Line AB passes through point (2,3) and intersects the positive x and y-axes at A(a,0) and B(0,b) respectively. If the area of `DeltaAOB` is 11. then the value of `4b^2+9a^2` is

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Correct Answer - 220
Area of `DeltaAOB,(ab)/(2)=11`
`therefore ab=22`,
But points `A(a,0),B(0,b)` and P(2,3) are collinear. `rArr` Slope of AB = Slope of AP
`rArr(b)/(a)=(3-0)/(2-a)`
`rArr-2b+ab=3a`
Squareing `9a^2+ab^2+12ab=484`
`rArr9a^2+4b^2=220`

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