All Exams Test series for 1 year @ ₹349 only
Question

Find a point on the X axis which is equidistant from A(2, -3) and B(-2, 1)

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
P(1, 0)

Point on X-axis Equidistant Calculation

The objective is to locate a specific point situated on the X-axis. This point must satisfy the condition of being equidistant from two given points, namely A(2, -3) and B(-2, 1).

Characteristics of a Point on the X-axis

A fundamental property of any point lying on the X-axis is that its y-coordinate is always zero. Consequently, we can represent the unknown point as P(x, 0).

Equidistant Condition Explained

The core requirement is that the distance from P to A is equal to the distance from P to B. This is expressed as $PA = PB$. Squaring both sides simplifies the calculation, giving us $PA^2 = PB^2$.

Applying the Distance Formula

The distance formula calculates the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ as $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Calculating the squared distance $PA^2$:

$PA^2 = (x - 2)^2 + (0 - (-3))^2$ $PA^2 = (x - 2)^2 + (3)^2$ $PA^2 = (x - 2)^2 + 9$

Calculating the squared distance $PB^2$:

$PB^2 = (x - (-2))^2 + (0 - 1)^2$ $PB^2 = (x + 2)^2 + (-1)^2$ $PB^2 = (x + 2)^2 + 1$

Solving the Equation for x

Equating the squared distances $PA^2$ and $PB^2$:

$(x - 2)^2 + 9 = (x + 2)^2 + 1$

Expanding the squared terms:

$(x^2 - 4x + 4) + 9 = (x^2 + 4x + 4) + 1$

Simplifying both sides:

$x^2 - 4x + 13 = x^2 + 4x + 5$

Subtract $x^2$ from both sides to eliminate it:

$-4x + 13 = 4x + 5$

Rearranging the terms to isolate x:

$13 - 5 = 4x + 4x$ $8 = 8x$

Solving for x:

$x = \frac{8}{8}$ $x = 1$

Final Point Coordinates

We determined that $x = 1$. Since the point P lies on the X-axis, its coordinates are P(x, 0). Substituting the value of x, we get P(1, 0).

Thus, the point P(1, 0) is equidistant from points A(2, -3) and B(-2, 1).

Was this answer helpful?

Similar Questions

  1. \(A(1, 2)\), \(B(a, -26)\), \(C(13, -14)\) and \(D(6, 14)\) are the vertices of a parallelogram, taken in order, find the value of \(a\).
  2. What would be the area of the triangle with \(A(8, 8)\), \(B(22, 8)\) and \(C(16, -10)\) as vertices?
  3. What would be the point which divides the line segment connecting the points \(P(10, 18), Q(5, 8)\) in the ratio \(2 : 3\) internally
  4. What is the midpoint of the line segment joining \((-2, -1)\) and \((-5, -3)\)?
  5. If point C is equidistant from A(5, -6) and B(7, 8), coordinate of C will be
  6. The area of the triangle whose vertices are given by (2, 4), (– 3, – 1) and (5, 3) is:

Important Questions from Coordinate Geometry

  1. Find the co-ordinates of the centroid of a triangle whose vertices are A(1, 4), B(7, 8) and C(10, 12).

  2. If x² + y² - 16x + 38y + 425 = 0, then the value of x² + y² is:
  3. If x² + y² - 12x + 18y + 117 = 0, then the value of x² + y² is:
  4. A geological survey team has established three research stations forming a triangular monitoring zone. The stations are located at coordinates P(2,5), Q(8,1), and R(4,9) on a coordinate grid where each unit represents 1 kilometer. What is the area of the triangular monitoring zone?
  5. \(A(1, 2)\), \(B(a, -26)\), \(C(13, -14)\) and \(D(6, 14)\) are the vertices of a parallelogram, taken in order, find the value of \(a\).
Need Expert Advice?
Upcoming Exams
RRB Technician
October 06, 2026
RRB JE
October 27, 2026
RRB ALP
November 03, 2026
Test Series
RRB ALP img
Railways
RRB ALP 2026 Mock Test series
1035 Tests 1 Tests Free
1085 Attempts
4.3(239)
English, Hindi

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App