Let ABC be a triangle formed by the lines 7x- | vedclass

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(A) First,find the vertices of the triangle by solving the equations of the lines pairwise:$1$) $7x-6y+3=0$ and $x+2y-31=0$ intersect at $A(9, 11)$.$2

Vedclass · Accepted Answer

$37$

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(A) First,find the vertices of the triangle by solving the equations of the lines pairwise:$1$) $7x-6y+3=0$ and $x+2y-31=0$ intersect at $A(9, 11)$.$2$) $7x-6y+3=0$ and $9x-2y-19=0$ intersect at $B(3, 4)$.$3$) $x+2y-31=0$ and $9x-2y-19=0$ intersect at $C(5, 13)$.The centroid $G$ of $\Delta ABC$ is given by $\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}ight) = \left(\frac{17}{3}, \frac{28}{3}ight)$.Let $(h, k)$ be the image of $G\left(\frac{17}{3}, \frac{28}{3}ight)$ in the line $3x+6y-53=0$. The formula for the image $(h, k)$ of a point $(x_1, y_1)$ in the line $ax+by+c=0$ is $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$.Here,$a=3, b=6, c=-53, x_1=\frac{17}{3}, y_1=\frac{28}{3}$.$\frac{h-17/3}{3} = \frac{k-28/3}{6} = -2\frac{3(17/3)+6(28/3)-53}{3^2+6^2} = -2\frac{17+56-53}{9+36} = -2\frac{20}{45} = -2\frac{4}{9} = -\frac{8}{9}$.$h - \frac{17}{3} = 3 	imes (-\frac{8}{9}) = -\frac{8}{3} \implies h = \frac{17-8}{3} = 3$.$k - \frac{28}{3} = 6 	imes (-\frac{8}{9}) = -\frac{16}{3} \implies k = \frac{28-16}{3} = 4$.Thus,$h^2+k^2+hk = 3^2+4^2+(3)(4) = 9+16+12 = 37$.

Let $ABC$ be a triangle formed by the lines $7x-6y+3=0$,$x+2y-31=0$,and $9x-2y-19=0$. Let the point $(h, k)$ be the image of the centroid of $\Delta ABC$ in the line $3x+6y-53=0$. Then $h^2+k^2+hk$ is equal to

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