Question

If three points (x1,y1),(x2,y2) and (x3,y3) lie on the same line,prove that
[y2-y3/x2x3 ]+[ y3 -y1 /x3x1]+[y1-y2/x1x2 ]= 0

Answer 1

Since the three points are colinear, you can write any given point as a "multiple" of another.
In other words, you can write
\[(x_1,y_1) = a(x_2,y_2) = b(x_3,y_3)\]where c and d are scalars.
So, let
\[x_1 = ax_2 = bx_3\]
and
\[y_1 = ay_2 = by_3.\]
You must show that
\[\dfrac{y_2-y_3}{x_2x_3}+\dfrac{y_3 -y_1}{x_3x_1}+\dfrac{y_1-y_2}{x_1x_2}= 0\]
Using the previous substitutions, the left side becomes
\[\dfrac{\frac{1}{a}y_1-\frac{1}{b}y_1}{(\frac{1}{a}x_1)(\frac{1}{b}x_1)}+\dfrac{\frac{1}{b}y_1-y_1}{(\frac{1}{b}x_1)x_1}+\dfrac{y_1-\frac{1}{a}y_1}{x_1(\frac{1}{a}x_1)}\]
\[\dfrac{(\frac{1}{a}-\frac{1}{b})y_1}{\frac{1}{ab}x_1^2}+\dfrac{(\frac{1}{b}-1)y_1}{\frac{1}{b}x_1^2}+\dfrac{(1-\frac{1}{a})y_1}{\frac{1}{a}x_1^2}\]
\[\dfrac{(\frac{1}{a}-\frac{1}{b})y_1}{\frac{1}{ab}x_1^2}+\dfrac{\frac{1}{a}(\frac{1}{b}-1)y_1}{\frac{1}{ab}x_1^2}+\dfrac{\frac{1}{b}(1-\frac{1}{a})y_1}{\frac{1}{ab}x_1^2}\]
\[\dfrac{\frac{1}{ab} [(\frac{1}{a}-\frac{1}{b})(\frac{1}{b}-1)(1-\frac{1}{a})]}{\frac{1}{ab}} \cdot \dfrac{y_1}{x_1^2}\]
\[(\frac{1}{a}-\frac{1}{b})(\frac{1}{b}-1)(1-\frac{1}{a}) \cdot \dfrac{y_1}{x_1^2}\]
The coefficient is 0 if
\[\frac{1}{a}-\frac{1}{b}=0,\\
\frac{1}{b}-1=0, or\\
1-\frac{1}{a}=0\]
This tells you that a = b, a = 1, or b = 1, which means that you only actually have one or two distinct points. So, if the three given points are distinct, the expression you are given is not necessarily equivalent to zero.

Answer 2

@SithsAndGiggles
Thank u so much