Question

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

Answer 1

In parallelogram, we know that, diagonals are bisects each other.

i.e., Mid-point of AC = Mid-point of BD

⇒ `((1 + a)/2, (-2 + 2)/2) = ((2 - 4)/2, (3 - 3)/2)`  ...`["Since, mid-point of a line segment having points"  (x_1, y_1)  "and"  (x_2, y_2)  "is" ((x_1 + x_2)/2, (y_1 + y_2)/2)]`

⇒ `(1 + a)/2 = (2 - 4)/2 = (-2)/2` = – 1

⇒ 1 + a = – 2

⇒ a = – 3

So, the required value of a is – 3.

Given that, AB as base of a parallelogram and drawn a perpendicular from D to AB which meet 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_1)/(x_2 - x_1)(x - x_1)`

⇒ (y + 2) = `(3 + 2)/(2 - 1)(x - 1)`

⇒ (y + 2) = 5(x – 1)

⇒ 5x – y = 7  ...(i)

Slope of AB, say m₁ = `(y_2 - y_1)/(x_2 - x_1) = (3 + 2)/(2 - 1)` = 5

Let the slope of DP be m₂.

Since, DP is perpendicular to AB.

By condition of perpendicularity,

m₁ · m₂ = – 1

⇒ 5 · m₂ = – 1

⇒ m₂ = `-1/5`

Now, equation of DP, having slope `(-1/5)` and passing the point (– 4, – 3) is

(y – y₁) = m₂(x – x₁)

⇒ (y + 3) = `-1/5(x + 4)`

⇒ 5y + 15 = – x – 4

⇒ x + 5y = – 19  ...(ii)

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

Put the value of y from equation (i) in equation (ii), we get

x + 5(5x – 7) = – 19   ...[Using equation (i)]

⇒ x + 25x – 35 = – 19

⇒ 26x = 16

∴ x = `8/13`

Put the value of x in equation (i), we get

y = `5(8/13) - 7 = 40/13 - 7`

⇒ y = `(40 - 91)/13`

⇒ y = `(-51)/13`

∴ Coordinates of point P ≡ `(8/13, (-51)/13)`

So, length of the height of a parallelogram,

DP = `sqrt((8/13 + 4)^2 + ((-51)/13 + 3)^2`   ...`[∵ "By distance formula, distance between two points"  (x_1, y_1)  "and"  (x_2, y_2)  "is"  d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)]`

⇒ DP = `sqrt((60/13)^2 + ((-12)/13)^2`

= `1/13 sqrt(3600 + 144)`

= `1/13 sqrt(3744)`

= `(12sqrt(26))/13`

Hence, the required length of height of a parallelogram is `(12sqrt(26))/13`.