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(C) Let $a$ and $b$ be the intercepts made by the line on the axes.We are given $a + b = 8$ and $ab = 15$.Solving the quadratic equation $t^2 - 8t + 1

Vedclass · Accepted Answer

$3x + 5y - 15 = 0, 5x + 3y - 15 = 0$

Vedclass · Answer

(C) Let $a$ and $b$ be the intercepts made by the line on the axes.We are given $a + b = 8$ and $ab = 15$.Solving the quadratic equation $t^2 - 8t + 15 = 0$,we get $(t - 3)(t - 5) = 0$,so $(a, b) = (3, 5)$ or $(5, 3)$.The intercept form of the equation of a line is $\frac{x}{a} + \frac{y}{b} = 1$.Case $1$: When $a = 3$ and $b = 5$,the equation is $\frac{x}{3} + \frac{y}{5} = 1$,which simplifies to $5x + 3y - 15 = 0$.Case $2$: When $a = 5$ and $b = 3$,the equation is $\frac{x}{5} + \frac{y}{3} = 1$,which simplifies to $3x + 5y - 15 = 0$.Thus,the equations are $5x + 3y - 15 = 0$ and $3x + 5y - 15 = 0$.

The equations of the lines which make intercepts on the axes whose sum is $8$ and product is $15$ are

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