Question

The area of the parallelogram formed by the lines $y=mx$, $y=mx+1$, $y=nx$ and $y=nx+1$ equals.

• A. $\frac{m+n}{{\left(m-n\right)}^2}$
• B. $\frac{2}{m+n}$
• C. $\frac{1}{m+n}$
• D. $\frac{1}{m-n}$

Answer 1

The correct option is D

$\frac{1}{m-n}$

Explanation for correct answer:

Step 1: Finding the coordinates of $A,B,O$:

Let the parallelogram is formed by the lines,

$OB=y=mx\to \left(i\right)$

$CA=y=mx+1\to \left(ii\right)$

$OC=y=nx\to \left(iii\right)$

$BA=y=nx+1\to \left(iv\right)$

Substitute the equation $\left(i\right)$ in $\left(iv\right)$ we get the point of intersection $B$

$$\begin{gathered} y=nx+1 \\ \Rightarrow mx=nx+1 \\ \Rightarrow mx-nx=1 \\ \Rightarrow x(m-n)=1 \\ \Rightarrow x=\frac{1}{m-n} \end{gathered}$$

Substitute $x=\frac{1}{m-n}$ in equation $\left(i\right)$ we get,

$\Rightarrow y=\frac{m}{m-n}$

So the coordinate of $B$ is $\left(\frac{1}{m-n},\frac{m}{m-n}\right)$

The intersection point ‘$O$’ has the coordinate $(0,0)$.

Now Substitute the equation $\left(ii\right)$ in $\left(iv\right)$ we get the point of intersection $A$. Hence,

$$\begin{gathered} y=nx+1 \\ \Rightarrow mx+1=nx+1 \\ \Rightarrow mx-nx=0 \\ \Rightarrow x(m-n)=0 \end{gathered}$$

$m-n\ne 0$because if $m-n=0\Rightarrow m=n$ which means the lines $y=mx$ and $y=nx$ will be identical.

Hence $x=0$.

Now put $x=0$ in equation $\left(ii\right)$ we get $y=1$

Therefore the coordinate $A$ is equal to $(0,1)$

Step 2: Finding the area of $∆OBA$:

Using matrix form to find the area of $∆OBA$

Area of the triangle$=\frac{1}{2}\times \left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

So Here,

$$\begin{gathered} =\frac{1}{2}\times \left|\begin{array}{ccc}0& 0& 1\\ 0& 1& 1\\ \frac{1}{m-n}& \frac{m}{m-n}& 1\end{array}\right| \\ =\frac{1}{2}\times \left|\left(-1\times \frac{1}{m-n}\right)\right| \\ =\frac{1}{2}\times \frac{1}{m-n} \end{gathered}$$

Area of $∆OBA=\frac{1}{2}\times \frac{1}{m-n}$

Step 3: Determine the area of the parallelogram.

Now the area of a parallelogram $OBAC$ is equal to twice the area of a triangle $OBA$.Hence,

Area of $OBAC$= $2\times$area of $∆OBA$

$$\begin{gathered} =2\times \frac{1}{2}\times \frac{1}{m-n} \\ =\frac{1}{m-n} \end{gathered}$$

Therefore the area of the parallelogram, formed by the given lines is equal to $\frac{1}{m-n}$.

Hence, the correct answer is option (D)