Find the equation to the sides of an isosceles right-angled triangled, the equation of whose hypotenuse is 3x + 4y = 4 and the opposite verter is the point (2, 2).

7y-x-12=0 and 7x+y=16

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7y+x-12=0 and 7x+y=16

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7y+x-12=0 and 7x-y=16

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7y-x+12=0 and 7x-y=16

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Solution

The correct option is A 7y-x-12=0 and 7x+y=16
$\begin{matrix} T h e e q u a t i o n o f h y p o t e n o u s i s 3 x + 4 y = 4 s l o p e o t h e l i n e = \frac{- 3}{4} s i n c e, i t i s i s o s c e l e s t r i a n g l e t h e a n g l e s h o u l d b e 45^{\circ}, 45^{\circ}, 90^{\circ} N o w, tan 45^{\circ} = ∣ ∣ ∣ ∣ \frac{m - (\frac{- 3}{4})}{1 + m (\frac{- 3}{4})} ∣ ∣ ∣ ∣ 1 = ∣ ∣ ∣ \frac{m + \frac{3}{4}}{1 - \frac{3 m}{4}} ∣ ∣ ∣ 1 - \frac{3 m}{4} = \pm (m + \frac{3}{4}) I f 1 - \frac{3 m}{4} = m + \frac{3}{4} ∴ m = \frac{1}{7} C o - o r d i a n t e s o f b i s (2, 2) E q u a t i o n o f l i n e A B (y - 2) = \frac{1}{7} (x - 2) ∴ x - 7 y = - 12 \Rightarrow 7 y - x - 12 = 0 m_{1} m_{2} = - 1 (r i g h t a n g l e) ∴ m_{2} = - 7 E q u^{n} o f l i n e B C (y - 2) = - 7 (x - 2) y - 2 = - 7 x + 14 y = 7 x = 16 H e n c e, o p t i o n A i s t h e c o r r e c t a n s w e r . \end{matrix}$

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