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

Solution:

Given, the vertices of the triangle with A(1, -4) and the mid-points of sides through A being (2, -1) and (0, -1)

We have to find the area of the triangle ABC.

Let the other vertices of the triangle B and C be (x, y) and (a, b).

The coordinates of the mid-point of the line segment joining the points P (x₁ , y₁) and Q (x₂ , y₂) are [(x₁ + x₂)/2, (y₁ + y₂)/2]

Given, the midpoint A(1, -4) and B(x, y) is (2, -1)

So, [(1 + x)/2, (-4 + y)/2] = (2, -1)

Now, (1 + x)/2 = 2

1 + x = 4

x = 4 - 1

x = 3

Also, (y - 4)/2 = -1

y - 4 = -2

y = -2 + 4

y = 2

The coordinates of B is (3, 2)

Given, the midpoint of A(1, -4) and C(a, b) is (0, -1)

So, [(1 + a)/2, (-4 + b)/2] = (0, -1)

Now, (1 + a)/2 = 0

1 + a = 0

a = -1

Also, (-4 + b)/2 = -1

b - 4 = -2

b = -2 + 4

b = 2

The coordinates of C are (-1, 2).

The area of a triangle with vertices A (x₁ , y₁) , B (x₂ , y₂) and C (x₃ , y₃) is

1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

The area of a triangle with vertices A (1, -4) , B (3, 2) and C (-1, 2)

= 1/2[1(2 - 2) + 3(2 - (-4)) + -1(-4 - 2)]

= 1/2[0 + 3(6) + (-1)(-6)]

= 1/2[18 + 6]

= 24/2

= 12 square units.

Therefore, the area of the triangle ABC is 12 square units.

✦ Try This: Find the area of the triangle PQR with P(2, -2) and the mid-points of sides through P being (1, 2) and (0, -2).

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Sample Problem 2

Summary:

The area of the triangle ABC with A (1, –4) and the mid-points of sides through A being (2, – 1) and (0, – 1) is 12 square units

☛ Related Questions: