JEE Mathematics 2024 Question with Solution

Given: $A(\alpha, \beta)$, $B(1,0)$, $C(\gamma, \delta)$, $D(1,2)$ are vertices of a parallelogram, $AB=\sqrt{10}$, and points $A$ and $C$ lie on $3y=2x+1$.

Find: $2(\alpha+\beta+\gamma+\delta)$.

Since $A$ and $C$ lie on the line, we have

$$3\beta = 2\alpha + 1$$

and

$$3\delta = 2\gamma + 1$$

Also, from $AB=\sqrt{10}$,

$$\sqrt{(\alpha-1)^2 + \beta^2} = \sqrt{10}$$

So,

$$(\alpha-1)^2 + \beta^2 = 10$$

Now use the parallelogram property. The vectors are

$$\overrightarrow{AB} = (1-\alpha,-\beta)$$

and

$$\overrightarrow{CD} = (1-\gamma,2-\delta)$$

For a parallelogram,

$$\overrightarrow{AB} = \overrightarrow{CD}$$

Hence,

$$1-\alpha = 1-\gamma$$

and

$$-\beta = 2-\delta$$

Therefore,

$$\alpha = \gamma$$

and

$$\beta + \delta = 2$$

Using $\alpha=\gamma$ in the two line equations gives

$$3\beta = 2\alpha + 1$$

and

$$3\delta = 2\alpha + 1$$

So,

$$3\beta = 3\delta$$

which implies

$$\beta = \delta$$

Together with $\beta+\delta=2$, we get

$$\beta = 1, \quad \delta = 1$$

Now from

$$3\beta = 2\alpha + 1$$

we obtain

$$3(1) = 2\alpha + 1$$

so

$$\alpha = 1$$

Thus,

$$\gamma = 1$$

Finally,

$$2(\alpha+\beta+\gamma+\delta) = 2(1+1+1+1) = 8$$

Therefore, the correct option is D.

Given: $A(\alpha, \beta)$, $B(1,0)$, $C(\gamma, \delta)$, $D(1,2)$ form a parallelogram.

Find: $2(\alpha+\beta+\gamma+\delta)$.

In a parallelogram, diagonals bisect each other. So the midpoint of $AC$ equals the midpoint of $BD$.

For diagonal $BD$, the midpoint is

$$\left(\frac{1+1}{2}, \frac{0+2}{2}\right) = (1,1)$$

Hence, for diagonal $AC$,

$$\frac{\alpha+\gamma}{2} = 1$$

and

$$\frac{\beta+\delta}{2} = 1$$

So,

$$\alpha+\gamma = 2, \quad \beta+\delta = 2$$

Therefore,

$$2(\alpha+\beta+\gamma+\delta) = 2\big((\alpha+\gamma)+(\beta+\delta)\big) = 2(2+2) = 8$$

This directly gives the correct option as D.

The shortcut works because the required expression depends only on the sums $\alpha+\gamma$ and $\beta+\delta$, which are obtained immediately from the midpoint property.