QJEEM23Ma1d02: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1d02 Solution: Let be the centroid of the triangle formed by the lines and .…

1. Solve the equations $15x - y = 82$ and $6x - 5y = -4$

Multiply the first equation by 5 and the second by 1 and then subtract the second from the first:

$75x - 5y = 410$

$6x - 5y = -4$

Subtracting these gives $69x = 414$ which leads to $x = 6$

Substitute $x = 6$ into the first equation to get $y = 8$

So, the first vertex is $(6,8)$ .

1. Solve the equations $15x - y = 82$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 1 and then add:

$60x - 4y = 328$

$9x + 4y = 17$

Adding these gives $69x = 345$ which leads to $x = 5$

Substitute $x = 5$ into the second equation to get $y = -7$

So, the second vertex is $(5,-7)$ .

1. Solve the equations $6x - 5y = -4$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 5 and then add:

$24x - 20y = -16$

$45x + 20y = 85$

Adding these gives $69x = 69$ which leads to $x = 1$

Substitute $x = 1$ into the first equation to get $y = 2$

So, the third vertex is $(1,2)$ .

So, the vertices of the triangle are $(6,8)$ , $(5,-7)$ , and $(1,2)$ .

Now that we have the vertices of the triangle, we can find the centroid. The centroid $(\alpha, \beta)$ of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ is given by :

$\alpha=\frac{x_1+x_2+x_3}{3}$ ,

$\beta=\frac{y_1+y_2+y_3}{3}$

Substituting the coordinates of the vertices into these equations, we get:

$\alpha=\frac{6+5+1}{3}=4$ ,

$\beta=\frac{8-7+2}{3}=1$

Then, we find $\alpha+2\beta$ and $2\alpha-\beta$ :

$\alpha+2\beta=4+2(1)=6 \2\alpha-\beta=2(4)-1=7$

So, $\alpha+2\beta=6$ and $2\alpha-\beta=7$ are the roots of the quadratic equation.

We can write the quadratic equation as

$x^2-(\alpha+2\beta+2\alpha-\beta)x+(\alpha+2\beta)(2\alpha-\beta)=0$

Substituting $\alpha=4$ , $\beta=1$ gives us:

$x^2-13x+42=0$