Question

The points (k, –2k), (–k + 1, 2k) and
(–4 –k, 6 –2k) can't be collinear for any value of k.

• A. True
• B. False

Answer 1

The correct option is A True

If the three points are collinear, then area of triangle formed by these points will be zero.
$i.e.,\frac{1}{2}\left(\begin{array}{ccc}1& k& -2k\\ 1& -k+1& 2k\\ 1& -4-k& 6-2k\end{array}\right|=0.......\left(1\right]i.e.,\frac{1}{2}\left(A\right|=0,\mathrm{say}\mathrm{On}\mathrm{applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-R_1\mathrm{on}A,\mathrm{we}\mathrm{get}\left(1\right]\mathrm{as}\left(\begin{array}{ccc}1& k& -2k\\ 0& -2k+1& 4k\\ 0& -4-2k& 6\end{array}\right|=0\mathrm{Expanding}\mathrm{along}{R}_1,\mathrm{we}\mathrm{get}-12k+6+4k\left(4+2k\right)=0\mathrm{or}8k^2+4k+6=0\mathrm{or}4k^2+2k+3=0\mathrm{For}\mathrm{given}\mathrm{equation},D={\left(2\right)}^2-4\times 4\times 3<0\mathrm{Hence},\mathrm{real}\mathrm{roots}\mathrm{of}\mathrm{this}\mathrm{equation}\mathrm{does}\mathrm{not}\mathrm{exist}.\mathrm{So},\mathrm{given}\mathrm{points}\mathrm{can}\text{'}t\mathrm{be}\mathrm{collinear}\mathrm{for}\mathrm{any}\mathrm{value}\mathrm{of}k.$