JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q10

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MathematicsSingle correct+4 / -1Question 10 of 90

Let the area of the triangle with vertices $A(1,\alpha)$ , $B(\alpha,0)$ and $C(0,\alpha)$ be 4 sq. units. If the points $(\alpha,-\alpha)$ , $(-\alpha,\alpha)$ and $\left(\alpha^2,\beta\right)$ are collinear, then $\beta$ is equal to

-8

-64

512

Answer and worked solution

Official answer: C

Area. $\frac12\left|1(0-\alpha)+\alpha(\alpha-\alpha)+0\right|=\frac{|\alpha|}2=4$ , so $\alpha^2=64$ .

Collinearity. $(\alpha,-\alpha)$ and $(-\alpha,\alpha)$ lie on $y=-x$ , so $(\alpha^2,\beta)$ must too: $\beta=-\alpha^2=-64$ .

(C)

Topics:Distance and section formulae

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 10