The point of concurrence of the lines $17 x + 2 y - 28 = 0, x + 5 y + 13 = 0$ and $7 x + 2 y - 8 = 0$ is

(- 2, 3)

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(2, - 3)

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(- 2, - 3)

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(2, 3)

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Solution

The correct option is B $(2, - 3)$
Let $(h, k)$ be the point of intersection of the first and second lines.
$17 h + 2 k - 28 = 0 h + 5 k + 13 = 0$
On solving the above equations, we get
$h = 2, k = - 3$

Substituting $(2, - 3)$ in third equation of line, we get
$7 (2) + 2 (- 3) - 8 = 0 \Rightarrow 14 - 14 = 0$
So, the three straight lines are concurrent.

Hence, the point of concurrence is $(h, k) = (2, - 3)$

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The equation of the line with slope $- \frac{3}{2}$ and which is concurrent with the lines $4 x + 3 y - 7 = 0$ and $8 x + 5 y - 1 = 0$ is

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x + 2 y - 2 = 0

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(1, \frac{1}{2})

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