JEE Mathematics 2023 Question with Solution

Given: The sides of triangle $\Delta ABC$ are $AB: 2x + y = 0$, $BC: x + py = 21a$, and $CA: x - y = 3$. The centroid is $P(2,a)$.

Find: The value of $(BC)^2$.

Assume

using the side equations $2x+y=0$ and $x-y=3$.

Using the centroid formula,

and

So,

and

From $\alpha+\beta=2$, we get

Substituting into the second relation,

Since $B$ lies on $BC: x+py=21a$,

Using $\alpha=-a$,

Since $a\neq 0$,

Now $C=(\beta+3,\beta)$ also lies on $BC$. Hence,

With $p=11$,

Using $\beta=2-\alpha$ and $\alpha=-a$,

Also from the solution working,

Substitute $\beta=2-\alpha$:

Therefore,

So,

Then

Therefore, the value of $(BC)^2$ is $122$.

Given: $AB: 2x+y=0$, $BC: x+py=21a$, $CA: x-y=3$, and centroid $P=(2,a)$.

Find: $(BC)^2$.

First find vertex $A$ as the intersection of $AB$ and $CA$:

From $x-y=3$, we get $y=x-3$. Substitute into the first equation:

Hence,

Let $B=(\gamma,\delta)$. Since $B$ lies on $AB$ and $BC$,

and

From the first equation,

Substitute into the second:

Let $C=(\epsilon,\zeta)$. Since $C$ lies on $CA$ and $BC$,

and

From $\epsilon-\zeta=3$, write

Substitute into the second:

Now use the centroid formula:

Thus,

and

Using the parameterization in the extracted working gives

so that

Finally,

Therefore, the value of $(BC)^2$ is $122$.