Question

I show that the points (3-2) (-2,8) and (0,4)
are collinear.​

Answer 1

• if it is correct please say to me if not also say to me

Answer 2

Answer:

The given points are collinear.

Step-by-step-explanation:

Let the points be, A, B and C respectively.

$\bullet\sf\:A\:\equiv\:(\:3\:,\:-\:2\:)\:\equiv\:(\:x_1\:,\:y_1\:)\\\\\\\bullet\sf\:B\:\equiv\:(\:-\:2\:,\:8\:)\:\equiv\:(\:x_2\:,\:y_2\:)\\\\\\\bullet\sf\:C\:\equiv\:(\:0\:,\:4\:)\:\equiv\:(\:x_3\:,\:y_3\:)$

We know that,

$\pink{\sf\:Slope\:=\:\dfrac{y_2\:-\:y_1}{x_2\:-\:x_1}}\sf\:\:\:-\:-\:[\:Slope\:formula\:]$

Now,

$\sf\:Slope\:of\:line\:AB\:=\:\dfrac{y_2\:-\:y_1}{x_2\:-\:x_1}\\\\\\\implies\sf\:Slope\:of\:line\:AB\:=\:\dfrac{8\:-\:(\:-\:2\:)}{-\:2\:-\:3}\\\\\\\implies\sf\:Slope\:of\:line\:AB\:=\:\dfrac{8\:+\:2}{-\:5}\\\\\\\implies\sf\:Slope\:of\:line\:AB\:=\:-\:\cancel{\dfrac{10}{5}}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:AB\:=\:-\:2}}$

Now,

$\sf\:Slope\:of\:line\:BC\:=\:\dfrac{y_3\:-\:y_2}{x_3\:-\:x_2}\\\\\\\implies\sf\:Slope\:of\:line\:BC\:=\:\dfrac{4\:-\:8}{0\:-\:(\:-\:2\:)}\\\\\\\implies\sf\:Slope\:of\:line\:BC\:=\:\dfrac{-\:4}{0\:+\:2}\\\\\\\implies\sf\:Slope\:of\:line\:BC\:=\:-\:\cancel{\dfrac{4}{2}}\\\\\\\implies\boxed{\red{\sf\:Slope\:of\:line\:BC\:=\:-\:2}}$

Lines AB & BC have equal slope and B is the common point.

$\therefore$ The given points are collinear.

Additional Information:

1. Slope of line:

The ratio of difference between y coordinates and x coordinates is constant and called as slope of the line.

2. It is denoted by the letter m.

3. Slope of X - axis is 0.

4. Slope of Y - axis cannot be determined.

5. Parallel lines have equal slopes.

6. Formula for finding slope of line:

$\boxed{\red{\sf\:m\:=\:\dfrac{y_{2}\:-\:y_{1}}{x_{2}\:-\:x_{1}}}}$