If A=(0,1), B=(1,2), C=(-2,1),then the equati | vedclass

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(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.

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$x^2-2xy-3y^2+2x+6y-3=0$

Vedclass · Answer

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. For $	riangle PAB$ with $P(x,y), A(0,1), B(1,2)$: Area $= \frac{1}{2} |x(1-2) + 0(2-y) + 1(y-1)| = \frac{1}{2} |-x + y - 1| = \frac{1}{2} |x - y + 1|$. For $	riangle PAC$ with $P(x,y), A(0,1), C(-2,1)$: Area $= \frac{1}{2} |x(1-1) + 0(1-y) + (-2)(y-1)| = \frac{1}{2} |0 + 0 - 2y + 2| = |1 - y|$. Equating the areas: $\frac{1}{2} |x - y + 1| = |1 - y| \implies |x - y + 1| = |2 - 2y|$. This gives two cases: Case $1$: $x - y + 1 = 2 - 2y \implies x + y - 1 = 0$. Case $2$: $x - y + 1 = -(2 - 2y) \implies x - y + 1 = -2 + 2y \implies x - 3y + 3 = 0$. Since the options are quadratic,we check the product of these lines: $(x + y - 1)(x - 3y + 3) = x^2 - 3xy + 3x + xy - 3y^2 + 3y - x + 3y - 3 = x^2 - 2xy - 3y^2 + 2x + 6y - 3 = 0$.

If $A=(0,1), B=(1,2), C=(-2,1)$,then the equation of the locus of a point $P(x,y)$ such that the area of triangle $PAB$ equals the area of triangle $PAC$ is:

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