BITSAT 2013 Paper

Equation of the line making intercepts a and b on the axes is $\frac{x}{a} + \frac{y}{b}$ = 1 Since, it passes through (1, 1) ⇒ $\frac{1}{a} + \frac{1}{b}$ = 1 ... (i) Also the area of the triangle formed by the line and the axes is A. ∴ $\frac{1}{2}$ ab = A ⇒ ab = 2a ... (ii) From eqs. (i) and (ii), we get, a + b = 2A Hence, a and b are the roots of the eq. $x^2$ - (a + b)x + ab = 0 ⇒ $x^2$ - 2Ax + 2A = 0