
If three points $({x_1},{y_1}),({x_2},{y_2})$ and $({x_3},{y_3})$ lie on the same line then, prove that
$\frac{{{y_2} - {y_3}}}{{{x_2}{x_3}}} + \frac{{{y_3} - {y_1}}}{{{x_3}{x_1}}} + \frac{{{y_1} - {y_2}}}{{{x_1}{x_2}}} = 0.$
Answer
513k+ views
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, ${x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2}) = 0$
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, ${x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2}) = 0$
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE
