Question:

The locus of a point which is equidistant from the points $(1,1)$ and $(3, 3)$ is

Updated On: Jun 8, 2024
  • $y = x + 4$
  • $x + y = 4$
  • $x = 2$
  • $y = 2$
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The Correct Option is B

Solution and Explanation

Let $P(h, k)$ be a point, which is equidistant from the points $A(1,1)$ and $B(3,3)$.
i.e., $ PA=PB $
$\Rightarrow\, (P A)^{2}=(P B)^{2}$
$\Rightarrow\,(h-1)^{2}+(k-1)^{2}=(h-3)^{2}+(k-3)^{2}$ (by distance formula)
$\Rightarrow 1-2 h+1-2 k=9-6 h+9-6 k$
$\Rightarrow\, 4 h+4 k=16$
$\Rightarrow\, h+k=4$
So, required locus is $x+y=4$.
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Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c