Question

Show that the points 2$\hat{i}$, − $\hat{i}$ − 4$\hat{j}$ and −$\hat{i}$ + 4$\hat{j}$ form an isosceles triangle.

Answer 1

Given:- The points $A,B,C$ with position vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ respectively.
Also, $\overrightarrow{a}=2\hat{i}$
$\overrightarrow{b}=-\hat{i}-4\hat{j}$
$\overrightarrow{c}=-\hat{i}+4\hat{j}$
Then,
$$\begin{gathered} \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a} \\ \Rightarrow \overrightarrow{AB}=\left(-\hat{i}-4\hat{j}\right)-2\hat{i} \\ \Rightarrow \overrightarrow{AB}=-3\hat{i}-4\hat{j} \\ \mathrm{Now},\left|\overrightarrow{AB}\right|=\sqrt{{\left(-3\right)}^2+{\left(-4\right)}^2}=\sqrt{9+16}=\sqrt{25}=5 \\ \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b} \\ \Rightarrow \overrightarrow{BC}=\left(-\hat{i}+4\hat{j}\right)-\left(-\hat{i}-4\hat{j}\right) \\ \Rightarrow \overrightarrow{BC}=-\hat{i}+4\hat{j}+\hat{i}+4\hat{j} \\ \Rightarrow \overrightarrow{BC}=8\hat{j} \end{gathered}$$
and

$$\begin{gathered} \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a} \\ \Rightarrow \overrightarrow{AC}=\left(-\hat{i}+4\hat{j}\right)-2\hat{i} \\ \Rightarrow \overrightarrow{AC}=-3\hat{i}+4\hat{j} \\ \mathrm{Now},\left|\overrightarrow{AC}\right|=\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2}=\sqrt{9+16}=\sqrt{25}=5 \end{gathered}$$

Since, the magnitude of AB and AC is equal.
Hence, the points 2$\hat{i}$, − $\hat{i}$ − 4$\hat{j}$ and −$\hat{i}$ + 4$\hat{j}$ form an isosceles triangle.