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

Open in App

Solution

In parallelogram, we know that, diagonals bisect each other.
$\Rightarrow$ Mid-point of AC = mid-point of BD.

$\Rightarrow (\frac{1 + a}{2}, \frac{- 2 + 2}{2}) = (\frac{2 - 4}{2}, \frac{3 - 3}{2}) \Rightarrow \frac{1 + a}{2} = \frac{2 - 4}{2} = \frac{- 2}{2} = - 1 ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} ∵ s i n c e, m i d - p o i n t o f l i n e s e g m e n t h a v i n g p o i n t s (x_{1}, y_{1}) a n d (x_{2}, y_{2}) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ \Rightarrow 1 + a = - 2 \Rightarrow a = - 3$
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
$\Rightarrow (y - y_{1}) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \Rightarrow (y + 2) = \frac{3 + 2}{2 - 1} (x - 1) \Rightarrow (y + 2) = 5 (x - 1) \Rightarrow 5 x - y = 7 S l o p e o f A B, s a y m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3 + 2}{2 - 1} = 5 . . . (i)$
Let the slope of DP be m_2.
Since, DP is perpendicular to AB.
By condition of perpendicularity,
$m_{1} m_{2} = - 1 \Rightarrow 5. m_{2} = - 1 m \Rightarrow m_{2} = - \frac{1}{5}$
Now, Eq. of DP, having slope $(- \frac{1}{5})$ and passing the point (-4,-3) is
$\Rightarrow (y + 3) = - \frac{1}{5} (x + 4) \Rightarrow 5 y + 15 = - x - 4 \Rightarrow x + 5 y = - 19 . . . (i i)$
On adding Eqs.(i) and (ii) then we get the intersection point P.
Put the value of y from Eq.(i) in Eq.(ii), we get
$x + 5 (5 x - 7) = - 19 [u s i n g E q . (i)] \Rightarrow x + 25 x - 35 = - 19 \Rightarrow 26 x = 16 ∴ x = \frac{18}{3}$
Put the value of x in Eq.(i), we get
$y = 5 (\frac{8}{13}) - 7 = \frac{40}{13} - 7 \Rightarrow y = \frac{40 - 91}{13} \Rightarrow y = \frac{- 51}{13} ∴ C o o r d i n a t e o f p o i n t P \equiv (\frac{8}{13}, \frac{- 51}{13})$
So, length of the height of a parallelogram
$D P = \sqrt{{(\frac{8}{13} + 4)}^{2} + {(\frac{- 51}{13} + 3)}^{2}} ⎡ ⎢ ⎢ ⎣ \begin{matrix} ∵ b y d i s t a n c e f o r m u l a, d i s t a n c e b e t w e e n t w o p o i n t s (x_{1}, y_{1}) a n d (x_{2}, y_{2}) = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \end{matrix} ⎤ ⎥ ⎥ ⎦ \Rightarrow D P = \sqrt{{(\frac{60}{13})}^{2} + {(\frac{- 12}{13})}^{2}} = \frac{1}{13} \sqrt{3600 + 144} = \frac{1}{13} \sqrt{3744} = \frac{12 \sqrt{26}}{13}$
Hence, the required length of height of a parallelogram is $\frac{12 \sqrt{26}}{13}$ .

Suggest Corrections

Similar questions

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

Q. If the points

A (1, - 2), B (2, 3), C (a, 2)

and

D (- 4, - 3)

from a parallelogram. Find the value of

^{'} a^{'}

and height of the parallelogram taking

A B

as base.

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

Q. 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

A B

as base.

Q. The points

A (1, - 2), B (2, 3), C (k, 2)

and

D (- 4, - 3)

are the vertices of a parallelogram. Find the value of

k

and the altitude of the parallelogram corresponding to the base

A B

Join BYJU'S Learning Program

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

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