right triangle is an isosceles triangle, because the altitude drawn from the point of intersection to the side (with a length of 5 since the y-intercept of y = -x + 5, is 5) that lies on y = 0, separates it in two equal parts of 2.5.
Since the altitude and the base of this right triangle are congruent, we can find their lengths by using the distance formula between two points (0, 0) and (2.5, 2.5) or using sin 45 degrees, which gives us the same result for a length of 3.536.
So the are of the triangle equals to 3.5362/2= 6.25 unit squares.
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