First slide

Cartesian plane

Question

The points $(p, 2 - 2 p), (1 - p, 2 p)$ and $(- 4 - p, 6 - 2 p)$ are collinear. Then $p =$

Moderate

A

$- 1$ or $\frac{1}{2}$
B

$\frac{- 1}{2}$ or 1
C

$- 1$ or 1
D

$\frac{- 1}{2}$ or $\frac{1}{2}$

Solution

$A (p, 2 - 2 p), B (1 - p, 2 p), C (- 4 - p, 6 - 2 p)$ are collinear
$\Rightarrow Δ A B C = 0$
$\Rightarrow \frac{1}{2} | \begin{matrix} 2 p - 1 & 2 - 4 p \\ 2 p + 4 & - 4 \end{matrix} | = 0$
$\Rightarrow | (- 8 p + 4) - (2 - 4 p) (2 p + 4) | = 0$
$\Rightarrow | - 8 p + 4 + 12 p + 8 p^{2} - 8 | = 0$
$\Rightarrow | 8 p^{2} + 4 p - 4 | = 0, \Rightarrow | 2 p^{2} + p - 1 | = 0$
$\Rightarrow 2 p^{2} + 2 p - p - 1 = 0$
$\Rightarrow 2 p (p + 1) - 1 (p + 1) = 0$
$\Rightarrow (2 p - 1) (p + 1) = 0$
$∴ p = - 1 or \frac{1}{2}$

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