If the points A(1, - 2), B(2, 3), C(a, 2) and D( - 4, - 3) form a parallelogram, then find the value of a and height of the parallelogram taking AB as base.

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 (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 – y_{1}) =

(y + 2) =

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

5x – y = 7

Slop of AB = m_{1} = ….. (i)

Let the slope of DP be m_{2}.

Since, DP is perpendicular to AB.

By condition of perpendicularity,

m_{1}.m_{2} = - 1 5.m_{2} = - 1

m_{2} = -

Now,

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

(y– y_{1}) = m_{2}(x – x_{1})

→ (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 (x_{1}, y_{1}) and (x_{2}, y_{2}) is;

D =

So, length of the height of a parallelogram,

DP =

DP =

Hence, the required length of height of a parallelogram is

4