). If P(3, 4), Q(4, 6) and R(5, 7) are the midpoints of AB, BC and CA. Then,

Adding (i), (iii) and (v), we get

2(x₁ + x₂ + x₃) = 6 + 8 + 10 = 24

⇒    x₁ + x₂ + x₃ = 12    ...(vii)

From (i) and (vii), we get x₃ = 12 - 6 = 6
From (iii) and (vii), we get v₁ = 12 - 8 = 4
From (v) and (vii), we get x₂ = 12 - 10 = 2
Now, adding (ii), (iv) and (vi), we get
20(y₁ + y₂ + y₃) = 8 + 12 + 14 = 34
⇒    y₁ + y₂ + y₃ = 17    (viii)
From (ii) and (viii), we get y₃ = 17 - 8 = 9
From (iv) and (viii), we get y₁ = 17 - 12 = 5
From (vi) and (viii), we get y₂ 17 - 14 = 3
Hence, the vertices of ∆ABC are A(4, 5), B(2, 3), C(6, 9)

Let the point A(x, y) divide the line segment joining the points P(3, 3) and Q(6, - 6) in the ratio 1 : 2. Then,



Now, the co-ordinates of P are given as

Here, we have, x₁ = 3, y₁ = 3;
x₂ = 6, y₂ = -6
So, co-ordinate of A be,

∴ The point A is (4, 0) which lies on the line
2x + y + k = 0
⇒    2(4) + 0 + k = 0
⇒    k = - 8. Ans.
Problems Based on Area of Triangle

Let the given points arc A(1, 3) and B(2, 7). Let the line 3x + 4y - 9 = 0 divides the line segment AB in the ratio K : 1 at point P. Then the co-ordinates of P are given as

 

Here, we have x₁ = 1,    y₁ = 3
x₂ = 2,    y₂ = 7
and    m₁ = k, m₂ = 1
So, co-ordinates of P are

But P lies on 3x + 4y - 9 = 0    ...(i)
Substituting the values of x (abscissa) and y (ordinate) in (i), we get
                     3x + 4y - 9 = 0


Hence, the required ratio is .