Let A B C be a triangle formed by lines 7 x-6 y+3=0, x+2 y-31=0 and 9 x-2 y-19=0 , Let point ( h , k

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9.Straight Line

hard

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

A$37$

B$47$

C$40$

D$36$

(JEE MAIN-2025)

Solution

image
$ \therefore \text { centroid of } \triangle ABC =\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right)$
$ =\left(\frac{17}{3}, \frac{28}{3}\right)$
image
Let image of centroid with respect to line mirror is $( h , k )$
$\therefore\left(\frac{ k -\frac{28}{3}}{h-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1$
$\& 3\left(\frac{h+\frac{17}{3}}{2}\right)+6 \cdot\left(\frac{\frac{ k +28}{3}}{2}\right)=53$
Solving (1) \ (2) we get $h =3, k =4$
$\therefore h^2+k^2+hk=37$

Standard 11

Mathematics

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

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