Question

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

• A. Square but not rhombus
• B. Rhombus
• C. Parallelogram
• D. Rectangle but not a square

Answer 1

The correct option is C

Parallelogram

Step 1: Find the slope of the given straight lines

Given equation are

$x^2-3xy+2y^2=0$ $..................$$\left(i\right)$

$x^2-3xy+2y^2+x-2=0$ $..................$$\left(ii\right)$

From equation $\left(i\right)$ we get

$$\begin{gathered} x^2-3xy+2y^2=0 \\ \Rightarrow x^2-2xy-xy+2y^2=0 \\ \Rightarrow x\left(x-2y\right)-y\left(x-2y\right)=0 \\ \Rightarrow \left(x-2y\right)\left(x-y\right)=0 \end{gathered}$$

From equation $\left(ii\right)$ we get

$$\begin{gathered} x^2-3xy+2y^2+x-2=0 \\ \Rightarrow \left(x-2y+2\right)\left(x-y-1\right)=0 \end{gathered}$$

Therefore the straight lines are

$$\begin{gathered} x-2y=0.......\left(iii\right) \\ x-y=0........\left(iv\right) \\ x-2y+2=0....\left(v\right) \\ x-y-1=0......\left(vi\right) \end{gathered}$$

Comparing $\left(iii\right)$ and $\left(v\right)$ we can say that they are parallel since the slope is the same

Also $\left(iv\right)$ and $\left(vi\right)$ are parallel, since the slope is same.

Consider the equation $\left(iii\right)$

Slope $m_1=\frac{1}{2}$ and from $\left(iv\right)$ slope $m_2=1$

Step 2: Find the angle using the slopes obtained.

The angle between the lines is ,

$$\begin{gathered} \tan \left(\theta \right)=\frac{\left|m_1-m_2\right|}{\left|1+m_1{m}_2\right|} \\ =\left|\frac{{\displaystyle \frac{1}{2}}-1}{1+{\displaystyle \frac{1}{2}}·1}\right| \\ =\frac{1}{3} \end{gathered}$$

So the angle is not $90°$

Thus, a parallelogram is formed.

Hence, the correct option is $\left(\mathrm{C}\right)$