The number of straight lines that can be draw | vedclass

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(C) Let the point be $P(-3, 4)$ and $Q(2, -8)$. The distance between $P$ and $Q$ is $d = \sqrt{(2 - (-3))^2 + (-8 - 4)^2} = \sqrt{5^2 + (-12)^2} = \sq

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let the point be $P(-3, 4)$ and $Q(2, -8)$. The distance between $P$ and $Q$ is $d = \sqrt{(2 - (-3))^2 + (-8 - 4)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ units.Any line passing through $P$ can be represented as a line at some angle $	heta$ with the line $PQ$. The perpendicular distance from $Q$ to such a line is $d \sin 	heta$,where $d = 13$.We want the distance to be $5$ units,so $13 \sin 	heta = 5$,which gives $\sin 	heta = \frac{5}{13}$.Since $|\sin 	heta| \le 1$ and $\frac{5}{13} Thus,there are $2$ such lines.

The number of straight lines that can be drawn through the point $(-3, 4)$ which are at a distance of $5$ units from the point $(2, -8)$ is

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