Question

# Find the equation of the straight line which passes through the point (−3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7.

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Solution

## The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$. Here, a + b = 7 $⇒$b = 7 − a ... (1) The line passes through (−3, 8). ∴ $\frac{-3}{a}+\frac{8}{b}=1$ ... (2) Substituting b = 7 − a in (2) we get, $\frac{-3}{a}+\frac{8}{7-a}=1\phantom{\rule{0ex}{0ex}}⇒-3\left(7-a\right)+8a=7a-{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+4a-21=0\phantom{\rule{0ex}{0ex}}⇒\left(a-3\right)\left(a+7\right)=0\phantom{\rule{0ex}{0ex}}⇒a=3,a\ne -7\left(\because a\mathrm{is}\mathrm{positive}\right)$ Substituting a = 3 in (1) we get, b = 7 − 3 = 4 Hence, the equation of the line is $\frac{x}{3}+\frac{y}{4}=1$ or 4x + 3y = 12

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