Question

Find the coordinates of the point equidistant from the points A (-2, -3), B (-1, 0) and C (7, -6)
A.(3, -3)
B.(-3, 3)
C.(-2, -3)
D.(-3, -3)

Answer 1

Hint: In this question remember to use that \[{d_1} = {d_2} = {d_3}\] and use the concept of distance formula between two points with coordinates (x, y) and (${x_1},{y_1}$) i.e. \[d = \sqrt {{{\left( {x - {x_1}} \right)}^2} + {{\left( {y - {y_1}} \right)}^2}} \] use these steps to find the solution of the given problem.

Complete step-by-step answer:
Let consider the point Q having coordinates (x, y) which is equidistant from the points A, B and C
Since we know that the distance of point Q from every point will be equal so if the distance of Q from A is equal to ${d_1}$ and distance from point B and C are ${d_2}$ and ${d_3}$ respectively
Therefore by the given information we can say that \[{d_1} = {d_2} = {d_3}\]
To find the distance from point P to point A let’s use the distance formula i.e. \[d = \sqrt {{{\left( {x - {x_1}} \right)}^2} + {{\left( {y - {y_1}} \right)}^2}} \]since we know the values of ${x_1},{y_1}$ is equal to (-2, -3)
Substituting the given values in the distance formula
\[{d_1} = \sqrt {{{\left( {x - \left( { - 2} \right)} \right)}^2} + {{\left( {y - \left( { - 3} \right)} \right)}^2}} \]
$ \Rightarrow $\[{d_1} = \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} \]
$ \Rightarrow $\[{d_1} = \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} \]
Now using the distance formula for Q and B here ${x_1},{y_1}$ is equal to (-1, 0)
\[{d_2} = \sqrt {{{\left( {x - \left( { - 1} \right)} \right)}^2} + {{\left( {y - \left( 0 \right)} \right)}^2}} \]
$ \Rightarrow $\[{d_2} = \sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} \]
For distance from point Q and C using the distance formula here ${x_1},{y_1}$is equal to (7, -6)
\[{d_3} = \sqrt {{{\left( {x - 7} \right)}^2} + {{\left( {y - \left( { - 6} \right)} \right)}^2}} \]
$ \Rightarrow $\[{d_3} = \sqrt {{{\left( {x - 7} \right)}^2} + {{\left( {y + 6} \right)}^2}} \]
By the given information in the question we can say that \[{d_1} = {d_2} = {d_3}\]
Comparing ${d_1}$ and ${d_2}$
\[\sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} = \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} \]
Squaring on both the sides
\[{\left( {\sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} } \right)^2} = {\left( {\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} } \right)^2}\]
$ \Rightarrow $\[{\left( {x + 1} \right)^2} + {y^2} = {\left( {x + 2} \right)^2} + {\left( {y + 3} \right)^2}\]
$ \Rightarrow $\[{x^2} + 2x + 1 + {y^2} = {x^2} + 4x + 4 + {y^2} + 6y + 9\]
$ \Rightarrow $\[2x + 6y + 12 = 0\]
$ \Rightarrow $\[x + 3y + 6 = 0\] (Equation 1)
Now comparing the ${d_2}$ and ${d_3}$
\[\sqrt {{{\left( {x - 7} \right)}^2} + {{\left( {y + 6} \right)}^2}} = \sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} \]
Squaring both sides
\[{\left( {\sqrt {{{\left( {x - 7} \right)}^2} + {{\left( {y + 6} \right)}^2}} } \right)^2} = {\left( {\sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} } \right)^2}\]
$ \Rightarrow $\[{\left( {x - 7} \right)^2} + {\left( {y + 6} \right)^2} = {\left( {x + 1} \right)^2} + {y^2}\]
$ \Rightarrow $\[{x^2} - 14{x} + 49 + {y^2} + 12y + 36 = {x^2} + 2x + 1 + {y^2}\]
$ \Rightarrow $\[16x - 12y = 84\]
$ \Rightarrow $\[4x - 3y = 21\]
$ \Rightarrow $\[4x - 3y - 21 = 0\] (Equation 2)
Now adding equation 1 and equation 2
\[x + 3y + 6 + 4x - 3y - 21 = 0\]
$ \Rightarrow $\[5x - 15 = 0\]
$ \Rightarrow $\[x = \dfrac{{15}}{5}\]
$ \Rightarrow $\[x = 3\]
Substituting the value of x in equation 1
\[3 + 3y + 6 = 0\]
$ \Rightarrow $\[y = - 3\]
So the coordinates of point Q are (3, -3)
Hence, option A is the correct option.

Note: In the above question we came across the term coordinates which can be explain as the couple of numbers showing the positions on the graph lying on the positions axis like X axis Y axis and Z axis, coordinates also help in determining the dimensions like 1D, 2D, 3D , etc. in which the point, body or any object lies.