Question

Find the length of AB

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  = $\sqrt{ (6- 2)^{2} + ( 2 - (-4))^{2} }$
               
                = $\sqrt{ (6- 2)^{2} + ( 2 +4))^{2} }$
    
                = $\sqrt{ (4)^{2} + ( 6))^{2} }$
 
                = $\sqrt{ 16+36}$
 
                = $\sqrt{52}$
  
               = $2 \sqrt{13}$

So, the length of AB is $2 \sqrt{13}$.




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 = $(\frac{ x_{1}+ x_{2} }{2} , \frac{ y_{1}+ y_{2} }{2} )$
        
$(1, 1)=(\frac{ 3+ a }{2} , \frac{ -2+b }{2} )$


Now, equaling the ordered pair, we have,

$1=\frac{ 3+ a }{2}$  .............equation (1)   

$1=\frac{ -2+b }{2}$  ................equation (2) 

Now, taking equation (1)
$1=\frac{ 3+ a }{2}$

$1*2=3+a$

$2-3=a$

$a=-1$

Now, taking equation (2)
$1=\frac{ -2+b }{2}$

$1*2=-2+b$

$2+2=b$

$b=4$

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