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Question

# Find the equation of the straight line which passes through the point of intersection of the straight lines 3x - 4y + 1 = 0 and 5x + y - 1 = 0 and cuts off equal intercepts from the axis.

A
32x + 32y + 11 = 0
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B
23x + 23y = 11
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C
9x + 18y + 5 = 0
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D
None of these
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Solution

## The correct option is B 23x + 23y = 11The equation of any line passing through the point of intersection of the lines 3x - 4y + 1 = 0 and 5x + y - 1 = 0 is (3x - 4y + 1) + k(5x+ y-1) = 0......(1) For intercept of this line with the x-axis, y = 0 ∴ 3x+ 1 + k (5x - 1) = 0 ⇒x=k−15k+3 For intercept of the line (1) on the y -axis, x = 0 ∴−4y+1+k(y−1)=0⇒y=k−1k−4 Since the intercepts on the axes are equal. ∴ k−15k+3=k−1k−4⇒k=1, or k=−74 But k ≠ 1, because if k = 1, the line (1) becomes 8x - 3y = 0 which passes through the origin and therefore cannot make non-zero intercepts on the axis. ∴k=−74 and from (1), we get 3x−4y+1−74(5x+y−1)=0 ⇒23x+23y=11, which is the required equation of the line.

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