Question

Which point on the x-axis is equidistant from \[(7,6)\] and \[( - 3,4)\]
A. \[(2,0)\]
B. \[(3,0)\]
C. \[( - 5,0)\]
D. \[(1,0)\]

Answer 1

Hint: We are given two points and we have to find a point on the $x$ axis which is equidistant from these two points. We will take a point on \[x\] and assume its coordinates as \[(p,0)\] where \[p\] will be any variable. If the point lies on x-axis, then \[p\] lies on$x$-axis and y-coordinate will be zero. Then apply distance formula for both sets of points and equate the distances.
Formula used: Distance between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by the formula \[D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

Complete step-by-step solution:
We are given two points namely \[A(7,6)\] and \[B( - 3,4)\]. We have to find out a point on the x-axis which is equidistant from both the points \[A(7,6)\] and \[B( - 3,4)\]. Let this point be denoted as \[C(x,y)\].
Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words, we have \[y = 0\]
Then the point C will be \[C(x,0)\]
Now let us find out the distances from A and B from point C one by one.
Distance from A to C:
We have to calculate distance between \[A(7,6)\] and \[C(x,0)\]
Here \[{x_1} = x,{y_1} = 0;{x_2} = 7,{y_2} = 6\]
Substituting the values in the formula we get:
$ \Rightarrow AC = \sqrt {{{\left( {7 - x} \right)}^2} + {{\left( {6 - 0} \right)}^2}} $
$ \Rightarrow AC = \sqrt {{{\left( {7 - x} \right)}^2} + {{\left( 6 \right)}^2}} $ … (1)
Distance from B to C:
We have to calculate distance between \[B( - 3,4)\] and \[C(x,0)\]
Here \[{x_1} = x,{y_1} = 0;{x_2} = - 3,{y_2} = 4\]
Substituting the values in the formula we get:
$ \Rightarrow BC = \sqrt {{{\left( { - 3 - x} \right)}^2} + {{\left( {4 - 0} \right)}^2}} $
$ \Rightarrow BC = \sqrt {{{\left( { - 3 - x} \right)}^2} + {{\left( 4 \right)}^2}} $ … (2)
We know that both these distances are the same. So, equating both distances from equations (1) and (2)
$ \Rightarrow AC = BC$
Substitute the values of AC and BC from equations (1) and (2)
$ \Rightarrow \sqrt {{{\left( {7 - x} \right)}^2} + {{\left( 6 \right)}^2}} = \sqrt {{{\left( { - 3 - x} \right)}^2} + {{\left( 4 \right)}^2}} $
Squaring both the sides of the equation
$ \Rightarrow {\left( {7 - x} \right)^2} + {\left( 6 \right)^2} = {\left( { - 3 - x} \right)^2} + {\left( 4 \right)^2}$
Using the identities: \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]
$ \Rightarrow 49 + {x^2} - 14x + 36 = 9 + {x^2} + 6x + 16$
Cancel same terms from both sides of the equation
$ \Rightarrow 49 - 14x + 36 = 9 + 6x + 16$
Shift all constants to RHS and all terms with variable to LHS
$ \Rightarrow - 14x - 6x = 9 + 16 - 49 - 36$
$ \Rightarrow - 20x = - 60$
Cancel same factors i.e. -20 from both sides of the equation
$ \Rightarrow x = 3$
Hence, the coordinate of the point C is \[(3,0)\]
\[\therefore \] Option B. is the correct answer.

Note: Many students assume the Y-coordinate another variable making the question lengthy and difficult to solve. Always remember if the point is on x-axis then y-coordinate will be zero. If the point is on the y-axis then the x-coordinate will be zero. Remove the roots by squaring both the sides and then equate the values which makes cancellation easier.