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
