Question

Using the slope concept determine whether the point P(1,2),Q(2,8/5),R(3,6/5) are collinear or not.

Answer 1

We have to show whether the points P(1,2), Q(2, 8/5) and R (3,6/5) are collinear or not that is they lie on the same straight line by using the concept of slope.

• we have 3 points,

        $P (1,2) , Q(2, \frac{8}{5} ) and R (3, \frac{6}{5})$

• Now, for P, Q and R to be collinear the slope m1 of the line from P to   Q , the slope m2 of line from point Q to R and the slope m3 of line from point P to R should be equal, therefore

• Solving for m1

            $m_{1} = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } \\= \frac{\frac{8}{5}-2 }{2-1} = \frac{8-10}{5} = -\frac{2}{5}$

           $m_{1} = -\frac{2}{5}$         - (1)

•   Solving for m2

            $m_{2} = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } \\= \frac{\frac{6}{5}-\frac{8}{5} }{3-2} = \frac{6-8}{5} = -\frac{2}{5}$

             $m_{2} = -\frac{2}{5}$          - (2)

•  Solving for m3

            $m_{2} = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } \\= \frac{\frac{6}{5}-2 }{3-1} = \frac{6-10}{(5)(2)} =\frac{-4}{10} = -\frac{2}{5}$

               $m_{3} = -\frac{2}{5}$          - (3)

• From (1), (2) and (3) we have

             $m_{1} =m_{2} =m_{3} = -\frac{2}{5}$

         Since m1 = m2 = m3 , therefore points P, Q and R are collinear.

Answer 2

P, Q, and R are collinear.

Step-by-step explanation:

The points P(1,2), Q($2,\frac{8}{5}$), and R($3,\frac{6}{5}$) are given three points on the coordinate plane.

Now, the slope of the straight line joining P and Q will be, $M_{1}$ = $\frac{2 - \frac{8}{5} }{1 - 2} = - \frac{2}{5}$

Again, the slope of the straight line joining Q and S will be, $M_{2} = \frac{\frac{8}{5} - \frac{6}{5}}{2 - 3} = - \frac{2}{5}$

Therefore, the line PQ and QR have the same slope and Q is their common point. Hence, P, Q, and R are collinear. (Answer)

We know the slope of a straight line passing through the two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given by the expression $M = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}$