Question

find the area of triangle formed by the points (p+1,1), (2p+1,3) and (2p+2,3p) and show that points are collinear if p=2 or -1/2

Answer 1

Answer:

Step-by-step explanation:

i want this answer please

Answer 2

Answer:

Step-by-step explanation:

Given,  

Consider $p=2$

Location of 1 st point - A is $(x_{1},y_{1})=(p+1,1)$

Location of 2nd point -B is (-2,0)

$(x_{2},y_{2})=(2p+1,3)$

Location of 3rd point -C is $(x_{3},y_{3})=(2p+2,3p)$

Area of a triangle formed by joining the three given points is

$A=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]\\\\\Rightarrow{A}=\frac{1}{2}[(p+1)(3-2p)+(2p+1)(2p-1)+(2p+2)(1-3)]\\\\\Rightarrow{A}=\frac{1}{2}[3p-2p^2+3-2p+4p^2-2p+2p-1+2p-6p+2-6]\\\\\Rightarrow{A}=\frac{1}{2}[2p^2-3p-2]$

If the area of the triangle is zero then we said that the three points are co linear.

Now consider the area of the triangle is zero,

$\frac{1}{2}[2p^2-3p-2]=0\\\\\Rightarrow[2p^2-3p-2]=0\\\\\Rightarrow[2p^2-4p+p-2]=0\\\\\Rightarrow[2p(p-2)+1(p-2]=0\\\\\Rightarrow(2p+1)(p-2)=0$

As the product of two numbers are zero then the then  

$(2p+1)=0\\\\\therefore p=-\frac{2}{2}$

and

$(p-2)=0\\\\\therefore p=2$

For the value of $p=2$ and $p=-\frac{1}{2}$ the area of the triangle is zero then it was proved that the said three points are co-linear.