The number of points having a distance of sqr | vedclass

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(C) Let the point be $P(h, k)$.Given the distance from the line $x-2y+1=0$ is $\sqrt{5}$,we have:$\left|\frac{h-2k+1}{\sqrt{1^2+(-2)^2}}\right| = \sqr

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$4$

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(C) Let the point be $P(h, k)$.Given the distance from the line $x-2y+1=0$ is $\sqrt{5}$,we have:$\left|\frac{h-2k+1}{\sqrt{1^2+(-2)^2}}ight| = \sqrt{5} \Rightarrow |h-2k+1| = \sqrt{5} 	imes \sqrt{5} = 5$.This gives two parallel lines: $h-2k+1 = 5$ or $h-2k+1 = -5$.Given the distance from the line $2x+3y-1=0$ is $\sqrt{13}$,we have:$\left|\frac{2h+3k-1}{\sqrt{2^2+3^2}}ight| = \sqrt{13} \Rightarrow |2h+3k-1| = \sqrt{13} 	imes \sqrt{13} = 13$.This gives two parallel lines: $2h+3k-1 = 13$ or $2h+3k-1 = -13$.Each pair of parallel lines intersects the other pair at $2 	imes 2 = 4$ distinct points.Thus,there are $4$ such points.

The number of points having a distance of $\sqrt{5}$ from the straight line $x-2y+1=0$ and a distance of $\sqrt{13}$ from the line $2x+3y-1=0$ is:

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