If three points (x1, y1) (x2, y2), (x3, y3) lie on the same line, prove that \[\frac{y_2 - y_3}{x_2 x_3} + \frac{y_3 - y_1}{x_3 x_1} + \frac{y_1 - y_2}{x_1 x_2} = 0\]
Solution
GIVEN: If three points (x1, y1) (x2, y2) and (x3, y3) lie on the same line
TO PROVE: \[\frac{y_2 - y_3}{x_2 x_3} + \frac{y_3 - y_1}{x_3 x_1} + \frac{y_1 - y_2}{x_1 x_2} = 0\]
PROOF:
We know that three points (x1, y1) (x2, y2) and (x3, y3) are collinear if
`x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 ) = 0`
⇒ `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`
⇒ \[\frac{x_1 (y_2 - y_3 ) }{x_1 x_2 x_3} + \frac{x_2 (y_3 - y_1 ) }{x_1x_2 x_3} + \frac{x_3 ( y_1 - y_2 ) }{x_1 x_2 x_3} = 0\]
⇒ \[\frac{(y_2 - y_3)}{x_2 x_3} + \frac{(y_3 - y_1)}{x_3 x_1} + \frac{(y_1 - y_2)}{x_1 x_2} = 0\]
Hence proved.
