Q. 43.7( 13 Votes )

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Answer :

In parallelogram, we know that, diagonals are bisects each other i.e., mid - point of AC = mid - point of BD




= - 1


Since, mid - point of a line segment having points (x1, y1) and (x2, y2) is


1 + a = - 2


a = - 3


So, the value of a is – 3


Given,


AB as base of a parallelogram and drawn a perpendicular from D to AB which meets AB at P.


So, DP is a height of a parallelogram.


Now, equation of base AB, passing through the points (1, - 2) and (2, 3) is;


(y – y1) =


(y + 2) =


(y + 2) = 5(x - 1)


5x – y = 7


Slop of AB = m1 = ….. (i)


Let the slope of DP be m2.


Since, DP is perpendicular to AB.


By condition of perpendicularity,


m1.m2 = - 1 5.m2 = - 1


m2 = -


Now,


Equation of DP, having slope - and passing the point ( - 4, - 3) is;


(y– y1) = m2(x – x1)


(y + 3) = - (x + 4)


5y + 15 = - x – 4


x + 5y = - 19 ……(ii)


On adding Equations (i) and (ii), then we get the intersection point P.


By putting the value of y from Eq. (i) to Eq. (ii),


We get;


X + 5(5x – 7) = - 19


X + 25x – 35 = - 19


26x = 16


x =


Put the value of x in Eq. (i);


We get;


y = 5


Coordinates of point P =


By distance formula,


Distance between two points (x1, y1) and (x2, y2) is;


D =


So, length of the height of a parallelogram,


DP =


DP =





Hence, the required length of height of a parallelogram is


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