if a and b are the roots of the equation x^2-px +(p+1)=0 then the area of the triangle whose vertices are (0,0),(p/a,p) and (-p/a,p) is

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Vijay Mukati · Answer

Hint. find the sum and product of roots from the given equation. Then find the area of triangle through the given coordinates (Area = (1/2)*base*altitute). Then relate the two. Thanks.

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if a and b are the roots of the equation x^2-px +(p+1)=0 then the area of the triangle whose vertices are (0,0),(p/a,p) and (-p/a,p) is

if a and b are the roots of the equation x^2-px +(p+1)=0 then the area of the triangle whose vertices are (0,0),(p/a,p) and (-p/a,p) is

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if a and b are the roots of the equation x^2-px +(p+1)=0 then the area of the triangle whose vertices are (0,0),(p/a,p) and (-p/a,p) is

if a and b are the roots of the equation x^2-px +(p+1)=0 then the area of the triangle whose vertices are (0,0),(p/a,p) and (-p/a,p) is

1 Answers

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