The line segment joining the points A(2, 1) and B(5, – 8) is trisected by the points P and Q, where P is nearer to A. If the point P also lies on the line 2x – y + k = 0. Find the value of k.

OR

Show that (a, a), (– a, – a) and (– √3 a, √3 a) are vertices of an equilateral triangle.

As line segment AB is trisected by the points P and Q.

So, AP : PB = 1 : 2.

∴, the coordinates of P, by applying the Section Formula are

…(i)

Here, m₁ = 1, m₂ = 2

(x₁, y₁) = (2, 1) and (x₂, y₂) = (5, -8)

Putting the above values in the above formula, we get

⇒ x = 3, y = -2

⇒ the coordinates of P are (3, –2)

Since the point P (3, – 2) lies on the line 2x – y + k = 0

∴ 2 × 3 – (-2) + k = 0

⇒ 6 + 2 + k = 0

⇒ 8 + k = 0

⇒ k = -8

OR

Given points are A(a, a), B(-a,-a) and C(-a√3 , a√3)

To show: ABC is an equilateral triangle

Now,

Since, AB = BC = CA = 2a√2

So, the triangle ABC is an equilateral triangle.