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Sia ? 6 years, 5 months ago
If a and b are odd numbers then it should be in 2q+1 or 2q+3 form where q is a positive integer.
Let a = 2q + 3 , b = 2q + 1 and a > b
Now, {tex}\frac{a + b } {2} = \frac{ 2q + 3 + 2q + 1}{2}{/tex}
{tex}= \frac { 4 q + 4 } { 2 }{/tex}
= 2q + 2
{tex}\frac{a+b}{2}{/tex}=2(q+1) = an even number..........(1)
Now
{tex}\frac { a - b } { 2 } = \frac { ( 2 q + 3 ) - ( 2 q + 1 ) } { 2 }{/tex}
{tex}= \frac { 2 q + 3 - 2 q - 1 } { 2 }{/tex}
{tex}\frac{a-b}{2}= \frac { 2 } { 2 }{/tex} = 1 = an odd number..........(2)
Hence From (1) and (2) {tex}\frac{a+b}{2}{/tex} and {tex}\frac{a-b}{2}{/tex} are even and odd numbers respectively
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Sia ? 6 years, 5 months ago
We have
A {tex}\rightarrow{/tex} (5, 1)
B {tex}\rightarrow{/tex} (-3, -7)
C {tex}\rightarrow{/tex} (7, -1)
Let the point P(x, y) be equidistant from the three given points A, B, and C.
Then PA = PB = PC
{tex}\Rightarrow{/tex} PA2 = PB2 = PC2
First two give
PA2 = PB2
{tex}\Rightarrow{/tex} (x - 5)2 + (y - 1)2 = (x + 3)2 + (y + 7)2
{tex}\Rightarrow{/tex} x2 - 10x + 25 + y2 - 2y + 1 = x2 + 6x + 9 + y2 + 14y + 49
{tex}\Rightarrow{/tex} 16x + 16y + 32 = 0
{tex}\Rightarrow{/tex} x + y + 2 = 0 ....Dividing throughout by 16 ....(1)
Last two give
PB2 = PC2
{tex}\Rightarrow{/tex} (x + 3)2 + (y + 7)2 = (x - 7)2 + (y + 1)2
{tex}\Rightarrow{/tex} x2 + 6x + 9 + y2 + 14y + 49 = x2 - 14x + 49 + y2 + 2y + 1
{tex}\Rightarrow{/tex} 20x + 12y + 8 = 0
{tex}\Rightarrow{/tex} 5x + 3y + 2 = 0 ....Dividing throughout by 4 ....(2)
Multiplying equation (1) by 3, we get
3x + 3y + 6 = 0 ....(3)
Subtracting equation (3) from equation (2), we get
2x - 4 = 0
{tex}\Rightarrow{/tex} 2x = 4
{tex}\Rightarrow x = \frac{4}{2} = 2{/tex}
Subtracting x = 2 in equation (1), we get
2 + y + 2 = 0
{tex}\Rightarrow{/tex} y + 4 = 0
{tex}\Rightarrow{/tex} y = -4
Hence, the required points is (2, -4).
Posted by Vijay Suryawanshi 7 years, 8 months ago
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