Question

pls solve fast!!!!!1

Answer 1

____Heya Mate Your solution is..!!

Given three points are (t, t-2) , (t+2, t+2) and (t+3, t)

use formula of traingle form of co-ordinate geometry .

area of traingle =1/2 {x1 (y2-y3)+x2 (y3-y1)+x3(y1-y2)}

now,

ar of triangle=1/2 {t (t+2-t)+(t+2)(t-t+2)+(t+3 (t-2-t-2)}

=1/2 {2t +2t +4 -4t -12}

=1/2 {-8}

=-4

but here mention only magnitude of area of triangle so , area of traingle =4 sq unit

you see in area of triangle not depend upon t

_____HOpe iT HelP YOu DEaR ♥

Answer 2

Answer:

$\huge\mathbb\blue{answer}$

`here is given ,

vertices are - (t, t–2 ) , (t+2,t+2) and (t+3,t)

equation of co - ordinate geometry .

formula of area of traingle

$=1/2 {x1 (y2-y3)+x2 (y3-y1)+x3(y1-y2)}$

now,

area of triangle=1/2 {t (t+2-t)+(t+2)(t-t+2)+(t+3 (t-2-t-2)}

solve -

=1/2 {2t +2t +4 -4t -12}

=1/2 {-8}

=-4 .

Hope it's helpful for you.