SATHEE: Straight Line Question 83

Straight Line Question 83

Question: The area of triangle formed by the lines $ x=0,y=0 $ and $ \frac{x}{a}+\frac{y}{b}=1 $ , is [RPET 1984]

Options:

A) $ \sqrt{2} $

B)1

C) $ \sqrt{3} $

D)2

Show Answer

Answer:

Correct Answer: D

Solution:

The given lines are $ \pm x\pm y=1 $ i.e. $ x+y=1,x-y=1,x+y=-1 $ and $ x-y=-1 $ These lines form a quadrilateral whose vertices are $ A(-1,0),B(0,-1),C(1,0) $ and $ D(0,1) $ Obviously ABCD is a square. Length of each side of this square is $ \sqrt{1^{2}+1^{2}}=\sqrt{2} $ Hence area of square is $ \sqrt{2}\times \sqrt{2}=2sq. $ unitsTrick: Required area = $ \frac{2c^{2}}{|ab|}=\frac{2\times 1^{2}}{|1\times 1|}=2 $ .

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Straight Line Question 83

Question: The area of triangle formed by the lines $ x=0,y=0 $ and $ \frac{x}{a}+\frac{y}{b}=1 $ , is [RPET 1984]

Options:

Answer:

Solution:

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