Are the points (1, 1), (2, 3) and (8, 11) collinear? - Vidyadip

MCQ

Are the points $(1, 1), (2, 3)$ and $(8, 11)$ collinear?

A
collinear
✓
Non collinear
C
coplaner
D
None of above

✓

Answer

Correct option: B.

Non collinear

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 $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R 1$
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\1&2&0\\7&10&0\end{vmatrix}=\frac{1}{2}(10-14)=2$
Hence points are non collinear

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Three Dimensional Geometry→Three Dimensional Geometry — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $f (x) = \frac{{\sin x}}{x}$ , then $\int\limits_0^{\frac{\pi }{2}} {f(x)\,\,f\left( {\frac{\pi }{2} - x} \right)\,dx} =$

The area bounded by the curve $\text{y}=\log_{\text{e}}\text{x}$ and $x-$ axis and the straight line $x = e$ is :

$\int {{e^x}(1 - \cot x + {{\cot }^2}x)\,\,dx} $ equals

If the function $f$ defined as $f(x)\, = \frac{1}{x} - \frac{{k - 1}}{{{e^{2x}} - 1}}$ ,$x\, \ne \,0,$ is continuous at $x = 0.$ then the ordered pair $(k,f(0))$ is equal to?

Let $'a'$ $(a < 0, a \notin I)$ be a constant and $'t'$ be a parameter, then set of values of $'t'$ for which the function $f(x)=\left( \frac{\left| \left[ t \right]+1 \right|+a}{\left| \left[ t \right]+1 \right|+1-a} \right)x$ ,is a decreasing function of $x$ ([.] denote greatest integer function) is

$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{ dx}$ is equal to:

$\int\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}=$

If $\alpha,\beta,\gamma$ are the angle which a half ray makes with the positive directions of the axis then $\sin^2\alpha + \sin^2\beta + \sin^2\gamma =$

The volume of a sphere is increasing at the rate of $4\pi\text{cm}^{3}/\text{sec}$. The rate of increase of the radius when the volume is $288\pi\text{cm}^{3}/\text{sec}$ is:

The value of $\int_{}^{} {x\sin kx\;dx} $is