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(D) The equation of the parabola is $y^2 = 4x$. Given the $y$-coordinates of the vertices are $y_1 = 1$,$y_2 = 2$,and $y_3 = 4$. Since the points lie

Vedclass · Accepted Answer

$7/4$

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(D) The equation of the parabola is $y^2 = 4x$. Given the $y$-coordinates of the vertices are $y_1 = 1$,$y_2 = 2$,and $y_3 = 4$. Since the points lie on the parabola,their $x$-coordinates are $x = y^2/4$. Thus,the vertices are $A(1/4, 1)$,$B(1, 2)$,and $C(4, 4)$. The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by: $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Substituting the values: $	ext{Area} = \frac{1}{2} |\frac{1}{4}(2 - 4) + 1(4 - 1) + 4(1 - 2)|$ $	ext{Area} = \frac{1}{2} |\frac{1}{4}(-2) + 1(3) + 4(-1)|$ $	ext{Area} = \frac{1}{2} |-\frac{1}{2} + 3 - 4| = \frac{1}{2} |-\frac{3}{2}| = \frac{3}{4}$ Note: The provided options in the prompt were incorrect; the calculated area is $3/4$.

What is the area of the triangle inscribed in the parabola $y^2 = 4x$ whose vertices have $y$-coordinates $1, 2,$ and $4$?

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