Question

D(5, 6) and E($-$5, $-$ 6).

Answer 1

The given points are D(5, 6) and E($-$5, $-$ 6).
Here, x₁ = 5, x₂ = $-$5, y₁ = 6 and y₂ = $-$6
Equation in two point form is given by

$\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}$

$$\begin{gathered} \Rightarrow \frac{x-5}{5-(-5)}=\frac{y-6}{6-\left(6\right)} \\ \Rightarrow \frac{x-5}{5+5}=\frac{y-6}{6+6} \\ \Rightarrow \frac{x-5}{10}=\frac{y-6}{12} \\ \Rightarrow \frac{x-5}{5}=\frac{y-6}{6} \end{gathered}$$
⇒ 6(x $-$ 5) = 5(y $-$ 6)
⇒ 6x $-$ 30 = 5y $-$ 30
∴ 6x $-$ 5y = 0, which is the required equation of the line.