Question

Using determinent, if area of triangle is 4, whose vertices are (2, 2), (6, 6), (5, k) then k=______.

• A. $5$
• B. $1$
• C. $7$
• D. $3$

Answer 1

The correct option is A $5$

For Three points to be collinear, the area of the triangle formed by these points will be zero.

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}2& 2& 1\\ 6& 6& 1\\ 5& k& 1\end{array}\right|$

Condition for points to be the vertices of the triangle, $\Delta =0$

$0=\frac{1}{2}\left|\begin{array}{ccc}2& 2& 1\\ 6& 6& 1\\ 5& k& 1\end{array}\right|$

$0=\left|\begin{array}{ccc}1& 1& \frac{1}{2}\\ 3& 3& \frac{1}{2}\\ \frac{5}{2}& \frac{k}{2}& \frac{1}{2}\end{array}\right|$

$0=\left|\begin{array}{ccc}0& 1& \frac{1}{2}\\ 0& 3& \frac{1}{2}\\ \frac{5-k}{2}& \frac{k}{2}& \frac{1}{2}\end{array}\right|$.................$R_1-R_2$

$0=\frac{5-k}{2}\times (-1)$

$0=\frac{k-5}{2}$

$\therefore k=5$