FIRST QUESTION
Given points are:
A(2, -4) and B(6, 2)
Now, Use the distance formula.
distance formula = [tex] \sqrt{ (x_{2}- x_{1})^{2} + ( y_{2} - y_{1} )^{2} }[/tex]
Now, plug the values into the formula, So,
distance = [tex] \sqrt{ (6- 2)^{2} + ( 2 - (-4))^{2} }[/tex]
= [tex] \sqrt{ (6- 2)^{2} + ( 2 +4))^{2} }[/tex]
= [tex] \sqrt{ (4)^{2} + ( 6))^{2} }[/tex]
= [tex] \sqrt{ 16+36} [/tex]
= [tex] \sqrt{52} [/tex]
= [tex]2 \sqrt{13} [/tex]
So, the length of AB is [tex]2 \sqrt{13} [/tex].
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 = [tex] (\frac{ x_{1}+ x_{2} }{2} , \frac{ y_{1}+ y_{2} }{2} )[/tex]
[tex](1, 1)=(\frac{ 3+ a }{2} , \frac{ -2+b }{2} )[/tex]
Now, equaling the ordered pair, we have,
[tex]1=\frac{ 3+ a }{2}[/tex] .............equation (1)
[tex]1=\frac{ -2+b }{2} [/tex] ................equation (2)
Now, taking equation (1)
[tex]1=\frac{ 3+ a }{2}[/tex]
[tex]1*2=3+a[/tex]
[tex]2-3=a[/tex]
[tex]a=-1[/tex]
Now, taking equation (2)
[tex]1=\frac{ -2+b }{2}[/tex]
[tex]1*2=-2+b[/tex]
[tex]2+2=b[/tex]
[tex]b=4[/tex]
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
