Question

prove that the area of triangle with vertices at (p-4,p+5) , (p+3,p-2) and (p,p) remains constant as p varies.​

Answer 1

Answer:

Step-by-step explanation:

Given vertices of the triangle are $\tt{(p-4,p+5),\,\,(p+3,p-2),\,\,(p,p)}$

So, the area of the triangle is given by,

$\sf{\triangle=\dfrac{1}{2}\left|\left|\begin{array}{ccc}p-4&p+5&1\\p+3&p-2&1\\p&p&1\end{array}\right|\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\left|(p-4)\left|\begin{array}{ccc}p-2&1\\p&1\end{array}\right|-(p+5)\left|\begin{array}{ccc}p+3&1\\p&1\end{array}\right|+1\left|\begin{array}{ccc}p+3&p-2\\p&p\end{array}\right|\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\left|(p-4)(p-2-p)-(p+5)(p+3-p)+1(p^2+3p-p^2+2p)\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\left|(p-4)(-2)-(p+5)(3)+1(3p+2p)\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\left|-2p+8-3p-15+5p\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\left|8-15\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\left|-7\right| }$

$\sf{\implies\triangle=\dfrac{7}{2}\,\,cm^2 }$