JEE Mathematics 2023 Question with Solution

Given: The triangle is formed by the lines $15x + y = 82$, $6x - 5y = -4$, and $9x + 4y = 17$.

Find: The quadratic equation whose roots are the required expressions obtained from the centroid coordinates.

First find the vertices of the triangle by taking pairwise intersections of the three lines.

From $15x + y = 82$ and $6x - 5y = -4$,

$$\begin{aligned} y &= 82 - 15x \\ 6x - 5(82 - 15x) &= -4 \\ 6x - 410 + 75x &= -4 \\ 81x &= 406 \\ x &= \text{not consistent with the listed vertex set in the provided solution} \end{aligned}$$

However, the provided solution explicitly uses the vertices $A(6, 8)$, $B(5, -7)$, and $C(1, 2)$.

Using those given vertices, the centroid $(\alpha, \beta)$ is

$$\alpha = \frac{6 + 5 + 1}{3} = 4, \qquad \beta = \frac{8 + (-7) + 2}{3} = 1$$

Now compute the two required quantities shown in the solution:

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

Therefore the quadratic equation whose roots are $6$ and $7$ is

$$x^2 - (6+7)x + (6)(7) = 0$$

So,

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

Therefore, the correct option is A. The solution labels option D, but its own working gives $x^2 - 13x + 42 = 0$, which matches option A.

Given: From the provided solution working, the two roots are $6$ and $7$.

Find: The quadratic equation having these roots.

For roots $r_1$ and $r_2$, the quadratic is

$$x^2 - (r_1+r_2)x + r_1r_2 = 0$$

Here,

$$r_1 + r_2 = 6 + 7 = 13$$

and

$$r_1r_2 = 6 \times 7 = 42$$

Hence,

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

Therefore, the correct option is A.