Question

show that (4, 5), (7, 6), (4, 3), (1, 2) are the vetices of Parallelogram​

Answer 1

Question :-

• Show that (4, 5) , (7, 6) , (4, 3) , (1, 2) are the vertices of parallelogram.

Solution :-

Let, the vertices are R(4, 5), A(7, 6), J(4, 3), S(1, 2).

Applying distance formula,

XY = √(x2 - x1)² + (y2 - y1)²

Case (I),

R(4, 5) & A(7 , 6)

⇝RA = √(7 - 4)² + (6 - 5)²

⇝RA = √(3)² + (1)²

⇝RA = √9 + 1

⇝RA = √10 __________(1)

Case (II),

A(7, 6) & J(4, 3)

⇝AJ = √(4 - 7)² + (3 - 6)²

⇝AJ = √(-3)² + (3)²

⇝AJ = √9 + 9

⇝AJ = √18 ___________(2)

Case (III),

J(4, 3) & S(1, 2)

⇝JS = √(1 - 4)² + (2 - 3)²

⇝JS = √(-3)² + (-1)²

⇝JS = √9 + 1

⇝JS = √10 ____________(3)

Case (IV),

R(4, 5) & S(1, 2)

⇝RS = √(1 - 4)² + (2 - 5)²

⇝RS = √(-3)² + (-3)²

⇝ RS = √9 + 9

⇝RS = √18 ___________(4)

From equation (1) & (3),

⇝ RA = JS

From equation (2) & (4),

⇝ AJ = RS

Now, check the diagonals,

R(4, 5) & J(4, 3)

⇝ RJ = √(4 - 4)² + (3 - 5)²

⇝ RJ = √(0)2 + (-2)²

⇝ RJ = √4

⇝ RJ = 2

A(7, 6) & S(1, 2)

⇝ AS = √(1 - 7)² + (2 - 6)²

⇝ AS = √(-6)² + (-4)²

⇝ AS = √36 + 16

⇝ AS = √52

.°. [] RAJS is a parallelogram.

Therefore,

• (4, 5), (7, 6), (4, 3), (1, 2) are the vetices of Parallelogram

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• Show that (4, 5), (7, 6), (4, 3), (1, 2) are the vertices of Parallelogram​

★═════════════════★  

♣ ᴀɴꜱᴡᴇʀ :

Let the Given points be :

A (4, 5)  

B (7, 6)

C (4, 3)

D (1, 2)

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\put(0.4,0.5){\bf D (1,2)}\qbezier(1,1)(1,1)(1.6,4)\put(6.2,0.5){\bf C (4,3)}\qbezier(1.6,4)(1.6,4)(6.6,4)\put(1,4.3){\bf A (4,5)}\qbezier(6,1)(6,1)(6.6,4)\put(6.9,3.8){\bf B (7,6)}\end{picture}$

Apply Distance Formula : $\sf{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

AB = $\sf{\sqrt{(7 - 4)^2 + (6 - 5)^2}}$

AB = $\sf{\sqrt{(3)^2 + (1)^2}}$

AB = $\sf{\sqrt{(3)^2 + (1)^2}}$

AB = $\sf{\sqrt{9+1}}$

AB = $\sf{\sqrt{10}}$ units

________________________________________

Apply Distance Formula : $\sf{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

BC = $\sf{\sqrt{(4 - 7)^2 + (3 - 6)^2}}$

BC = $\sf{\sqrt{(-3)^2 + (3)^2}}$

BC = $\sf{\sqrt{9+9}}$

BC = $\sf{\sqrt{18}}$ units

________________________________________

Apply Distance Formula : $\sf{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

CD = $\sf{\sqrt{(1 - 4)^2 + (2 - 3)^2}}$

CD = $\sf{\sqrt{(-3)^2 + (-1)^2}}$

CD = $\sf{\sqrt{9+1}}$

CD = $\sf{\sqrt{10}}$ units

________________________________________

Apply Distance Formula : $\sf{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

AD = $\sf{\sqrt{(1 - 4)^2 + (2 - 5)^2}}$

AD = $\sf{\sqrt{(-3)^2 + (-3)^2}}$

AD = $\sf{\sqrt{9+9}}$

AD = $\sf{\sqrt{18}}$ units

________________________________________

Here BC = AD and AB = CD

So it must be a Rectangle or Parallelogram

Let's check the Diagnols :

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf D}\put(6,0.5){\sf C}\put(1.4,4.3){\sf A}\put(6.6,4.3){\sf B}\end{picture}$

Apply Distance Formula : $\sf{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

AC = $\sf{\sqrt{(4- 4)^2 + (3 - 5)^2}}$

AC = $\sf{\sqrt{(0)^2 + (-2)^2}}$

AC = $\sf{\sqrt{0+4}}$

AC = $\sf{\sqrt{4}}$

AC = $\sf{2}$ units

________________________________________

Apply Distance Formula : $\sf{\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

BD = $\sf{\sqrt{(1- 7)^2 + (2 - 6)^2}}$

BD = $\sf{\sqrt{(-6)^2 + (-4)^2}}$

BD = $\sf{\sqrt{36+16}}$

BD = $\sf{\sqrt{52}}$ units

________________________________________

From the data we got from this calculations , it is a Paralleogram

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\put(0.4,0.5){\bf D (1,2)}\qbezier(1,1)(1,1)(1.6,4)\put(6.2,0.5){\bf C (4,3)}\qbezier(1.6,4)(1.6,4)(6.6,4)\put(1,4.3){\bf A (4,5)}\qbezier(6,1)(6,1)(6.6,4)\put(6.9,3.8){\bf B (7,6)}\end{picture}$

∴ (4, 5), (7, 6), (4, 3), (1, 2) are the vetices of Parallelogram​