Question

Please help me regarding question no. 22​

Answer 1

Cos 60 cos 30 + sin 60 sin 30

Answer 2

Step-by-step explanation:

Given:-

ABCD is a Parallelogram where A(x,y) ,B(5,8) ,C(4,7) and D(2,-4)

To find:-

I) The coordinates of A

ii) The equation of the diagonal BD

Solution:-

Given that

ABCD is a Parallelogram where A(x,y) ,B(5,8) ,C(4,7) and D(2,-4)

We know that

In a Parallelogram the diagonals bisecting each other

=>Mid point of AC = Mod point of BD

i)Mid point of AC:-

Let (x1, y1)=A(x,y)=>x1=x and y1 = y

Let (x2, y2)=C(4,7)=>x2=4 and y2=7

We know that

A(x1, y1) and (x2, y2) are the points then the mid point of the linesegment joining by the two points is M=[(x1+x2)/2 ,(y1+y2)/2]

=>[(x+4)/2 ,(y+7)/2]------------(1)

ii)Mid point of BD:-

Let (x1, y1)=B(5,8)=>x1=5 and y1 = 8

Let (x2, y2)=D(2,-4)=>x2=2 and y2=-4

We know that

A(x1, y1) and (x2, y2) are the points then the mid point of the linesegment joining by the two points is M=[(x1+x2)/2 ,(y1+y2)/2]

=>[(5+2)/2 ,(8+(-4))/2]

=>[(7/2 , (8-4)/2]

=>(7/2 ,4/2)

=>(7/2 ,2)-------------(2)

From (1) & (2)

Mid point of AC= Mid point of BD

=>[(x+4)/2 ,(y+7)/2] = (7/2 ,2)

On comparing both sides then

=>(x+4)/2 = 7/2 and (y+7)/2=2

=>x+4 = 7 and y+7 = 4

=>x=7-4 and y=4-7

=>x = 3 and y= -3

The coordinates of A = (3,-3)

iii) Equation of the diagonal BD:-

Let (x1, y1)=B(5,8)=>x1=5 and y1 = 8

Let (x2, y2)=D(2,-4)=>x2=2 and y2=-4

We know that

A(x1, y1) and (x2, y2) are the points then the equation is (y-y1)/(y2-y1) = (x-x1)/(x2-x1)

=>(y-8)/(-4-8) = (x-5)/(2-5)

=>(y-8)/(-12) = (x-5)/(-3)

=>(y-8)/12 = (x-5)/3

On applying cross multiplication then

=>12(x-5)=3(y-8)

=>12x-60=3y-24

=>12x-60-3y+24=0

=>12x-3y-36 =0

=>3(4x-y-12) = 0

=>4x-y-12=0

The equation is 4x-y-12=0

Answer:-

i) The Coordinates of A = (3,-3)

ii)The equation of the diagonal BD = 4x-y-12=0

Used formulae:-

• In a Parallelogram the Diagonals bisect each other.

• A(x1, y1) and (x2, y2) are the points then the equation is (y-y1)/(y2-y1) = (x-x1)/(x2-x1)

• A(x1, y1) and (x2, y2) are the points then the mid point of the linesegment joining by the two points is M=[(x1+x2)/2 ,(y1+y2)/2]