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Question

Question: If three points \(({x_1},{y_1}),({x_2},{y_2})\) and \(({x_3},{y_3})\) lie on the same line then, pro...

If three points (x1,y1),(x2,y2)({x_1},{y_1}),({x_2},{y_2}) and (x3,y3)({x_3},{y_3}) lie on the same line then, prove that
y2y3x2x3+y3y1x3x1+y1y2x1x2=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.

Explanation

Solution

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 (x1,y1),(x2,y2)({x_1},{y_1}),({x_2},{y_2}) and (x3,y3)({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:
x1(y2y3)+x2(y3y1)+x3(y1y2)=0{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2}) = 0
Dividing by x1x2x3{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:
y2y1x2x1=y3y2x3x2\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, x1(y2y3)+x2(y3y1)+x3(y1y2)=0{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2}) = 0