Question

The intersection of three lines $x-y=0$, $x+2y=3$ and $2x+y=6$ is a

• A. Equilateral triangle.
• B. Right-angled triangle.
• C. Isosceles triangle.
• D. None of the above.

Answer 1

The correct option is C

Isosceles triangle.

Step 1: Solve the equation $\left(\mathrm{ii}\right)-$Equation $\left(\mathrm{i}\right)$.

Given equations of the straight lines are as follows,

$x-y=0\_\left(\mathrm{i}\right)$

$x+2y=3\_\left(\mathrm{ii}\right)$

$2x+y=6\_\left(\mathrm{iii}\right)$

Equation $\left(\mathrm{ii}\right)-$ Equation $\left(\mathrm{i}\right)$.

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

Substitute $y$ with $1$ in equation $\left(\mathrm{i}\right)$.

$$\begin{gathered} x-y=0 \\ \Rightarrow x-1=0 \\ \Rightarrow x=1 \end{gathered}$$

Thus, the point is $A\left(1,1\right)$.

Step 2: Solve the equation $\left(iii\right)-$equation $\left(\mathrm{ii}\right)$.

Equation $\left(\mathrm{iii}\right)\times 2-$ Equation $\left(\mathrm{ii}\right)$.

$$\begin{gathered} 2\times \left(2x+y\right)-\left(x+2y\right)=\left(2\times 6\right)-3 \\ \Rightarrow 3x=9 \\ \Rightarrow x=3 \end{gathered}$$

Substitute $x$ with $3$ in equation $\left(\mathrm{ii}\right)$.

$$\begin{gathered} x+2y=3 \\ \Rightarrow 3+2y=3 \\ \Rightarrow y=0 \end{gathered}$$

Thus, the point is $B\left(3,0\right)$.

Step 3: Solve the equation $\left(iii\right)-$equation $\left(i\right)$

Equation $\left(\mathrm{iii}\right)+$ Equation $\left(\mathrm{i}\right)$.

$$\begin{gathered} \left(2x+y\right)+\left(x-y\right)=6+0 \\ \Rightarrow 3x=6 \\ \Rightarrow x=2 \end{gathered}$$

Substitute $x$ with $2$ in equation $\left(\mathrm{i}\right)$.

$$\begin{gathered} x-y=0 \\ \Rightarrow 2-y=0 \\ \Rightarrow y=2 \end{gathered}$$

Thus, the point is $C\left(2,2\right)$.

Step 4: Check which type of triangle formed.

Apply the distance formula, $\sqrt{{\left(x_1-x_2\right)}^{\mathbf{2}}\mathbf{+}{\left(y_1-y_2\right)}^{\mathbf{2}}}$.

$AB=\sqrt{{\left(1-3\right)}^2+{\left(1-0\right)}^2}\mathrm{units}=\sqrt{4+1}\mathrm{units}=\sqrt{5}\mathrm{units}$.

$BC=\sqrt{{\left(3-2\right)}^2+{\left(0-2\right)}^2}\mathrm{units}=\sqrt{1+4}\mathrm{units}=\sqrt{5}\mathrm{units}$.

$CA=\sqrt{{\left(2-1\right)}^2+{\left(2-1\right)}^2}\mathrm{units}=\sqrt{1+1}\mathrm{units}=\sqrt{2}\mathrm{units}$.

So, the given triangle has $AB=BC\ne CA$, thus the triangle is an isosceles triangle.

Hence, the intersection forms an isosceles triangle.