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Slope of a line passing through P(2,3) and intersecting the line , $x+y=7$ at a distance of $4$ units from $P,$ is :

Option 1)
$\frac{1-\sqrt{5}}{1+\sqrt{5}}$

Option 2)
$\frac{1-\sqrt{7}}{1+\sqrt{7}}$

Option 3)
$\frac{7-\sqrt{1}}{7+\sqrt{1}}$

Option 4)
$\frac{5-\sqrt{1}}{5+\sqrt{1}}$

Answers (1)

best_answer

$x+y=7$

$P\left ( 2,3 \right )$

Let point $Q$ be $(a,7-a)$

distance b/w $PQ=\sqrt{\left ( a-2 \right )^{2}+\left ( 7-a-3 \right )^{2}}=4$

$\left ( a-2 \right )^{2}+\left ( 4-1 \right )^{2}=16$

$a=3\pm \sqrt{7}$

When $a=3- \sqrt{7}$

slope of $PQ=M=\frac{1+\sqrt{7}}{1-\sqrt{7}}$

When $a=3+\sqrt{7}$

Slope of $PQ=\frac{1-\sqrt{7}}{1+\sqrt{7}}$

Option 1)

$\frac{1-\sqrt{5}}{1+\sqrt{5}}$

Option 2)

$\frac{1-\sqrt{7}}{1+\sqrt{7}}$

Option 3)

$\frac{7-\sqrt{1}}{7+\sqrt{1}}$

Option 4)

$\frac{5-\sqrt{1}}{5+\sqrt{1}}$

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