Question

Find the linear relation between x and y such that P(x, y) is equidistant from the points A(1, 4) and B(–1, 2).

Answer 1

Answer:

x + y = 3

Step-by-step explanation:

Given that-

P(x,y) is equidistant form the points A(1, 4) and B(-1, 2).

PA = PB

Distance between two points X(a, b) and Y(c, d) is-

$XY = \sqrt{(a-c)^{2}+(b-d)^{2}  }$

So, distance between P(x, y) and A(1, 4) is-

PA = $\sqrt{(x-1)^{2}+(y-4)^{2}  }$

     = $\sqrt{x^{2}+1-2x+y^{2}+16-8y  }$

PA = $\sqrt{x^{2}+y^{2}-2x-8y+17  }$

Similarly, distance between P(x, y) and B(-1, 2) is-

PB = $\sqrt{(x+1)^{2}+(y-2)^{2}  }$

     = $\sqrt{x^{2}+1+2x+y^{2}+4-4y  }$

PB = $\sqrt{x^{2}+y^{2}+2x-4y+5  }$

So, $\sqrt{x^{2}+y^{2}-2x-8y+17  } = \sqrt{x^{2}+y^{2}+2x-4y+5  }$

Squaring on both sides-

$x^{2}+y^{2}-2x-8y+17 = x^{2}+y^{2}+2x-4y+5$

$x^{2}+y^{2}-2x-8y+17-x^{2}-y^{2}-2x+4y-5 = 0$

-4x - 4y + 12 = 0

4x + 4y = 12

Divided by 4-

x + y = 3

Hence, this is the required linear between x and y.

Answer 2

$\textbf{\underline {Solution:-}}$

$\textbf{For find mid point formula:-}$

$c (\frac{ + 1( - 1}{ 2} ) = \frac{4 + 2}{2} \\ \\ 2,3$

$\textbf{Therefore final answer is 2,3}$