If three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ lie on the same line, prove that$ \frac{y_{2}-y_{3}}{x_{2} x_{3}}+\frac{y_{3}-y_{1}}{x_{3} x_{1}}+\frac{y_{1}-y_{2}}{x_{1} x_{2}}=0 $.
Given:
Points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ lie on the same line.
To do:
We have to prove that
\( \frac{y_{2}-y_{3}}{x_{2} x_{3}}+\frac{y_{3}-y_{1}}{x_{3} x_{1}}+\frac{y_{1}-y_{2}}{x_{1} x_{2}}=0 \).
Solution:
Let $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ be the vertices of $\triangle ABC$.
We know that,
If the points $A, B$ and $C$ are collinear then the area of $\triangle ABC$ is zero.
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})] \)
\( 0=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})] \)
\( x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})=0 \)
Dividing by \( x_{1} x_{2} x_{3} \), we get
\( \frac{x_{1}\left(y_{2}-y_{3}\right)}{x_{1} x_{2} x_{3}}+\frac{x_{2}\left(y_{3}-y_{1}\right)}{x_{1} x_{2} x_{3}}+\frac{x_{3}\left(y_{1}-y_{2}\right)}{x_{1} x_{2} x_{3}}=0 \)
\( \frac{y_{2}-y_{3}}{x_{2} x_{3}}+\frac{y_{3}-y_{1}}{x_{3} x_{1}}+\frac{y_{1}-y_{2}}{x_{1} x_{2}}=0 \)
Hence proved.
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