Answer to Question #198516 in Analytic Geometry for prince

Question

Use the fact that

 x y 1

a1 b1 1 =0

a2 b2 1

to determine the equation of the line passing through the distinct points (a1, b1) and (a2, b2), where |·| stands for det(·), the determinant.

Answer 1

We have,

$\begin{vmatrix} x & x&1 \\ a_1 & b_1&1\\ a_2&b_2&1 \end{vmatrix}=0\\\ \\\Rightarrow x(b_1-b_2)-y(a_1-a_2)+b_2a_1-a_2b_1=0\\\Rightarrow x(b_2-b_1)-y(a_2-a_1)+a_2b_1-b_2a_1=0\\\Rightarrow x(b_2-b_1)+a_2b_1=y(a_2-a_1)+b_2a_1\\\text{Add }a_2b_2\ on\ both\ sides,\ we\ get\\\Rightarrow x(b_2-b_1)-a_2b_2+a_2b_1=y(a_2-a_1)-b_2(a_2-a_1)\\\Rightarrow (b_2-b_1)(x-a_2)=(a_2-a_1)(y-b_2)\\ \ \\ \Rightarrow\dfrac{y-b_2}{b_2-b_1}=\dfrac{x-a_2}{a_2-a_1}$

So, from the above equation, we can conclude that

the equation of the line will pass through the points $(a_1,b_1)\ \ and\ (a_2,b_2)$