The area of the triangle formed by the intersection of a line parallel to x - axis and passing through P(h,k) with the lines y = x and x + y = 2 is $4 h^{2}$ . Find the locus of the point P.

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Solution

$a r (△ A B C) = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 2 - k & k & 1 k & k & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 4 h^{2}$
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 2 - k & k & 1 k & k & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 8 h^{2}$
$\Rightarrow (2 - k) [k - 1 - 1] - k (k - 1) + 1 (k - k) = 4 h^{2}$
$(2 - h) (k - 2) - k^{2} + k = 4 h^{2}$
$2 k - k^{2} - 4 + 2 k - k^{2} + k = 4 h^{2}$
$5 k - 2 k^{2} - 4 - 4 h^{2}$
$2 k^{2} - 5 k + 4 h^{2} + 4 = 0$
$2 y^{2} - 5 y + 4 x^{2} + 4 = 0$
$∴$ Locus P is $4 x^{2} + 2 y^{2} - 5 y + 4 = 0$
Hence, solved.

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The area of the triangle formed by the intersection of a line parallel to x - axis and passing through P(h,k) with the lines y = x and x + y = 2 is 4h2. Find the locus of the point P.

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