Solveeit Logo
Login

Question

Question: Find a point on the x axis which is equidistant from the point \(A(2,-5)\) and \(B(-2,9)\)....

Find a point on the x axis which is equidistant from the point A(2,−5)A(2,-5) and B(−2,9)B(-2,9).

Explanation

Solution

In this question, we are given two points A and B and we have to find point P on x axis which is equidistant from A and B. For this, we will suppose the point on x axis and then use distance formula between to find distance AP and AB and then equate them to find value of variable supposed to be the coordinate of P.

Distance formula between two points, P(x1,y1)P(x_1,y_1) and Q(x2,y2)Q(x_2,y_2) is given by PQ=(x2−x1)2+(y2−y1)2PQ= \sqrt{{(x_2-x_1)}^2 + {(y_2-y_1)}^2 }

Complete step-by-step answer:
The given points are A(2,-5) and B(-2,9).
Since, we have to find point P on x axis therefore, y coordinate of point will be zero and let x coordinate be equal to 'a'. Hence, point P becomes (a,0).
Now, we have to find the value of a.
As per question, point P is equidistant from A and B. Therefore, AP = AB.
Let us calculate the distance between A and P.
Coordinates of A are (2,-5) and P are (a,0).

Applying distance formula, that is the Distance between two points, P(x1,y1)P(x_1,y_1) and Q(x2,y2)Q(x_2,y_2) is given by PQ=(x2−x1)2+(y2−y1)2PQ= \sqrt{{(x_2-x_1)}^2 + {(y_2-y_1)}^2 }

& AP=\sqrt{{{\left( a-2 \right)}^{2}}+{{\left( 0-\left( -5 \right) \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( a-2 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{a}^{2}}+4-4a+25} \\\ & \Rightarrow \sqrt{{{a}^{2}}-4a+29} \\\ \end{aligned}$$ Now, let us calculate the distance between B and P. Coordinates of B are (-2,9) and P are (a,0). Applying distance formula, we get: $$\begin{aligned} & BP=\sqrt{{{\left( a-\left( -2 \right) \right)}^{2}}+{{\left( 0-9 \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( a+2 \right)}^{2}}+{{\left( -9 \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{a}^{2}}+4+4a+81} \\\ & \Rightarrow \sqrt{{{a}^{2}}+4a+85} \\\ \end{aligned}$$ As we know, AP = BP therefore, putting value of both we get: $$\sqrt{{{a}^{2}}-4a+29}=\sqrt{{{a}^{2}}+4a+85}$$ Squaring both sides we get: $${{a}^{2}}-4a+29={{a}^{2}}+4a+85$$ Taking variable on one side and constant on other side we get: $$\begin{aligned} & {{a}^{2}}-{{a}^{2}}-4a-4a=85-29 \\\ & \Rightarrow -8a=56 \\\ \end{aligned}$$ Dividing both sides by -8, we get: $$a=-7$$ **Since the -7 was x coordinate of point P. Therefore, point P becomes (-7,0) which lies on the x axis.** **Note:** Students should note that, for points of x axis, y coordinate is always zero and for points of y axis, x coordinate is always zero. Students should carefully apply the distance formula between two points. Students should not try to solve square roots when calculating distance AP and BP as they will be cancelled automatically when we take square of the terms while equating them.