Question

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 :

• A. \(\left(1,\frac{3}{4}\right)\)
• B. \(\left(1,-\frac{3}{4}\right)\)
• C. \(\left(2,\frac{1}{2}\right)\)
• D. \(\left(2,-\frac{1}{2}\right)\)

Hint:

Find k from the area condition (k = 2 is the only integer), then intersect two altitudes.

Answer 1

The Correct Option is C

Area \(=\frac{1}{2}\left|k(k-2)+5(2+3k)+(-k)(-3k-k)\right|=\frac{1}{2}\left|5k^2+13k+10\right|=28\).
\(5k^2+13k+10=56\Rightarrow5k^2+13k-46=0\Rightarrow k=\frac{-13\pm33}{10}=2\) or \(-4.6\); the value \(-56\) gives \(5k^2+13k+66=0\), which has no real root. So \(k=2\).
Vertices: \(A(2,-6)\), \(B(5,2)\), \(C(-2,2)\). \(BC\) is horizontal (\(y=2\)), so the altitude from \(A\) is \(x=2\).
Slope of \(AC\) \(=\frac{2+6}{-2-2}=-2\), so the altitude from \(B\) has slope \(\frac{1}{2}\): \(y-2=\frac{1}{2}(x-5)\). At \(x=2\): \(y=2-\frac{3}{2}=\frac{1}{2}\).
Orthocentre \(=\left(2,\frac{1}{2}\right)\).
Answer: (3) \(\left(2,\frac{1}{2}\right)\)