Question

Find the length of AB

A (2,-4) B (6,2)

Solve for x and y
2x+3y+14
-4x+2y+4

Given that B is the midpoint of segment AC and A (3,-2) and B (1,1), find C

Answer 1

FIRST QUESTION
Given points are:
A(2, -4) and B(6, 2)
Now, Use the distance formula.
distance formula =
`\sqrt{ (x_(2)- x_(1))^(2) + ( y_(2) - y_(1) )^(2) }`

Now, plug the values into the formula, So,
distance =
`\sqrt{ (6- 2)^(2) + ( 2 - (-4))^(2) }`

=
`\sqrt{ (6- 2)^(2) + ( 2 +4))^(2) }`

=
`\sqrt{ (4)^(2) + ( 6))^(2) }`

=
`√( 16+36)`

=
`√(52)`

=
`2 √(13)`

So, the length of AB is
`2 √(13)`.




THIRD QUESTION
Two points given are:
A(3, -2) and B(1, 1)
Also given that B is the midpoint of AC.

Let, the co-ordinates of C be C(a, b).
Now, using midpoint formula,
Midpoint =
`(( x_(1)+ x_(2) )/(2) , ( y_(1)+ y_(2) )/(2) )`


`(1, 1)=(( 3+ a )/(2) , ( -2+b )/(2) )`


Now, equaling the ordered pair, we have,


`1=( 3+ a )/(2)` .............equation (1)


`1=( -2+b )/(2)` ................equation (2)

Now, taking equation (1)

`1=( 3+ a )/(2)`


`1*2=3+a`


`2-3=a`


`a=-1`

Now, taking equation (2)

`1=( -2+b )/(2)`


`1*2=-2+b`


`2+2=b`


`b=4`

So, the co ordinates of C are (a, b) which is (-1 , 4)




SECOND QUESTION:
Given equations are:
2x + 3y = 14.....................equation (1)
-4x + 2y = 4 .....................equation (2)
Taking equation (2)
-4x + 2y = 4
2y = 4 + 4x
y = (4 + 4x) / 2
y = 2 + 2x .......................equation (3)
Now, Taking equation (1)
2x + 3y = 14
Substituting the value of y from equation (3), we get,
2x + 3(2 + 2x) = 14
2x + 6 + 6x = 14
8x = 14 - 6
x = (14 - 6) / 8
x = 1

Taking equation (3)
y = 2 + 2x
Now, substituting the value of x in equation (3), we get,
y= 2 + 2 (1)
y = 2 + 2
y = 4

So, x=1 and y=4