The points A (1, - 2) , B(2,3), C (k,2) and D( - 4, - 3) are the vertices of a parallelogram. Find the value of k.

OR

Find the value of k for which the points (3k - 1, k - 2), (k,k - 7) and (k - 1, - k - 2) are collinear.

The figure of the above condition is as shown below

We know in a parallelogram, the diagonals bisect each other, so O is the midpoint of AC and BD (see the above figure)

Let the coordinates of O be x and y,

As O is midpoint of line BD, so the coordinates of O can be written as

Here x₁ and y₁ are the coordinates of point B(2,3)

And x₂ and y₂ are the coordinates of point D( - 4, - 3)

Substituting the values, we get

So, x = - 1, y = 0

So, the point O is ( - 1, 0)

Now we will consider the next diagonal AC,

Here also O ( - 1,0) is the midpoint

So, applying the midpoint formula

Here x₁ and y₁ are the coordinates of point A (1, - 2).

And x₂ and y₂ are the coordinates of point C(k,2).

Substituting the values, we get

⇒ 1 + k = 2 × ( - 1)

⇒ k = - 2 - 1

⇒ k = - 3

So, the required value of k is - 3.

OR

Given: points (3k - 1, k - 2), (k,k - 7) and (k - 1, - k - 2) are collinear

To find: the value of ‘k’

Explanation: collinear means all the points lie on same line.

Let the three points be A(3k - 1,k - 2), B(k,k - 7) and C(k - 1, - k - 2)

So, A, B, C will form a straight line and not a triangle

So, the area should be equal to 0

i.e., ar(ΔABC) = 0

now we know area of any triangle is given by

This should be equal to zero,

From the given points,

x₁ = 3k - 1, y₁ = k - 2

x₂ = k, y₂ = k - 7

x₃ = k - 1, y₃ = - k - 2

Now substituting these values in area formula, we get

⇒[(3k - 1) ((k - 7) - ( - k - 2)) + k(( - k - 2) - (k - 2) ) + (k - 1)((k - 2) - (k - 7))] = 0

⇒[(3k - 1) (k - 7 + k + 2) + k ( - k - 2 - k + 2) + (k - 1) (k - 2 - k + 7)] = 0

⇒[(3k - 1) (2k - 5) + k( - 2k) + (k - 1) (5)] = 0

⇒(3k(2k - 5) - 1(2k - 5)) - 2k² + 5k - 5 = 0

⇒ 6k² - 15k - 2k + 5 - 2k² + 5k - 5 = 0

⇒ 4k² - 12k = 0

⇒ 4k(k - 3) = 0

⇒ 4k = 0 or k - 3 = 0

⇒ k = 0 or k = 3

Hence the required value of k is 0 and 3.