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Question

# Find the equation of the line which passes through the point (3, 4) and the sum of its intercepts on the axes is 14.

A

2x+y7=0

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B

x+y7=0

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C

xy7=0

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D

x+y+7=0

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Solution

## The correct option is B x+y−7=0 Given a+b=14, Coordinates are A=(0,b),B=(a,0) Since A=(0,b)=(0,14−a) So equation of AB y−(14−a)=0−(14−a)(a−0)(x−0)⇒y−14+a=a−14a(x),…(i) It passes through (3, 4) So, 4−14+a=a−14a×3⇒ a−10=3a−42a⇒ a2−10a−3a+42=0⇒ a2−13a+42=0⇒ (a−7)(a−6)=0 So, a=7 or a=6 Put a=6 equation (i), we get y−14+a=a−14a(x)⇒ y−14+6=6−146x⇒ y−8=−43x⇒3y−24=−4x⇒ 4x+3y−24=0 Put a=7 equation (i), we get y−14+a=a−14a(x)⇒ y−14+7=7−147x⇒ y−7=−11x⇒ x+y−7=0

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