Question #293608

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


1
Expert's answer
2022-02-08T11:36:24-0500

Solution:

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

Slope =m=y2y1x2x1=m=\dfrac{y_2-y_1}{x_2-x_1}

Intercept=c=x1y2x2y1x2x1=c=\dfrac{x_1y_2-x_2y_1}{x_2-x_1}

Now, equation is y=mx+cy=mx+c

y=y2y1x2x1x+x1y2x2y1x2x1 ...(i)\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 xy1x1y11x2y21=0\begin{vmatrix} x&y&1\\x_1&y_1&1\\x_2&y_2&1\end {vmatrix}=0

On expanding,

x(y1y2)y(x1x2)+(x1y2x2y1)=0x(y1y2)(x1x2)y+(x1y2x2y1)(x1x2)=0y=y2y1x2x1x+x1y2x2y1x2x1 ...(ii)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.


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