Question

13. Find the area of the triangle formed by the collinear points P(-1.5,3) ,Q(6,2) and R (-3,4).
a) 0 Square units
b) 2 Square units
c) 9 square units
d) 8 squre units​

Answer 1

Answer:

a) 0 Square Units

Step-by-step explanation:

Given:

P(−1.5,3),Q(6,−2) and R(−3,4)

We know that the area of the triangle whose vertices are (x₁ , y₁) , (x₂ , y₂) , (x₃ , y₃) is :

$\frac{1}{2}$ × [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)

∴The area of the triangle is:

$\frac{1}{2}$ × [{(−1.5) × (−2 −4)}+ {6 × (4−3)}+ {(−3) × (3+2)}]

= $\frac{1}{2}$ × (9 + 6 − 15)

= 0

∴ The area of the triangle is 0 square units