∣ x y 1 a 1 b 1 1 a 2 b 2 1 ∣ = 0 \begin{vmatrix}
x & y & 1 \\
a_1 & b_1 & 1 \\
a_2 & b_2 & 1
\end{vmatrix}=0 ∣ ∣ x a 1 a 2 y b 1 b 2 1 1 1 ∣ ∣ = 0
x ∣ b 1 1 b 2 1 ∣ − y ∣ a 1 1 a 2 1 ∣ + 1 ∣ a 1 b 1 a 2 b 2 ∣ = 0 x\begin{vmatrix}
b_1& 1 \\
b_2 & 1
\end{vmatrix}-y\begin{vmatrix}
a_1 & 1 \\
a_2 & 1
\end{vmatrix}+1\begin{vmatrix}
a_1 & b_1 \\
a_2 & b_2
\end{vmatrix}=0 x ∣ ∣ b 1 b 2 1 1 ∣ ∣ − y ∣ ∣ a 1 a 2 1 1 ∣ ∣ + 1 ∣ ∣ a 1 a 2 b 1 b 2 ∣ ∣ = 0
x ( b 1 − b 2 ) − y ( a 1 − a 2 ) + a 1 b 2 − a 2 b 1 = 0 x(b_1-b_2)-y(a_1-a_2)+a_1b_2-a_2b_1=0 x ( b 1 − b 2 ) − y ( a 1 − a 2 ) + a 1 b 2 − a 2 b 1 = 0 The equation of the line passing through the distinct points ( a 1 , b 1 ) (a_1, b_1) ( a 1 , b 1 ) and ( a 2 , b 2 ) (a_2, b_2) ( a 2 , b 2 ) in slope-intercept form
y = b 1 − b 2 a 1 − a 2 x + a 1 b 2 − a 2 b 1 a 1 − a 2 , a 1 ≠ a 2 y=\dfrac{b_1-b_2}{a_1-a_2}x+\dfrac{a_1b_2-a_2b_1}{a_1-a_2}, a_1\not=a_2 y = a 1 − a 2 b 1 − b 2 x + a 1 − a 2 a 1 b 2 − a 2 b 1 , a 1 = a 2
Comments