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Question Number 190075 by Spillover last updated on 26/Mar/23

	Answered by PowerMaths last updated on 27/Mar/23

	$(a) p_{1} = ⊥^{r} f r o m A (1, 3) t o l i n e = \frac{∣ 3 \times 1 + 4 \times 3 - 9 ∣}{\sqrt{3^{2} + 4^{2}}} = \frac{6}{5} = 1.2$ $p_{2} = ⊥^{r} f r o m B (2, 7) t o l i n e = \frac{∣ 3 \times 2 + 4 \times 7 - 9 ∣}{\sqrt{3^{2} + 4^{2}}} = \frac{25}{5} = 5$ $r a t i o o f d i v i s i o n = \frac{1.2}{5} = \frac{12}{50} = \frac{6}{25} f r o m A$ $[n o t i c : t h e r a t i o o f d i v i s i o n o f s e g m e n t = t h e r a t i o b e t w e e n$ $p e r p e n d i c u l a r s d u e t o s i m i l a r i t y]$ $(b) e q u a t i o n o f l i n e :$ $\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow b x + a y = a b$ $p = \frac{∣ b \times 0 + a \times 0 - a b ∣}{\sqrt{a^{2} + b^{2}}} \Rightarrow p = \frac{a b}{\sqrt{a^{2} + b^{2}}} s q u a r i n g$ $p^{2} = \frac{a^{2} b^{2}}{a^{2} + b^{2}} \Rightarrow \frac{1}{p^{2}} = \frac{a^{2} + b^{2}}{a^{2} b^{2}} \Rightarrow \frac{1}{p^{2}} = \frac{1}{b^{2}} + \frac{1}{a^{2}}$
	Commented by Spillover last updated on 27/Mar/23

	$good . thanks$
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