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Solution of Linear Equation in 2 Variables

The point of ...

Question

The point of intersection of the above line with both the coordinates axes is

A

(13/3, 0) and (0, 13)

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B

(0, 13/3) and (13, 0)

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C

(13, 0) and (0, 1/3)

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D

None of these

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Solution

The correct option is C (13, 0) and (0, 1/3)
$T h e e q u a t i o n o f t h e l i n e p a s s i n g t h r o u g h (x_{1}, y_{1}) a n d (x_{2}, y_{2}) i s y - y_{1} = (\frac{x - x_{1}}{x_{2} - x_{1}}) (y_{2} - y_{1}) . H e r e (x_{1}, y_{1}) = (4, 1) & (x_{2}, y_{2}) = (5, - 2) S o t h e e q u a t i o n o f t h e l i n e p a s s i n g t h r o u g h (x_{1}, y_{1}) a n d (x_{2}, y_{2}) i s y - 1 = \frac{x - 4}{5 - 4} (- 2 - 1) ⟹ 3 x + y = 13.......... (i) T h i s l i n e (i) i n t e r s e c t s t h e X - a x i s a t (x, 0) . W e c a n g e t x b y p u t t i n g y = 0 i n e q u a t i o n (i) ∴ 3 x + 0 = 13 ⟹ x = \frac{13}{3} u n i t s . A g a i n t h e l i n e (i) i n t e r s e c t s t h e Y - a x i s a t (0, y) . W e c a n g e t y b y p u t t i n g x = 0 i n e q u a t i o n (i) ∴ 3 \times 0 + y = 13 ⟹ y = 13 u n i t s . S i n c e t h e Δ h a s t h e a x e s a s i t s t w o s i d e s ∴ i t i s a r i g h t t r i a n g l e . I t s a r e a = \frac{1}{2} \times b a s e \times h e i g h t I t s b a s e = x = \frac{13}{3} u n i t s, h e i g h t = y 13. u n i t s A r e a o f t h e Δ = \frac{1}{2} \times \frac{13}{3} \times 13 s q u n i t s = \frac{169}{6} s q u n i t s . A n s - O p t i o n C$

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