Circles Question - None

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

Accepted Answer

Given, vertices of triangle are (k, - 3k), (5, k) and (- k, 2)
$\begin{aligned} & herefore \quad \frac{1}{2}\left|\begin{array}{ccc} {k} & {-3 k} & {1} \ {5} & {k} & {1} \ {-k} & {2} & {1} \end{array} ight|=\pm 28\ &\Rightarrow \quad\left|\begin{array}{ccc} {k} & {-3 k} & {1} \ {5} & {k} & {1} \ {-k} & {2} & {1} \end{array} ight|=\pm 56 \end{aligned}$
$\Rightarrow$ k(k - 2) + 3k (5 + k) + 1(10 + K²) = $\pm 56$
$\Rightarrow$ k² - 2k + 15k + 3k² + 10 + k² = $\pm 56$
$\Rightarrow$ 5k² + 13k + 10 = $\pm 56$
$\Rightarrow$ 5k² + 13 - 66 = 0
or 5k² + 13k - 46 = 0
$\Rightarrow$ k = 2 [$\because$ k $\in$ I]
Thus, the coordinates of vertices of triangle are
A(2, - 6), B(5, 2) and C (- 2, 2).
Now, equation of altitude from vertex A is
$y-(-6)=\frac{-1}{\left(\frac{2-2}{-2-5} ight)}(x-2) \Rightarrow x=2$ ...(i)

Equation of altitude from vertex C is
$y-2=\frac{-1}{\left[\frac{2-(-6)}{5-2} ight]}[x-(-2)]$
$\Rightarrow$ 3x + 8y - 10 = 0 ...(ii)
On solving Eqs. (i) and (ii), we get x = 2 and $y=\frac{1}{2}$
$ herefore$ Orthocentre = $\left(2, \frac{1}{2} ight)$

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