# The points A (1,

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 the 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.

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