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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 xa+yb=1.

Here, a + b = 7
b = 7 − a ... (1)

The line passes through (−3, 8).

-3a+8b=1 ... (2)

Substituting b = 7 − a in (2) we get,

-3a+87-a=1-37-a+8a=7a-a2a2+4a-21=0a-3a+7=0a=3, a-7 a is positive

Substituting a = 3 in (1) we get,

b = 7 − 3 = 4

Hence, the equation of the line is x3+y4=1 or 4x + 3y = 12

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