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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Aicharya · Accepted Answer

Clearly, by figure, line $x=-3$ is 5 units apart from point $Q(2,-8)$.

Now let ' $m$ ' be the slope of a line passing through $(-3,4)$ and 5 units apart from point $Q(2,-8)$. Hence equation of line can be given as $(y-4)=m(x+3)$ $ m x-y+(3 m+4)=0 $ Since equation (i) is 5 unit apart from $Q(2,-8)$ Hence $ \begin{aligned} & \Rightarrow \frac{m(2)-(-8)+(3 m+4)}{\sqrt{m^2+(-1)^2}}=5 \\ & \Rightarrow 5 m+12=5 \sqrt{m^2+1} \\ & \Rightarrow 25 m^2+144+120 m=25 m^2+25 \\ & \Rightarrow m=-\frac{119}{120} \end{aligned} $ Hence $\left(-\frac{119}{120}\right) x-y+\left(\frac{123}{120}\right)=0$ will be $2^{\text {nd }}$ equation from $(-3,4)$ which is 5 units apart from $(2,-8)$

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

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