FIRST QUESTION
Given points are:
A(2, -4) Â and B(6, 2)
Now, Â Use the distance formula.
distance formula =Â
![\sqrt{ (x_{2}- x_{1})^{2} + ( y_{2} - y_{1} )^{2} }](/tpl/images/0176/0353/c9645.png)
Â
Now, plug the values into the formula, So,
distance  =Â
![\sqrt{ (6- 2)^{2} + ( 2 - (-4))^{2} }](/tpl/images/0176/0353/0c363.png)
       Â
        =Â
![\sqrt{ (6- 2)^{2} + ( 2 +4))^{2} }](/tpl/images/0176/0353/64aae.png)
  Â
        =Â
![\sqrt{ (4)^{2} + ( 6))^{2} }](/tpl/images/0176/0353/9b429.png)
Â
        =Â
![\sqrt{ 16+36}](/tpl/images/0176/0353/88f2e.png)
Â
        =Â
![\sqrt{52}](/tpl/images/0176/0353/5c3c7.png)
 Â
        =Â
![2 \sqrt{13}](/tpl/images/0176/0353/48f00.png)
So, the length of AB isÂ
![2 \sqrt{13}](/tpl/images/0176/0353/48f00.png)
.
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 =Â
![(\frac{ x_{1}+ x_{2} }{2} , \frac{ y_{1}+ y_{2} }{2} )](/tpl/images/0176/0353/5472e.png)
    Â
![(1, 1)=(\frac{ 3+ a }{2} , \frac{ -2+b }{2} )](/tpl/images/0176/0353/ecd06.png)
Now, equaling the ordered pair, we have,
![1=\frac{ 3+ a }{2}](/tpl/images/0176/0353/b17e9.png)
 .............equation (1) Â
![1=\frac{ -2+b }{2}](/tpl/images/0176/0353/9e449.png)
 ................equation (2)Â
Now, taking equation (1)
![1=\frac{ 3+ a }{2}](/tpl/images/0176/0353/b17e9.png)
![1*2=3+a](/tpl/images/0176/0353/20500.png)
![2-3=a](/tpl/images/0176/0353/2d318.png)
![a=-1](/tpl/images/0176/0353/346ac.png)
Now, taking equation (2)
![1=\frac{ -2+b }{2}](/tpl/images/0176/0353/c0855.png)
![1*2=-2+b](/tpl/images/0176/0353/9da68.png)
![2+2=b](/tpl/images/0176/0353/e5216.png)
![b=4](/tpl/images/0176/0353/53753.png)
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