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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+y7=0





Given a+b=14,
Coordinates are A=(0,b),B=(a,0)
Since A=(0,b)=(0,14a)
So equation of AB
y(14a)=0(14a)(a0)(x0)y14+a=a14a(x),(i)
It passes through (3, 4) So,
      414+a=a14a×3  a10=3a42a  a210a3a+42=0  a213a+42=0  (a7)(a6)=0
So, a=7 or a=6
Put a=6 equation (i), we get
      y14+a=a14a(x)  y14+6=6146x  y8=43x3y24=4x  4x+3y24=0
Put a=7 equation (i), we get
      y14+a=a14a(x)  y14+7=7147x  y7=11x  x+y7=0


Mathematics

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