Q. 164.4( 8 Votes )

The points A (1,

Answer :

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.


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