Straight LineHardBloom L3

Question

The perpendicular distance $p_{1},p_{2},p_{3}$ of points $\left( a^{2},2a \right),(ab,a + b),\left( b^{2},2b \right)$ respectively from straight line $x + ytan\theta + \tan^{2}\theta = 0$ are in:

Options

A.$AP$
B.$GP$
C.$HP$
D.$AGP$

Solution

Sol. [Key Concept: Compute the perpendicular distances $p_{1,}p_{2,}p_{3}$ from the three points to the line and test for AP / GP / HP.]

$$p_{1} = \frac{\left| a^{2} + 2atan\theta + \tan^{2}\theta \right|}{|sec\theta|}$$

$$\begin{matrix} & \ = \frac{(a + tan\theta)^{2}}{\left| \sec\theta \right|} \\ p_{3} & \ = \frac{(b + tan\theta)^{2}}{\left| \sec\theta \right|} \\ p_{2} & \ = \frac{\left| ab + (a + b)\tan\theta + \tan^{2}\theta \right|}{\left| \sec\theta \right|} \\ & \ = \frac{\mid \left( a + \tan\theta \right)\left( b + \tan\theta \right)}{\left| \sec\theta \right|} \end{matrix}$$

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