Answer to Question #293608 in Analytic Geometry for Elley

Question

Prove that the equation of a line through (𝑥₁ , 𝑦₁) and (𝑥₂ , 𝑦₂) can be expressed in the form determinant

x y 1

x₁ y₁ 1

x₂ y₂ 1

= 0

Answer 1

Solution:

Two points are (𝑥₁ , 𝑦₁) and (𝑥₂ , 𝑦₂).

Slope $=m=\dfrac{y_2-y_1}{x_2-x_1}$

Intercept$=c=\dfrac{x_1y_2-x_2y_1}{x_2-x_1}$

Now, equation is $y=mx+c$

$\Rightarrow y=\dfrac{y_2-y_1}{x_2-x_1}x+\dfrac{x_1y_2-x_2y_1}{x_2-x_1}\ ...(i)$

Now, given determinant is $\begin{vmatrix} x&y&1\\x_1&y_1&1\\x_2&y_2&1\end {vmatrix}=0$

On expanding,

$x(y_1-y_2)-y(x_1-x_2)+(x_1y_2-x_2y_1)=0 \\\Rightarrow x\dfrac{(y_1-y_2)}{(x_1-x_2)}-y+\dfrac{(x_1y_2-x_2y_1)}{(x_1-x_2)}=0 \\ \Rightarrow y=\dfrac{y_2-y_1}{x_2-x_1}x+\dfrac{x_1y_2-x_2y_1}{x_2-x_1}\ ...(ii)$

From (i) & (ii), it is proved.