Question

The diagonals of a square are along the pair of lines whose equation is $2{x^2} - 3xy - 2{y^2} = 0$. If $\left( {2,1} \right)$ is a vertex of the square, then the vertex of the square adjacent to it may be
(This question has multiple correct options)
A.$\left( {1, - 4} \right)$
B.$\left( { - 1,4} \right)$
C.$\left( {1, - 2} \right)$
D.$\left( { - 1,2} \right)$

Answer 1

Here, we will first replace 2 for $x$ in the equation $2{x^2} - 3xy - 2{y^2} = 0$ and then replace 1 for$y$ in the given equation to find the value of $y$ and $x$ respectively. Then we will find the uncommon vertices to find the required values.

Complete step-by-step answer:
We are given the diagonals of a square along the pair of lines whose equation is $2{x^2} - 3xy - 2{y^2} = 0$.
Replacing 2 for $x$ in the above equation, we get

$$\Rightarrow 2{\left( 2 \right)^2} - 3\left( 2 \right)y - 2{y^2} = 0 \\\ \Rightarrow 8 - 6y - 2{y^2} = 0 \\\$$

Dividing the above equation by 2 on both sides, we get

$$\Rightarrow \dfrac{{8 - 6y - 2{y^2}}}{2} = 0 \\\ \Rightarrow 4 - 3y - {y^2} = 0 \\\ \Rightarrow {y^2} + 3y - 4 = 0 \\\$$

Factoring the above equation, we get

$$\Rightarrow {y^2} + 4y - y - 4 = 0 \\\ \Rightarrow y\left( {y + 4} \right) - \left( {y + 4} \right) = 0 \\\ \Rightarrow \left( {y + 4} \right)\left( {y - 1} \right) = 0 \\\$$

$\Rightarrow y + 4 = 0$ or $y - 1 = 0$
$\Rightarrow y = - 4,1$
Thus, the two vertices are $\left( {2,1} \right)$ and $\left( {2, - 4} \right)$.
Now consider replacing 1 for$y$ in the given equation, we get

$$\Rightarrow 2{x^2} - 3x\left( 1 \right) - 2{\left( 1 \right)^2} = 0 \\\ \Rightarrow 2{x^2} - 3x - 2 = 0 \\\$$

Factoring the above equation, we get

$$\Rightarrow 2{x^2} - 4x + x - 2 = 0 \\\ \Rightarrow 2x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0 \\\ \Rightarrow \left( {2x + 1} \right)\left( {x - 2} \right) = 0 \\\$$

$\Rightarrow 2x + 1 = 0$ or $x - 2 = 0$
$\Rightarrow x = 2, - \dfrac{1}{2}$
Thus, the two vertices are $\left( {2,1} \right)$ and $\left( { - \dfrac{1}{2},1} \right)$.
From the above, we get the two uncommon vertices as $\left( {2, - 4} \right)$ and $\left( { - \dfrac{1}{2},1} \right)$.
Now, decreasing first vertex by a factor of 2 and increasing the second by the same, we get
$\left( {1, - 2} \right)$ and $\left( { - 1,2} \right)$
Hence, option C and D is correct.

Note: In solving these types of questions, Heron’s formula should be used to compute the area of a triangle where sides of the triangle are given. Students can also find the area of triangle of using the formula, ${\text{Area}} = \dfrac{1}{2} \times {\text{base}} \times {\text{height}}$, but this way could take a little longer as we have to find the height of the triangle first.