JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q12

← JEE Main 2022 · 27 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 12 of 90

In an isosceles triangle ABC, the vertex A is $(6,1)$ and the equation of the base BC is $2x+y=4$ . Let the point B lie on the line $x+3y=7$ . If $(\alpha,\beta)$ is the centroid of $\Delta ABC$ , then $15(\alpha+\beta)$ is equal to :

Answer and worked solution

Official answer: C

Point B. $2x+y=4$ and $x+3y=7$ meet at $B(1,2)$ .

Foot from A. The foot of the perpendicular from $A(6,1)$ to BC is the midpoint of BC: $(6,1)-\frac95(2,1)=(2.4,-0.8)$ , so $C=(3.8,-3.6)$ .

Centroid. $(3.6,-0.2)$ , so $15(\alpha+\beta)=15\times3.4=51$ .

(C)

Topics:Distance and section formulae Equations of a line

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JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 12