Question

The triangle formed by $x^2-9y^2=0$ and $x=4$ is

• A. Isosceles
• B. Equilateral
• C. Right angled
• D. None of these

Answer 1

The correct option is A

Isosceles

Explanation for correct option

Given $x^2-9y^2=0$

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

Equation of the line is

$x–3y=0$ …(i)

$x+3y=0$ …(ii)

$x=4$ …(iii)

Solving (i) and (iii)

We get the intersecting points $x=4,y=\frac{4}{3}$

Let $A=\left(4,\frac{4}{3}\right)$

Solving (ii) and (iii)

We get $x=4$ and $y=-\frac{4}{3}$

Let$B=\left(4,-\frac{4}{3}\right)$

Solving (i) and (ii)

We get $x=0$and $y=0$

Let$C=(0,0)$

$$\begin{gathered} AB=\sqrt{{\left(4-4\right)}^2+{\left(-\frac{4}{3}–\frac{4}{3}\right)}^2} \\ =\frac{8}{3} \end{gathered}$$ (using distance formula)

$BC=\sqrt{{(0–4)}^2+{\left(0+\frac{4}{3}\right)}^2}$ (using distance formula)

$$\begin{gathered} =\sqrt{16+\frac{16}{9}} \\ =\sqrt{\frac{160}{9}} \\ =\frac{4\sqrt{10}}{3} \end{gathered}$$

$AC=\sqrt{{(0–4)}^2+{\left(0+\frac{4}{3}\right)}^2}$ (using distance formula)

$$\begin{gathered} =\sqrt{16+\frac{16}{9}} \\ =\sqrt{\frac{160}{9}} \\ =\frac{4\sqrt{10}}{3} \end{gathered}$$

Here $AC=BC$

So $ABC$ is an isosceles triangle.

Hence, the correct option is $o\mathrm{ption}\left(\mathrm{A}\right)$