The number of lines that can be drawn through | vedclass

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(B) Let the given point be $P(4, -5)$ and the fixed point be $Q(1, 3)$.The distance $d$ between $P$ and $Q$ is given by the distance formula:$d = \sqr

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let the given point be $P(4, -5)$ and the fixed point be $Q(1, 3)$.The distance $d$ between $P$ and $Q$ is given by the distance formula:$d = \sqrt{(4 - 1)^2 + (-5 - 3)^2} = \sqrt{3^2 + (-8)^2} = \sqrt{9 + 64} = \sqrt{73}$.Since $\sqrt{73} \approx 8.54$,the distance between the point $P$ and the point $Q$ is less than $10$ units.For a line to be at a distance of $10$ units from point $Q$,the point $P$ must be at a distance of at least $10$ units from $Q$ (if $P$ lies on the line) or the line must be tangent to a circle of radius $10$ centered at $Q$.Since the distance $PQ Any line passing through a point inside a circle must intersect the circle at two points,meaning the distance from the center $Q$ to any line passing through $P$ will always be less than the radius $10$.Therefore,it is impossible to draw a line through $P$ such that its distance from $Q$ is $10$ units.The number of such lines is $0$.

The number of lines that can be drawn through the point $(4, -5)$ at a distance of $10$ units from the point $(1, 3)$ is

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