Question

If A(-2, 1), B(a, 0), C (4,b)and D(1,2)are vertices of a parallelogram ABCD, find the value of a and b. Hence find length of its side.​

Answer 1

AnswEr :

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.1,3){\large{A(-2,1)}}\put(7.2,0.9){\large{B(a,0)}}\put(11.1,0.9){\large{C(4,b)}}\put(9.9,2.1){\large{O}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D(1,2)}}\end{picture}$

• Let's Head to the Question Now :

• Diagonals of Parallelogram Bisects each other, Here at O.
• Mid Points of Both Diagonal will be Equal.

$\implies \tt Mid \:Point \:of \:Diagonal \:AC=Mid \:Point \:of \:Diagonal \:BD \\ \\\implies \tt\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg) = \bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\ \\\implies \tt\bigg(\dfrac{-2+4}{2},\dfrac{1+b}{2}\bigg) = \bigg(\dfrac{a+1}{2},\dfrac{0+2}{2}\bigg) \\ \\\implies \tt\bigg( \cancel\dfrac{2}{2},\dfrac{1+b}{2}\bigg) = \bigg(\dfrac{a+1}{2},\cancel\dfrac{2}{2}\bigg) \\ \\\implies \tt\bigg( 1,\dfrac{1+b}{2}\bigg) = \bigg(\dfrac{a+1}{2},1\bigg) \\ \\\implies \tt1 = \dfrac{a+1}{2} \quad and \quad\dfrac{1+b}{2} = 1 \\ \\\implies \tt2 = a + 1\quad and \quad1 + b = 2 \\ \\\implies \red{\tt a = 1\quad} \tt and \quad \red{ b = 1}$

⠀

∴ Therefore, Value of a and, b will be 1.

$\rule{300}{2}$

• CALCULATION⠀OF⠀SIDES :

$\longrightarrow \tt AB = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{(a-( - 2))^{2}+(0 - 1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{(1 + 2)^{2}+(1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{(3)^{2}+(1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{9 + 1} \\ \\\longrightarrow \green{\tt AB = \sqrt{10} \:unit}$

$\rule{100}{2}$

$\longrightarrow \tt BC = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{(4-a)^{2}+(b-0)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{(4-1)^{2}+(1-0)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{(3)^{2}+(1)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{9 + 1} \\\\\longrightarrow \green{\tt BC = \sqrt{10} \:unit}$

◗ In Parallelogram opposite sides are equal

⇒ AB = CD = √10 Unit

⇒ BC = AD = √10 Unit

⠀

∴ Therefore, All Sides will be √10 unit.

#answerwithquality #BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\bf{\blue{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

If A(-2, 1) , B(a, 0), C(4,b) & D(1,2) are vertices of a parallelogram ABCD.

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The value of a & b and also the length of its side.

$\bf{\Large{\boxed{\sf{\green{Explanation\::}}}}}$

$\sf{Attachment\:a\:figure,\:here;}$

We have,

$\sf{Vertices\:of\:parallelogram\begin{cases}\sf{A(\:-2,1)}\\ \sf{B\:(a,0)}\\ \sf{C\:(4,b)}\\ \sf{D\:(1,2)}\end{cases}}$

$\leadsto\rm{Midpoint\:of\:AC\:=\:Midpoint\:of\:BD}$

So,

R is the mid point of AC & BD.

• $\bf{\large{\boxed{\sf{\red{Using\:midpoint\:formula\::}}}}}$

$\mapsto\rm{(\frac{x1+x2}{2} \:,\:\frac{y1+y2}{2} )}$

So,

$\longmapsto\sf{Midpoint\:of\:AC\:=\:Midpoint\:of\:BD}$

$\longmapsto\sf{(\frac{x1+x2}{2} )\:,\:(\frac{y1+y2}{2}) \:\:=\:\:(\frac{x1+x2}{2} )\:,\:(\frac{y1+y2}{2} )}$

$\longmapsto\sf{(\frac{-2+4}{2} \:,\:\frac{1+b}{2} )\:\:=\:\:(\frac{a+1}{2} \:,\:\frac{2+0}{2} )}$

$\longmapsto\sf{(\frac{2}{2} \:,\frac{1+b}{2} )\:=\:(\frac{a+1}{2} \:,\:\frac{2}{2})}$

Therefore,

$\longmapsto\sf{\frac{2}{\cancel{2}} =\frac{a+1}{\cancel{2}} }$

$\longmapsto\sf{2=a+1}$

$\longmapsto\sf{a\:=2-1}$

$\longmapsto\sf{\red{a\:=\:1}}$

&

$\longmapsto\sf{\frac{2}{\cancel{2}} =\frac{1+b}{\cancel{2}} }$

$\longmapsto\sf{2=b+1}$

$\longmapsto\sf{b\:=2-1}$

$\longmapsto\sf{\red{b\:=\:1}}$

Thus,

$\sf{\boxed{\sf{\pink{The\:value\:of\:a\:=1\:&\:also\:the\:value\:of\:b\:=1}}}}}}$

______________________________________________

$\bf{\large{\underline{\sf{\orange{Distance\:between\:two\:points\:A(x1,y1) and B(x2,y2)\:is\:given\:by\::}}}}}}$

$\leadsto\sf{AB\:=\:\sqrt{(x2-x1)^{2}\:+\:(y2-y1)^{2}}}$

So,

$\longmapsto\sf{AB\:=\:\sqrt{(a-(-2)^{2} \:+(0-1)^{2} } }$

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

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

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

$\longmapsto\sf{\red{AB\:=\:\sqrt{10} }}$

&

$\longmapsto\sf{CD\:=\:\sqrt{(x2-x1)^{2} +(y2-y1)^{2} } }$

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

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

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

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

$\longmapsto\sf{\red{CD\:=\:\sqrt{10} }}$

Now,

We know that in parallelogram opposite sides & angles are equal.

$\mapsto\sf{AB\:=\:CD\:=\sqrt{10} \:unit}$

$\mapsto\sf{AD\:=\:BC\:=\sqrt{10} \:unit}$

Thus,

$\bf{\large{\boxed{\sf{\pink{The\:length\:of\:side\:of\:parallelogram\:is\:\sqrt{10} \:unit}}}}}$