Question

find the area of the triangle formed by the points (p+1+1) (2p +1, 3)and (3p+2,3p) and show that the points are collinear if p=2or - 1/3 please answer fast brother and sister ☺️☺️☺️​

Answer 1

Answer:

Area of triangle = $\frac{1}{2}\left|3p^2-5p-2\right|$

Step-by-step explanation:

Given Points are say $A(x_1,y_1)=(p+1,1)\:,\:B(x_2,y_2)=(2p+1,3)\:and\:C(x_3,y_3)=(3p+2,3p)$

To find: Area of ΔABC

To prove: Points are collinear if p = 2 or $\frac{-1}{3}$

Formula used to find Area of triangle is given by,

$Area\,of\,triangle=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$\implies Area\,of\,\Delta ABC=\frac{1}{2}\left|(p+1)(3-3p)+(2p+1)(3p-1)+(3p+2)(1-3)\right|$

$Area\,of\,\Delta ABC=\frac{1}{2}\left|-3p^2+3+6p^2+p-1-6p-4\right|$

$Area\,of\,\Delta ABC=\frac{1}{2}\left|3p^2-5p-2\right|$ ....... (1)

Now first put p = 2 in eqn (1) , we get

$Area\,of\,\Delta ABC=\frac{1}{2}\left|3\times2^2-5\times2-2\right|$

                               $=\frac{1}{2}\left|12-12\right|$

                               = 0

Now, put $p=\frac{-1}{3}$ in eqn (1)

$Area\,of\,\Delta ABC=\frac{1}{2}\left|3\times(\frac{-1}{3})^2-5\times(\frac{-1}{3})-2\right|$

                               $=\frac{1}{2}\left|\frac{3}{9}+\frac{15}{9}-\frac{18}{9}\right|$

                                = 0

Therefore, for p = 2 or  $\frac{-1}{3}$ points are collinear. and Area of triangle =  $\frac{1}{2}\left|3p^2-5p-2\right|$