A quadrilateral ABCD is divided by the diagon | vedclass

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Question

(D) Given points are $A(3, 4), B(-3, 6), C(-5, 1), D(x, y)$.Area of $	riangle ABC = \frac{1}{2} |3(6 - 1) + (-3)(1 - 4) + (-5)(4 - 6)|$$= \frac{1}{2}

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$(3x - 8y - 11)(3x - 8y + 57) = 0$

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(D) Given points are $A(3, 4), B(-3, 6), C(-5, 1), D(x, y)$.Area of $	riangle ABC = \frac{1}{2} |3(6 - 1) + (-3)(1 - 4) + (-5)(4 - 6)|$$= \frac{1}{2} |3(5) + (-3)(-3) + (-5)(-2)| = \frac{1}{2} |15 + 9 + 10| = 17 \dots (1)$Area of $	riangle ACD = \frac{1}{2} |3(1 - y) + (-5)(y - 4) + x(4 - 1)|$$= \frac{1}{2} |3 - 3y - 5y + 20 + 3x| = \frac{1}{2} |3x - 8y + 23| \dots (2)$Since area of $	riangle ABC = $ area of $	riangle ACD$,we have $\frac{1}{2} |3x - 8y + 23| = 17$$|3x - 8y + 23| = 34$$3x - 8y + 23 = 34$ or $3x - 8y + 23 = -34$$3x - 8y - 11 = 0$ or $3x - 8y + 57 = 0$Thus,the locus of $D$ is $(3x - 8y - 11)(3x - 8y + 57) = 0$.

$A$ quadrilateral $ABCD$ is divided by the diagonal $AC$ into two triangles of equal areas. If $A, B, C$ are respectively $(3, 4), (-3, 6), (-5, 1)$,then the locus of $D$ is

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