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If three points $({x_1},{y_1}),({x_2},{y_2})$ and $({x_3},{y_3})$ lie on the same line then, prove that
$\frac{{{y_2} - {y_3}}}{{{x_2}{x_3}}} + \frac{{{y_3} - {y_1}}}{{{x_3}{x_1}}} + \frac{{{y_1} - {y_2}}}{{{x_1}{x_2}}} = 0.$

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Hint: The given three points lie on the same line i.e., they are collinear. Use the condition of collinearity of three points.
According to question, three points $({x_1},{y_1}),({x_2},{y_2})$ and $({x_3},{y_3})$ lie on the same line. So we can say that the points are collinear. And we know that for three points to be collinear, following condition will hold:
${x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2}) = 0 $
Dividing by ${x_1}{x_2}{x_3}$ on both sides of this equation, we’ll get:
$
   \Rightarrow \frac{{{x_1}({y_2} - {y_3})}}{{{x_1}{x_2}{x_3}}} + \frac{{{x_2}({y_3} - {y_1})}}{{{x_1}{x_2}{x_3}}} + \frac{{{x_3}({y_1} - {y_2})}}{{{x_1}{x_2}{x_3}}} = 0 \\
   \Rightarrow \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 \\
\ $
This is the required proof.
Note: Since the points are collinear (lying on the same line), we can equate the slope of line formed using any two pair of points:
$ \Rightarrow \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}}$
We will get the same condition as we have used earlier, ${x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2}) = 0$