Question

please answer this please............

Answer 1

Answer:

Correct option is ( B ) .

Step-by-step explanation:

By Distance Formula,

$\Rightarrow \text{Length\:of\:AB}=\sqrt{(4-1)^2+(4-2)^2}$

$\Rightarrow \text{Length\:of\:AB}= \sqrt{(3)^2+(2)^2}$

$\Rightarrow \text{Length\:of\:AB} =\sqrt{9+4}$

$\Rightarrow \text{Length\:of\:AB}=\sqrt{13}$

Now,

As the given figure is a square, length of AB is equal to all the other sides.

So, AD = BC = DC = AB = √13

Then,

$\Rightarrow Slope\:of\:AB=\dfrac{4-1}{4-2}$

$\Rightarrow Slope\:of\:AB=\dfrac{3}{2}$

As the lines BC and AD are perpendicular to AB, slope of BC and AD should be negative reciprocal of AB.

Therefore,

Slope of BC or AD =$-\dfrac{2}{3}$    ...( i )

Let the coordinates of D are ( c , d ).

By Formula ( for slope ),

$\text{Slope\:of\:AD}=\dfrac{d-4}{c-4}$

From ( i ) slope of AD is $-\dfrac{2}{3}$,

$\Rightarrow -\dfrac{2}{3}=\dfrac{d-4}{c-4}$

= >  - 2( c - 4 ) = 3( d - 4 )

= >  - 2c + 8 = 3d - 12

= >  8 + 12 - 3d = 2c

= >  20 - 3d = 2c

= >  ( 20 - 3d ) / 2 = c

= >  10 - ( 3d / 2 ) = c       ...( ii )

Now, by distance formula,

⇒ Length of AD = length of AB

⇒ Length of AD = √13

$\Rightarrow\sqrt{(c-4)^2+(d-4)^2}=\sqrt{13}$

$\Rightarrow (c-4)^2+(d-4)^2=13$

Substitute the value of c from ( ii ),

$\Rightarrow \bigg( 10 - \dfrac{3d}{2}-4\bigg)^2 + (d-4)^2=13$

$\Rightarrow \bigg( 6 - \dfrac{3d}{2}\bigg)^2 +(d-4)^2 = 13$

$\Rightarrow \bigg( \dfrac{12-3d}{2}\bigg)^2 + d^2 + (4)^2 - 8d =13$

$\Rightarrow \dfrac{144+9d^2-72d}{4}+d^2 + 16-8d=13$

$\Rightarrow \dfrac{144+9d^2-72d+4d^2+64-32d}{4}=13$

$\Rightarrow 13d^2 - 104d + 208 = 72$

= >  13d^2 - 104d + 208 - 72 = 0

= >  13d^2 - 104d + 156 = 0

= >  d^2 - 8d + 12 = 0

= >  d^2 - ( 6 + 2 )d + 12 = 0

= >  d^2 - 6d - 2d + 12 = 0

= >  d( d - 6 ) - 2( d - 6 ) = 0

= >  ( d - 6 )( d - 2 ) = 0

= > d = 6 or d = 2

Substitute the value of d in ( ii ),

= >  10 - ( 3d / 2 ) = c

= >  10 - 3( 6 / 2 ) = c  OR  10 - 3( 2 / 2 ) = c

= >  10 - 9 = c  OR   10 - 3 = c

= >  1 = c   OR   7 = c

Since the A , B , C and D are in anticlock wise manner, x co ordinate of D can't be more the x - co ordinate of A.

So,

Co ordinates of D = ( c , d ) = ( 1 , 6 ).

As there is only one option with ( 1 , 6 ), that is the correct answer. But I am doing the whole to make the confusions cleared.

Now, repeating the same process for BC by taking its co ordinates.

Slope of BC = $-\dfrac{2}{3}$

Let the coordinates of C are ( a , b ) ,

So,

$\Rightarrow \dfrac{b-1}{a-2}=-\dfrac{2}{3}$

= >  3( b - 1 ) = - 2( a - 2 )

= >  3b - 3 = - 2a + 4

= >  3b - 3 - 4 = - 2a

= >  3b - 7 = - 2a

= >  7 - 3b = 2a

= >  ( 7 - 3b ) / 2 = a        ...( iii )

Then, by distance formula for the length of BC.

$\Rightarrow\sqrt{(b-1)^2+(a-2)^2}=\sqrt{13}$

$\Rightarrow (b-1)^2 + (a-2)^2 = 13$

Substitute the value of a from ( iii ),

$\Rightarrow (b-1)^2 + \bigg( \dfrac{7-3b}{2}-2\bigg)^2 = 13$

$\Rightarrow b^2 + 1 -2b + \bigg( \dfrac{7-3b-4}{2}\bigg)^2 = 13$

$\Rightarrow b^2 + 1 - 2b + \bigg( \dfrac{3-3b}{2}\bigg)^2=13$

$\Rightarrow b^2 + 1 - 2b + \dfrac{9+9b^2-18b}{4}=13$

$\Rightarrow \dfrac{4b^2 + 4 - 8b + 9 + 9b^2- 18b }{4}=13$

= >  13b^2 - 26b + 13 = 52

= >  13b^2 - 26b + 13 - 52 = 0

= >  13b^2 - 26b - 39 = 0

= >  b^2 - 2b - 3 = 0

= >  b^2 - ( 3 - 1 )b - 3 = 0

= >  b^2 - 3b + b - 3 = 0

= >  b( b - 3 ) + ( b - 3 ) = 0

= >  ( b - 3 )( b + 1 ) = 0

= > b = 3 or b = - 1

Substitute the value of b in ( iii ),

= >  ( 7 - 3b ) / 2 = a

= >  7 - 3( 3 ) = 2a OR 7 - 3( - 1 ) = 2a

= >  7 - 9 = 2a   OR   7 + 3 = 2a

= >  - 2 = 2a    OR   10 = 2a

= >  - 2 / 2 = a   OR   10 / 2 = a

= >  - 1 = a   OR    5 = a

Since the A , B , C and D are in anticlock manner, x co ordinate of C can't be more the x - co ordinate of B.

So,

Co ordinates of C = ( a , b ) = ( - 1 , 3 ) .

Therefore,

Co ordinates of D = ( 1 , 6 )

Co ordinates of C = ( - 1 , 3 ).

Hence, option is the correct option.

Answer 2

Answer:

Option(B).

Step-by-step explanation:

Given points are A(4,4) and B(2,1).

Here (x₁,y₁) = (4,4) and (x₂,y₂) = (2,1)

Finding AB:

AB = √(x₂ - x₁)² + (y₂ - y₁)²

     = √(2 - 4)² + (1 - 4)²

    = √13.

Since, all the sides are equal. Therefore, AB = BC = CD = DA = √13.

Slope of AB = (1 - 4)/(2 - 4)

                    = -3/-2

                    = 3/2

Coordinates of C:

Let the coordinates of C be (a,b)

Finding BC:

√(b - 1)² + (a - 2)² = √13

On squaring both sides, we get

⇒ (b - 1)² + (a - 2)² = 13   ---- (i)

∴ Slope of BC = -1/Slope of AB

(b - 1)/(a - 2) = -2/3

3(b - 1) = -2(a-2)

3b - 3 = -2a + 4

3b - 7 = -2a

(3b - 7)/-2 = a  

(7 - 3b)/2 = a  ---- (ii)

Substitute (ii) in (i), we get

⇒ (b - 1)² + (7-3b/2 -2)² = 13

⇒ (13b²/4) - (13b/2) + (49/4) - 9 = 13

⇒ (13b²/4) - (13b/2) + (49/4) - 22 = 9

⇒ 13b² - 26b - 39 = 0

⇒ 13(b² - 2b - 3) = 0

⇒ 13(b² + b - 3b - 3) = 0

⇒ 13(b(b + 1) - 3(b + 1)) = 0

⇒ 13(b + 1)(b - 3) = 0

⇒ (b + 1)(b - 3) = 0

⇒ b = -1,3.

Given that the vertices are marked in anti-clock wise sense.{x-coordinate of  C should not be greater than B}

Consider b = 3 and put in (ii), we get:

⇒ (7 - 3b) = 2a

⇒ -2 = 2a

⇒ a = -1.

Hence, the coordinates of C are (-1,3)

Coordinates of D:

Let the coordinates of D be (p,q)

Finding AD:

√(p - 4)² + (q - 4)² = √13

On squaring both sides, we get

(p - 4)² + (q - 4)² = 13       ------- (iii)

∴ Slope of AD = -1/slope of AB

(q - 4)/(p - 4) = -2/3

3(q - 4) = -2(p - 4)

3q - 12 = -2p + 8

3q - 12 - 8 = -2p

3q - 20 = -2p

3q - 20/-2 = p

(20 - 3q)/2 = p     -------- (iv)

Substitute (iv) in (iii), we get

⇒ (20-3q/2 - 4)² + (q - 4)² = 13

⇒ (10 - 3q/2 - 4)² + (q - 4)² = 13

⇒ (6 - 3q/2)² + (q - 4)² = 13

⇒ (13q²/4) - 26q + 39 = 0

⇒ 13q² - 104q + 156 = 0

⇒ 13(q² - 8q + 12) = 0

⇒ 13(q² - 2q - 6q + 12) = 0

⇒ 13(q(q - 2) - 6(q - 2)) = 0

⇒ 13(q - 2)(q - 6) = 0

⇒ (q - 2)(q - 6) = 0

⇒ q = 2,6

Given that the vertices are marked in anti-clock wise sense.{x-coordinate of D Should not be greater than A}

Substitute q = 6 in (iv), we get

⇒ (20 - 3q) = 2p

⇒ (20 - 3(6)) = 2p

⇒ 20 - 18 = 2p

⇒ 2 = 2p

⇒ p = 1

Hence, coordinates of D are (1,6).

Therefore:

Coordinates of A = (4,4)

Coordinates of B = (2,1)

Coordinates of C = (-1,3)

Coordinates of D = (1,6)

Hope it helps!