Question

The equation ${\left(x-5\right)}^2+\left(x-5\right)\left(y-6\right)-2{\left(y-6\right)}^2=0$ represents

• A. A circle
• B. Two straight lines passing through the origin
• C. Two straight lines passing through the point $(5,6)$
• D. None of these

Answer 1

The correct option is C

Two straight lines passing through the point $(5,6)$

Explanation for the correct option:

Step 1: Simplify the given equation.

An equation ${\left(x-5\right)}^2+\left(x-5\right)\left(y-6\right)-2{\left(y-6\right)}^2=0$ is given.

Simplify the given equation as follows:

$$\begin{gathered} {\left(x-5\right)}^2+\left(x-5\right)\left(y-6\right)-2{\left(y-6\right)}^2=0 \\ \Rightarrow {\left(x-5\right)}^2+\left(x-5\right)\left(y-6\right)+\left(x-5\right)\left(y-6\right)-\left(x-5\right)\left(y-6\right)-2{\left(y-6\right)}^2=0 \\ \Rightarrow {\left(x-5\right)}^2+2\left(x-5\right)\left(y-6\right)-\left(x-5\right)\left(y-6\right)-2{\left(y-6\right)}^2=0 \\ \Rightarrow (x-5)\left[\left(x-5\right)+2\left(y-6\right)\right]-\left(y-6\right)\left[\left(x-5\right)+2\left(y-6\right)\right]=0 \\ \Rightarrow (x-5)\left[x+2y-17\right]-\left(y-6\right)\left[x+2y-17\right]=0 \\ \Rightarrow \left[x-y+1\right]\left[x+2y-17\right]=0 \end{gathered}$$

So, the given line represents the pair of straight lines $x-y+1=0$ and $x+2y-17=0$.

Step 2: Find the point of intersection of both lines.

Since, $x-y+1$

Therefore, $x=y-1...1$

Also, $x+2y-17=0...2$

From equation $1$ and equation $2$, we get

$$\begin{gathered} y-1+2y-17=0 \\ \Rightarrow 3y=18 \\ \Rightarrow y=6 \end{gathered}$$

Now, substitute $y=6$ in equation $1$ and solve for $x$.

$$\begin{gathered} x=6-1 \\ \Rightarrow x=5 \end{gathered}$$

Therefore, the point of intersection is $(5,6)$.

That is, the equation ${\left(x-5\right)}^2+\left(x-5\right)\left(y-6\right)-2{\left(y-6\right)}^2=0$ represents a pair of straight lines passing through the point $(5,6)$.

Hence, the option $C$ is correct.