A line through (2, 2) and the axes form a triangle of area A α unit

1.	A line through (2, 2) and the axes form a triangle of area A α units. Then, the intercepts on the axes made by the line are roots of the equation(a) x2 - αx + α = 0(b) x2 + αx + 2α = 0(c) x2 -αx + 2α = 0(d) x2 + αx - 2α = 0
Answer» Correct option (c),(d) Explanation : Let x/b + y/b = 1 be the equation of the line which forms a triangle with the coordinate axes of area α sq. unit. Since the line passes through (2, 2), we have 2/a + 2/b = 1 ...(1) We have 1/2\|ab\| = α so that ab = ±2α ....(2) now, from Eq.(2) 2a + 2b = ab = 2α Therefore a + b = α ...(3) Hence, from Eqs. (2) and (3), a and b are the roots of the equation x² - αx + 2α = 0. Therefore 2a + 2b = -2α a + b = -α Hence a and b are roots of the equation x² + αx - 2α = 0

A line through (2, 2) and the axes form a triangle of area A α units. Then, the intercepts on the axes made by the line are roots of the equation(a) x2 - αx + α = 0(b) x2 + αx + 2α = 0(c) x2 -αx + 2α = 0(d) x2 + αx - 2α = 0

Answer»

Correct option (c),(d)

Explanation :

Let x/b + y/b = 1

be the equation of the line which forms a triangle with the coordinate axes of area α sq. unit. Since the line passes through (2, 2), we have

2/a + 2/b = 1 ...(1)

We have 1/2|ab| = α so that

ab = ±2α ....(2)

now, from Eq.(2)

2a + 2b = ab = 2α

Therefore

a + b = α ...(3)

Hence, from Eqs. (2) and (3), a and b are the roots of the equation x² - αx + 2α = 0. Therefore

2a + 2b = -2α

a + b = -α

Hence a and b are roots of the equation

x² + αx - 2α = 0

Discussion

No Comment Found

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