Question

Find the area of the triangle formed by the points (p+1, 1) , (2p+1, 3) and (2p+2, 2p) and show that the points are collinear if p=2 or \[ - \dfrac{1}{2}\]

Answer 1

Hint: To handle such types of questions we will use the underneath steps with the goal that we can make our answer as easy as could be expected so that it will be helpful to save our critical time. We use below formula to get the area of triangle when 3 pints are given:
\[Area = \dfrac{1}{2}\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right]\]

Given: the points (p+1,1) , (2p+1,3) and (2p+2,2p) are given on x-y plane

Complete step-by-step answer:
When we are solving this type of question, we need to follow the steps provided in the hint part above.
\[\begin{array}{l}
Area = \dfrac{1}{2}\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right]\\
Area = \dfrac{1}{2}\left[ {(p + 1)(3 - 2p) + (2p + 1)(2p - 1) + (2p + 2)(1 - 3)} \right]\\
Area = \dfrac{1}{2}\left[ {3p - 2{p^2} + 3 - 2p + 4{p^2} - 1 + 2p - 6p + 2 - 6} \right]\\
Area = \dfrac{1}{2}\left[ {p + 3 + 2{p^2} - 1 - 4p - 4} \right]\\
Area = \dfrac{1}{2}\left[ {2{p^2} - 3p - 2} \right]\\
Area = \dfrac{1}{2}(p - 2)(2p + 1)
\end{array}\]
Hence the area of the triangle is obtained.
As we can see that for p = 2 and p = \[ - \dfrac{1}{2}\] area will be zero it means these three points will be collinear because collinear points have zero area.
Hence after following the each and every step given in the hint part, we obtained our final answer.

Additional Information:
Here we can clearly see that in this solution we did not use any complicated process because we followed basic and simple things in the right order as per given in the above hint section.

Note: In this sort of example we need to use the right formula to avoid confusion and to get the right answer precisely. The general formula used to find out the area of the triangle should not be applied here.