The equation of the straight line passing thr | vedclass

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(B) Let the line intersect the $x$-axis at $A(-a, 0)$ and the $y$-axis at $B(0, b)$.The area of the triangle formed by the axes is given by $\frac{1}{

Vedclass · Accepted Answer

$2Tx - a^2y + 2aT = 0$

Vedclass · Answer

(B) Let the line intersect the $x$-axis at $A(-a, 0)$ and the $y$-axis at $B(0, b)$.The area of the triangle formed by the axes is given by $\frac{1}{2} 	imes |OA| 	imes |OB| = T$.Since $OA = |-a| = a$,we have $\frac{1}{2} 	imes a 	imes |b| = T$,which implies $|b| = \frac{2T}{a}$.Taking $b = \frac{2T}{a}$ (assuming the intercept is positive),the intercept form of the line is $\frac{x}{-a} + \frac{y}{2T/a} = 1$.Multiplying by $2T$,we get $-\frac{2Tx}{a} + ay = 2T$.Rearranging gives $2Tx - a^2y + 2aT = 0$.

The equation of the straight line passing through $(-a, 0)$ and forming a triangle with the coordinate axes of area $T$ is

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