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
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 = -
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),
X + 5(5x – 7) = - 19
X + 25x – 35 = - 19
26x = 16
Put the value of x in Eq. (i);
y = 5
∴ Coordinates of point P =
By distance formula,
Distance between two points (x1, y1) and (x2, y2) is;
So, length of the height of a parallelogram,
Hence, the required length of height of a parallelogram is
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