A line passes through (2,2) and has x-intercept and y-intercept as α units and β units respectively. It makes a triangle of area A with co-ordinate axes. Then the quadratic equation whose roots are α and β is :
(α>0,β>0)
A
x2−2Ax+2A=0
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B
x2−Ax+2A=0
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C
x2+2Ax+2A=0
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D
x2+Ax+2A=0
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Solution
The correct option is Bx2−Ax+2A=0 Let the equation of line be xα+yβ=1
it passes through (2,2) so 2α+2β=1 ⇒2(α+β)=αβ
Here, α and β are intercepts
So the area of triangle =12αβ=A ⇒αβ=2A⇒α+β=A
So, Quadratic equation will be x2−(α+β)x+αβ=0 ⇒x2−Ax+2A=0