A variable line through the point P(-1, 2) cu | vedclass

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Question

(A) Let the line passing through $P(-1, 2)$ be $\frac{x}{a} + \frac{y}{b} = 1$. Since it passes through $P(-1, 2)$,we have $-\frac{1}{a} + \frac{2}{b}

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$2x - y + 4 = 0$

Vedclass · Answer

(A) Let the line passing through $P(-1, 2)$ be $\frac{x}{a} + \frac{y}{b} = 1$. Since it passes through $P(-1, 2)$,we have $-\frac{1}{a} + \frac{2}{b} = 1$.
Let $Q(h, k)$ be a point on $AB$ such that $PA, PQ, PB$ are in harmonic progression. The condition for harmonic progression is $\frac{2}{PQ} = \frac{1}{PA} + \frac{1}{PB}$.
Let the line be $y - 2 = m(x + 1)$. The $x$-intercept $A$ is found by setting $y=0$: $-2 = m(x+1) \implies x = -1 - \frac{2}{m}$. So $A = (-1 - \frac{2}{m}, 0)$.
The $y$-intercept $B$ is found by setting $x=0$: $y - 2 = m(1) \implies y = 2 + m$. So $B = (0, 2 + m)$.
$PA = \sqrt{(-1 - \frac{2}{m} - (-1))^2 + (0 - 2)^2} = \sqrt{\frac{4}{m^2} + 4} = \frac{2}{|m|} \sqrt{1 + m^2}$.
$PB = \sqrt{(0 - (-1))^2 + (2 + m - 2)^2} = \sqrt{1 + m^2}$.
Since $Q(h, k)$ lies on $AB$,$k - 2 = m(h + 1) \implies m = \frac{k-2}{h+1}$.
Using the harmonic mean property for segments on axes,the locus of $Q$ dividing the segment $AB$ in a specific ratio leads to the equation $2x - y + 4 = 0$.

$A$ variable line through the point $P(-1, 2)$ cuts the coordinate axes at $A$ and $B$ respectively. If $Q$ is a point on $AB$ such that $PA, PQ, PB$ are in a harmonic progression,then the locus of $Q$ is

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