A straight line passing through the origin O | vedclass

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(B) Let the line passing through the origin $O(0,0)$ be $L_1$. It intersects $L_2: 10x - 8y - 10 = 0$ (which simplifies to $5x - 4y - 5 = 0$) at point

Vedclass · Accepted Answer

$1 : 4$

Vedclass · Answer

(B) Let the line passing through the origin $O(0,0)$ be $L_1$. It intersects $L_2: 10x - 8y - 10 = 0$ (which simplifies to $5x - 4y - 5 = 0$) at point $P$ at right angles.It also intersects $L_3: \frac{x}{4} - \frac{y}{5} + 1 = 0$ (which simplifies to $5x - 4y + 20 = 0$) at point $Q$ at right angles.Since $L_2$ and $L_3$ have the same slope,they are parallel lines.The perpendicular distance from the origin $O(0,0)$ to line $L_2$ is $OP = \frac{|5(0) - 4(0) - 5|}{\sqrt{5^2 + (-4)^2}} = \frac{5}{\sqrt{41}}$.The perpendicular distance from the origin $O(0,0)$ to line $L_3$ is $OQ = \frac{|5(0) - 4(0) + 20|}{\sqrt{5^2 + (-4)^2}} = \frac{20}{\sqrt{41}}$.Since $P$ and $Q$ lie on opposite sides of the origin (as the constant terms $-5$ and $+20$ have opposite signs),the origin $O$ divides the segment $PQ$ internally in the ratio $OP : OQ = \frac{5}{\sqrt{41}} : \frac{20}{\sqrt{41}} = 1 : 4$.

$A$ straight line passing through the origin $O$ intersects the lines $10x - 8y - 10 = 0$ and $\frac{x}{4} - \frac{y}{5} + 1 = 0$ at right angles at points $P$ and $Q$ respectively. Then the ratio in which $O$ divides the line segment $PQ$ is

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