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If a pair of perpendicular straight lines drawn through the origin forms an isosceles triangle with the line `2x+3y=6` , then area of the triangle so formed is (a) `36/13` (b) `12/17` (c) `13/5` (d) `17/14`

A

`(36)/(13) " sq. unit"`

B

`(12)/(17) " sq. unit"`

C

`(13)/(5) " sq. unit"`

D

`(17)/(13) " sq. unit"`

Text Solution

Verified by Experts

As shown in the following figure, pair of perpendicular lines through origin meets given line 2x+3y=6 at A and B such that AOB is right angled isosceles triangle.

Now, OM=BM=AM, where OM `=(|0+0-6|)/(sqrt(2^(2) + 3^(2))) = (6)/(sqrt(13))`
`therefore AB=2AM=2OM = (12)/(sqrt(13))`
`therefore " Area of triangle "AOB = (1)/(2) xx (12)/(sqrt(13)) xx (6)/(sqrt(13)) = (36)/(13)` sq. units.
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