A straight line passing through a point (3, 2 | vedclass

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Question

(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.The equation of the line passing through $A$ and $B$ is $\frac{x}{a} + \frac{y}{b} = 1$

Vedclass · Accepted Answer

$4x + 9y = 5xy$

Vedclass · Answer

(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.The equation of the line passing through $A$ and $B$ is $\frac{x}{a} + \frac{y}{b} = 1$.Since the line passes through $(3, 2)$,we have $\frac{3}{a} + \frac{2}{b} = 1$.Point $P(h, k)$ divides $AB$ in the ratio $2: 3$. Using the section formula:$h = \frac{2(0) + 3(a)}{2 + 3} = \frac{3a}{5} \implies a = \frac{5h}{3}$.$k = \frac{2(b) + 3(0)}{2 + 3} = \frac{2b}{5} \implies b = \frac{5k}{2}$.Substitute $a$ and $b$ into the line equation:$\frac{3}{5h/3} + \frac{2}{5k/2} = 1 \implies \frac{9}{5h} + \frac{4}{5k} = 1$.Multiplying by $5$,we get $\frac{9}{h} + \frac{4}{k} = 5$,or $\frac{9}{x} + \frac{4}{y} = 5$ is incorrect; let's re-evaluate.Actually,$\frac{9}{5h} + \frac{4}{5k} = 1 \implies \frac{9}{h} + \frac{4}{k} = 5$. Wait,the options suggest $\frac{9}{x} + \frac{4}{y} = 1$ is not matching. Let's re-check the ratio.If $P$ divides $AB$ in $2:3$,$P = (\frac{2(0)+3(a)}{5}, \frac{2(b)+3(0)}{5}) = (\frac{3a}{5}, \frac{2b}{5})$.$a = \frac{5h}{3}, b = \frac{5k}{2}$.$\frac{3}{5h/3} + \frac{2}{5k/2} = 1 \implies \frac{9}{5h} + \frac{4}{5k} = 1 \implies \frac{9}{h} + \frac{4}{k} = 5$.Given the options,let's check if the ratio was $3:2$ instead. If $P$ divides $AB$ in $3:2$,$P = (\frac{3(0)+2(a)}{5}, \frac{3(b)+2(0)}{5}) = (\frac{2a}{5}, \frac{3b}{5})$.$a = \frac{5h}{2}, b = \frac{5k}{3}$.$\frac{3}{5h/2} + \frac{2}{5k/3} = 1 \implies \frac{6}{5h} + \frac{6}{5k} = 1 \implies \frac{6}{h} + \frac{6}{k} = 5$. Still not matching.Re-reading: $P$ divides $AB$ in $2:3$. $P = (\frac{3a}{5}, \frac{2b}{5})$. $a = 5h/3, b = 5k/2$. $\frac{3}{5h/3} + \frac{2}{5k/2} = 1 \implies \frac{9}{5h} + \frac{4}{5k} = 1$. This simplifies to $9k + 4h = 5hk$. Thus $4x + 9y = 5xy$.

$A$ straight line passing through a point $(3, 2)$ cuts the $X$ and $Y$-axes at points $A$ and $B$ respectively. If a point $P(h, k)$ divides $AB$ in the ratio $2: 3$,then the equation of the locus of point $P$ is

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