$p_{1}$ and $p_{2}$ are the distances of a variable point from the
two lines $2 x + y - 1 = 0$ and $x + 2 y + 1 = 0$ satisfying the
condition that $p_{1}^{2} = p_{2}^{2} + 3$ . Prove that the product of
the distance of the same point from the lines
$x + y + 2 = 0$ and $x - y - 3 = 0$ varies as its distance from the line $x - 3 y - 1 = 0$ .

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Solution

$p_{1}^{2} - p_{2}^{2} = 3 \Rightarrow (h + k) (h - k - 2) = 5$
Or $h^{2} - k^{2} = 5 + 2 h + 2 k$
Now product of perpendicular from $(h, k)$ to other two lines is $d_{1} . d_{2} = \frac{h + k + 2}{\sqrt{2}} . \frac{h - k - 3}{\sqrt{2}}$
$\frac{1}{2} [(h^{2} - k^{2}) + (2 h - 2 k) - 3 (h + k) - 6]$
$\frac{1}{2} [5 + 2 h + 2 k + 2 h - 2 k - 3 h - 3 k - 6]$
$= \frac{h - 3 k - 1}{2}$ by $(1)$
Again if $p$ be distance of $(h, k)$ from
$x - 3 y - 1 = 0$
Then $p = \frac{h - 3 k - 1}{\sqrt{10}}$
$∴ \frac{d_{1} . d_{2}}{p} = \frac{\sqrt{10}}{2} =$ constant, by $(2)$ and $(3)$
Hence $d_{1} . d_{2}$ varies as $p$

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p1 and p2 are the distances of a variable point from the two lines 2x+y−1=0 and x+2y+1=0 satisfying the condition that p21=p22+3. Prove that the product of the distance of the same point from the lines x+y+2=0 and x−y−3=0 varies as its distance from the line x−3y−1=0.