Question

Prove that the area of a triangle with vertices (t.t-2), (t +2,t+2) and (t+3) is independent of t.​

Answer 1

Correct question:

Prove that the area of a triangle whose vertices are (t, t - 2), (t + 2, t + 2), and (t + 3, t), is independent of t.

Given:

• (t, t - 2)
• (t + 2, t + 2)
• (t + 3, t)

To prove: That the area by the so formed triangle is independent of t.

Answer:

Formula to find the area of a triangle:

$\tt \dfrac{1}{2}\ \times\ \bigg[x_1\bigg(y_2\ -\ y_3\bigg)\ +\ x_2\bigg(y_3\ -\ y_1\bigg)\ +\ x_3\bigg(y_1\ -\ y_2\bigg)\bigg]$

From the given points, we have,

$\tt x_1\ =\ t\\\\x_2\ =\ t\ +\ 2\\\\x_3\ =\ t\ +\ 3\\\\y_1\ =\ t\ -\ 2\\\\y_2\ =\ t\ +\ 2\\\\y_3\ =\ t$

Using these values in the formula,

$\tt \dfrac{1}{2}\ \times\ \bigg[t\bigg(t\ +\ 2\ -\ t\bigg)\ +\ t\ +\ 2\bigg(t\ -\ t\ +\ 2\bigg)\ +\ t\ +\ 3\bigg(t\ -\ 2\ -\ t\ -\ 2\bigg)\bigg]\\\\\\\\\\=\ \dfrac{1}{2}\ \times\ \bigg[2t\ +\ 2t\ +\ 4\ -\ 4t\ -\ 12\bigg]\\\\\\=\ \dfrac{1}{2}\ \times\ - 8\\\\\\=\ 4\ units\ (sign\ not\ considered)$

Therefore, the area of the triangle is independent of t.

Answer 2

$\bf{\Large{\underline{Correct\:Question\::}}}}}$

Prove that the area of a triangle whose vertices (t,t-2), (t+2, t+2) & (t+3, t) is independent of t.

$\bf{\Huge{\boxed{\underline{\underline{\bf{ANSWER\::}}}}}}$

$\bf{\red{Given\:Vertices}}\begin{cases}\sf{(t,t-2)}\\ \sf{(t+2,t+2)}\\ \sf{(t+3,t)}\end{cases}}$

$\bf{\Large{\underline{\underline{\tt{\red{Explanation\::}}}}}}$

Let the points be:

• (x1 , y1)
• (x2 , y2)
• (x3 , y3)

$\bf{\large{\boxed{\sf{Area\:of\:triangle\:=\:\frac{1}{2} |x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|}}}}}$

A/q

$|\implies\sf{Area\:of\:triangle\:=\:\frac{1}{2} |t(\cancel{t}+2\cancel{-t})+(t+2)(\cancel{t-t}+2)+(t+3)(\cancel{t}-2\cancel{-t}-2)|}\\\\\\\\|\implies\sf{Area\:of\:triangle\:=\:\frac{1}{2} |t(2)+2(t+2)-4(t+3)|}\\\\\\\\|\implies\sf{Area\:of\:triangle\:=\:\frac{1}{2} |2t+2t+4-4t-12|}\\\\\\\\|\implies\sf{Area\:of\:triangle\:=\:\frac{1}{2} |\cancel{4t}+4\cancel{-4t}-12|}\\\\\\\\|\implies\sf{Area\:of\:triangle\:=\:\frac{1}{2} |-8|}\\\\\\\\|\implies\sf{Area\:of\:triangle\:=\:\frac{1}{2} *8}$

$|\implies\sf{Area\:of\:triangle\:=\:\cancel{\dfrac{8}{2} }}\\\\\\\\|\implies\sf{\red{Area\:of\:triangle\:=\:4\:unit}}$

∴ Independent of t                          [hence Prove]