Question

If the points A (6, 1), B (8, 2), C (9, 4) and D (p, 3) are the vertices of a parallologram, taken in order,

find the value of p.​

Answer 1

➤Given :

The given vertices of a parallelogram are

•A=(6,1)

•B=(8,2)

•C=(9,4)

•D=(p,3)

➤To Find :

The values of p =?

➤Solution :

$\red\bigstar\green{\bf{Given \: points }} \begin{cases} \: \sf{A=(6,1) } \\ \\ \sf{B=(8,2)} \\ \\ \sf{C=(9,4)} \\ \\ \sf{D=(p, 3)} \end{cases}$

As we know that,

⇒The diagonals of a parallelogram bisect each other.

Consider,

• O is the midpoint of AC and BD

>> We find X-coordinates of O from both AC and BD.

Finding the midpoint of AC.

We have to find x-coordinate of O

$\\ \bullet\bf{ x-coordinate \: of \: O=\dfrac{x_{1}+x_{2}}{2}}$

Where,

$\bullet\tt{ x_{1} =6 \: and \: x_{2}=9}$

Putting values of x -coordinates :

$\\ \implies\sf{x-coordinate \: of \: O=\dfrac{6+9}{2}}$

$\\ \implies\sf{x-coordinate \: of \: O=\dfrac{15}{2}}$ ......(1)

Finding midpoint of BD,

We have to find x-coordinates of O

$\\ \implies\sf{ x-coordinate \: of \: O =\dfrac{x_{1}+x_{2}}{2}}$

Where,

$\bullet\tt{x_{1}=8 \: and \: x_{2}=p}$

Putting the values of x-coordinates :

$\\ \implies\sf{x-coordinates \: of \: O=\dfrac{x_{1}+x_{2}}{2}}$

$\\ \implies\sf{x-coordinates \: of \: O=\dfrac{8+p}{2}}$

$\\ \implies\sf{x-coordinates \: of \: O=\dfrac{8+p}{2}}$ ......(2)

Comparing equations (1) and (2) :

$\\ \\ \implies \sf \: \frac{15}{2} = \frac{8 + p}{2} \\ \\ \\ \implies \sf \: \frac{15}{ \cancel2} = \frac{8 + p}{ \cancel2} \\ \\ \\ \implies \sf \: 15 = 8 + p \\ \\ \\ \implies \sf \: 15 - 8 = p \\ \\ \\ \implies \sf \: 7 = p \\ \\ \\ \implies \boxed{ \bf{p = 7}}$

Hence,

$\red \bigstar \boxed{ \sf{The \: value \: of \: p = 7}}$