JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q78

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MathematicsSingle correct+4 / -1Question 78 of 90

Let k be an integer such that the triangle with vertices $(k,-3k)$ , $(5,k)$ and $(-k,2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point :

\left(1,\dfrac34\right)

\left(1,-\dfrac34\right)

\left(2,\dfrac12\right)

\left(2,-\dfrac12\right)

Answer and worked solution

Official answer: C

The area is $\frac12|5k^2+13k+10|=28$ . That gives $5k^2+13k-46=0$ , so $k=2$ (an integer).

The vertices are $(2,-6)$ , $(5,2)$ and $(-2,2)$ . The altitude from $(2,-6)$ to the horizontal side is $x=2$ . The altitude from $(5,2)$ , perpendicular to the side of slope $-2$ , is $y-2=\frac12(x-5)$ . They meet at $\left(2,\frac12\right)$ .

Topics:Distance and section formulae Equations of a line

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 78