Question
Find a point which is equidistant from the points A(-5, 4) and B(-1, 6)? How many such points are there?
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Answer
Let P (h, k) be the point which is equidistant from the points A (- 5, 4) and B (-1, 6).
$\Big[\because$ distance between two point $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\therefore$ PA = PB
$\Rightarrow (PA)^2 = (PB)^2$
$\Rightarrow (-5 - h)^2 + (4 - k)^2 = (-1 - h)^2 + (6 - k)^2$
$\Rightarrow 25 + h^2+ 10h + 16 + k^2 - 8k = 1 + h^2 + 2h + 36 + k^2 - 12k$
$\Rightarrow 25 + 10h + 16 - 8k = 1 + 2h + 36 - 12k$
$\Rightarrow 8h + 4k + 4 = 0$
$\Rightarrow 2h + k + 1 = 0$
Mid-point $\text{AB}=\Big(\frac{-5-1}{2},\frac{4+6}{2}\Big)=(-3, 5)$
$\Big[\because\text{mid-point}=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
At point (-3, 5), from EQ. (i).
$\Rightarrow 2h + k = 2(-3) + 5$
$\Rightarrow 2h + k = -6 + 5$
$\Rightarrow 2h + k = -1$
$\Rightarrow 2h + k +1 = 0$
So, the mid-point of AB satisfy the Eq. (i). Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the points A and B. Replacing h, kby, x, y in above equation, we have 2 x + y + 1= 0.
$\Big[\because$ distance between two point $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\therefore$ PA = PB
$\Rightarrow (PA)^2 = (PB)^2$
$\Rightarrow (-5 - h)^2 + (4 - k)^2 = (-1 - h)^2 + (6 - k)^2$
$\Rightarrow 25 + h^2+ 10h + 16 + k^2 - 8k = 1 + h^2 + 2h + 36 + k^2 - 12k$
$\Rightarrow 25 + 10h + 16 - 8k = 1 + 2h + 36 - 12k$
$\Rightarrow 8h + 4k + 4 = 0$
$\Rightarrow 2h + k + 1 = 0$
Mid-point $\text{AB}=\Big(\frac{-5-1}{2},\frac{4+6}{2}\Big)=(-3, 5)$
$\Big[\because\text{mid-point}=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
At point (-3, 5), from EQ. (i).
$\Rightarrow 2h + k = 2(-3) + 5$
$\Rightarrow 2h + k = -6 + 5$
$\Rightarrow 2h + k = -1$
$\Rightarrow 2h + k +1 = 0$
So, the mid-point of AB satisfy the Eq. (i). Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the points A and B. Replacing h, kby, x, y in above equation, we have 2 x + y + 1= 0.
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