Question

Find the area of the triangle ABC with A(1, −4) and mid-points of sides through A being (2, −1) and (0, −1).

 

Answer 1

Let the coordinates of points B and C of the ∆ABC be (a_1, b₁) and (a_2, b₂), respectively.

Q is the midpoint of AB.

Using midpoint formula, we have

`(0,-1)=((a_1+1)/2,(b_1-4)/2)`

`=>(a_1+1)/2=0 `

⇒ a₁=−1 and b₁=2

Therefore, the coordinates of B are (−1, 2).

P is the midpoint of AC.

Now,

`(2,-1)=((a_2+1)/2,(b_2-4)/2)`

`=>(a_2+1)/2=2`

⇒ a₂=3 and b₂=2

Therefore, the coordinates of C are (3, 2)

Thus, the vertices of ∆ABC are A(1,−4), B(−1, 2) and C(3, 2).

Now,

Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) =  `1/2`[x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)]

∴ Area of ∆ABC = `1/2`[x₁(y₂−y₃)+x₂(y₃−y₁)+x3(y₁−y₂)]

= `1/2`×[1(2−2)+(−1)(2+4)+3(−4−2)]

= `1/2`(−6−18)

= `1/2` (−24)

=−12

Since area is a measure that cannot be negative, we will take the numerical value of −12, that is, 12. Thus, the area of ∆ABC is 12 square units