MCQ
Are the points (1, 1), (2, 3) and (8, 11) collinear?
- Acollinear
- BNon collinear
- Ccoplaner
- DNone of above
Solution:
Area of triangle formed by these vertices is,
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\2&3&1\\8&11&1\end{vmatrix}$
Applying R2 → R2 − R1, R3 → R3 − R1
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\1&2&0\\7&10&0\end{vmatrix}=\frac{1}{2}(10-14)=2$
Hence points are non collinear
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$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$
has a non-trivial solution, then the value of $\theta$ is :
$\frac{5}{3}$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{9}$