MCQ
If the points $(a,b),\,(a',b')$and $(a - a',b - b')$are collinear, then
  • $ab' = a'b$
  • B
    $ab = a'b'$
  • C
    $aa' = bb'$
  • D
    ${a^2} + {b^2} = 1$

Answer

Correct option: A.
$ab' = a'b$
a
(a) $\frac{{a - a' - a'}}{{a' - a}} = \frac{{b - b' - b'}}{{b' - b}}$

$ \Rightarrow \,\,\,\frac{{a - 2a'}}{{a' - a}} = \frac{{b - 2b'}}{{b' - b}}$

$ \Rightarrow \,\,\frac{a}{{a'}} = \frac{b}{{b'}}\,\,$

$\Rightarrow \,\,ab' = a'b.$

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

Similar questions

The coordinates of the join of trisection of the points $(-2, 3)$, $(3, -1)$ nearer to $(-2, 3)$, is
The number of non - zero terms in the expansion of $\big(1+3\sqrt{2​}\text{x}\big)^{9}+\big(1-3\sqrt{2}​\text{x}\big)^{9}$ is:
If $A (2, 2), B(-4, -4), C(5, -8)$ are vertices of any triangle, then the length of median passes through $C$ will be
If $\frac{3+i \sin \theta}{4-i \cos \theta}, \theta \in[0,2 \pi],$ is a real number, then an argument of $\sin \theta+\mathrm{i} \cos \theta$ is
The equation of radical axis of the circles ${x^2} + {y^2} + x - y + 2 = 0$ and $3{x^2} + 3{y^2} - 4x - 12 = 0,$ is
A quadratic polynomial $ y = f (x)$  with absolute term  $3$  neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$ . The coefficient of the leading term of the polynomial is unity. A point $A(x_1, y_1)$  with abscissa $x_1 = 1$  and a point $B(x_2, y_2) $ with ordinate $y_2 = 11 $ are given in a cartisian rectangular system of co-ordinates $OXY $ in the first quadrant on the curve $y = f (x)$  where $ 'O'$  is the origin. The scalar product of the vectors $\vec {OA}$ and $\vec {OB}$ is
Points $P (-3,2), Q (9,10)$ and $R (\alpha, 4)$ lie on a circle $C$ with $P R$ as its diameter. The tangents to $C$ at the points $Q$ and $R$ intersect at the point $S$. If $S$ lies on the line $2 x - ky =1$, then $k$ is equal to $.........$.
$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$ is equal to$......$.
The expansion of $\frac{1}{{{{(4 - 3x)}^{1/2}}}}$ binomial theorem will be valid, if
If $\omega ( \ne 1)$is a cube root of unity and ${(1 + \omega )^7} = A + B\omega $, then $A$ and $B$ are respectively, the numbers