ABCD is a square.
A is the point (-2, 0) and C is the point (6,4).
AC and BD are diagonals of the square, which intersect at M.
a Find the coordinates of M, B and D.
b Find the area of ABCD.
Answers
Answer :
Hope it helps you
ABCD is a square.
A is the point (-2, 0) and C is the point (6,4).AC and BD are diagonals of the square, which intersect at M.
To find :-
a)Find the coordinates of M, B and D.
b )Find the area of ABCD.
Solution :-
Given that
ABCD is a square
The coordinates of A = (-2,0)
The coordinates of C = ( 6,4)
Given that
AC and BD are diagonals of the square, which intersect at M.
So M is the mid point of AC and BD
Mid Point of AC :-
Let (x1, y1)=(-2,0)=> x1=-2 and y1=0
Let (x2, y2)=(6,4)=> x2=6 and y2=4
We know that
Mid Point formula =( {x1+x2}/2 , {y1+y2}/2)
Mid point of AC = ({-2+6}/2,{0+4}/2)
=>(4/2,4/2)
=> (2,2)
Mid point of AC = Mid point of BD
Coordinates of M = (2,2)
Let B =(x1,y1)
Let D=(x2, y2)
Mid point of BD
= ({x1+x2}/2 , {y1+y2}/2) = (2,2)
=>( x1+x2)/2 = 2 and (y1+y2)/2 = 2
=> x1+x2 = 4 and y1+y2 = 4
=> x2 = 4-x1 ----(1)and y2 = 4-y1--------(2)
We know that
All sides are equal in a square
=> AB = BC = CD = DA
We know that
Distance formula=√[(x2-x1)²+(y2-y1)²] units
AB = √[(x1+2)²+(y1-0)²]
=> AB = √[(x1+2)²+y1²]
DA = √[(x2+2)²+(y2-0)²]
=> DA =√[(x2+2)²+y2²]
From (1)&(2)
AB = AD
=> √[(x1+2)²+y1²] = [(x2+2)²+y2²]
On squaring both sides then
=> (x1+2)²+y1² = (x2+2)²+y2²
=> x1²+4x1+4+y1²=(4-x1+2)²+(4-y1)²
=> x1²+4x1+4+y1²=(6-x1)²+(4-y1)²
=> x1²+4x1+y1²+4=36+x1²-12x1+16+y1²-8y1
=> 16x1+8y1=48
=> 8(2x1+y1)=6(8)
=> 2x1+y1 = 6
=> y1 = 6-2x1------(4)
The diagonal = √2 × side
=> AC = √2 AD
=> √[(6+2)²+(4-0)²] = √2√[(x2+2)²+(y2-0)²
=> √(64+16)=√2(x2+2)²+y2²
=>√80=√2(x2+2)²+y2²
On squaring both sides
=> 80=2((x2+2)²+y2²)
=> 80/2 = (x2+2)²+y2²
=> 40 = (4-x1+2)²+(4-y1)²
=> 40 = (6-x1)²+(4-y1)²
=> 40 = 36+x1²-12x1+16-8y1+y1²
=> 40 = 52-12x1-8y1+x1²+y1²
from (4)
=> 40 = 52-12x1-8(6-2x1)+x1²+(6-2x1)²
=> 40-52=-12x1-48+16x1+x1²+36-24x1+4x1²
=> -12 = 5x1²-20x1-12
=> -12+12 = 5x1²-20x1
=> 0 = 5x12-20x1
=> 5x1² = 20x1
=> x1= 20/5
=> x1 = 4
from (4)
y1 = 6-2(4)
=> y1 = 6-8
=> y1=-2
The coordinates of B=(x1, y1)=(4,-2)
and
From (1)&(2)
x2 = 4-x1
=> x2 = 4-4
=> x2 = 0
and
y2 = 4-y1
=> y2 = 4-(-2)
=> y2 = 4+2
=> y2 = 6
The coordinates of D = (x2, y2)=(0,6)
Area of a square = side × side
=> AB²= BC²=CD²=DA²
By distance formula
AB = √[(4+2)²+(-2-0)²]
=> AB =√[6²+(-2)²]
=> AB = √(36+4)
=> AB =√40
=>AB²=(√40)²
=> AB = 40 sq.units
Answer:-
The coordinates of M = (2,2)
The other vertices are (4,-2) and (0,6)
Area of the square = 40 sq.units
Used formulae:-
In a square
All sides are equal.
Diagonals are equal
Area of a square = side×side
Mid Point formula = ( {x1+x2}/2 , {y1+y2}/2)
Distance formula
=√[(x2-x1)²+(y2-y1)²] units
Answer:
a) (5,6)
b) k = 8
Step-by-step explanation:
this is the proper answer.
from book