If the vertices of a triangle are $(t, t - 2),$ $(t + 2, t + 2)$ and $(t + 3, t),$ then the area (in sq. units) of the triangle is

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Dependent on the value of

t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4$
Given : $(t, t - 2), (t + 2, t + 2)$ and $(t + 3, t)$
The area of $△$ formed by given vertices
$= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = \frac{1}{2} | t (t + 2 - t) + (t + 2) (t - t + 2) + (t + 3) (t - 2 - t - 2) | = \frac{1}{2} | 2 t + 2 t + 4 - 4 t - 12 |$
$= \frac{1}{2} | - 8 | = 4$ sq. units

Suggest Corrections

Similar questions

(t, t - 2),

(t + 2, t + 2)

(t + 3, t),

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

| t | = 3,

| 2 t - 1 |

A, B, C

P

(2, 3), (1, 2), (4, 3)

(t, t^{2})

t > 0

△ A B C

△ P A B

t

[a t_{1} t_{2}, a (t_{1} + t_{2})], [a t_{2} t_{3}, a (t_{2} + t_{3})], [a t_{3} t_{1}, a (t_{3} + t_{1})]

Join BYJU'S Learning Program