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Question

What is the distance between the points (4, 3) and (3, -2)?

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is \(\sqrt {26} \)

Finding the Distance Between Two Points

To find the distance between two points in a coordinate plane, we use the distance formula. This formula is derived from the Pythagorean theorem and is essential in coordinate geometry.

Understanding the Distance Formula

The distance formula states that the distance \(d\) between two points with coordinates \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

Here, \(x_1\) and \(y_1\) are the coordinates of the first point, and \(x_2\) and \(y_2\) are the coordinates of the second point.

Applying the Formula to the Given Points

We are given two points: \((4, 3)\) and \((3, -2)\).

Let the first point be \((x_1, y_1) = (4, 3)\).

Let the second point be \((x_2, y_2) = (3, -2)\).

Now, we substitute these values into the distance formula:

\[d = \sqrt{(3 - 4)^2 + (-2 - 3)^2}\]

Next, we perform the subtractions inside the parentheses:

  • \(3 - 4 = -1\)
  • \(-2 - 3 = -5\)

Substitute these results back into the formula:

\[d = \sqrt{(-1)^2 + (-5)^2}\]

Now, square the results:

  • \((-1)^2 = 1\)
  • \((-5)^2 = 25\)

Substitute the squared values back into the formula:

\[d = \sqrt{1 + 25}\]

Finally, add the numbers under the square root:

\[d = \sqrt{26}\]

Conclusion

The distance between the points \((4, 3)\) and \((3, -2)\) is \(\sqrt{26}\).

Checking the Options

Let's compare our calculated distance with the given options:

  • Option 1: \(\sqrt{26}\)
  • Option 2: \(\sqrt{24}\)
  • Option 3: 6
  • Option 4: 5

Our calculated distance, \(\sqrt{26}\), matches Option 1.

Revision Table: Distance Formula Key Points

Concept Description Formula
Distance Formula Used to find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a coordinate plane. \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Pythagorean Theorem Connection The formula is derived from applying the Pythagorean theorem to a right triangle formed by the two points and their coordinate differences. \(a^2 + b^2 = c^2\)

Additional Information: Visualizing Distance

You can visualize the distance between points \((4, 3)\) and \((3, -2)\) on a coordinate plane. Imagine moving from \((4, 3)\) to \((3, -2)\). The horizontal change is \(3 - 4 = -1\) (1 unit to the left), and the vertical change is \(-2 - 3 = -5\) (5 units down). These changes form the legs of a right triangle with lengths 1 and 5. The distance between the points is the hypotenuse of this triangle. Using the Pythagorean theorem (\(1^2 + 5^2 = d^2\)), we get \(1 + 25 = d^2\), so \(d^2 = 26\), and \(d = \sqrt{26}\). This confirms the distance formula result.

The distance is always a non-negative value, representing a length.

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Similar Questions

  1. Find the slope of the line joining the points (3, -4) and (5, 2).

  2. The graph of 2x = 5 - 3y cuts the x-axis at the point P (α, β) . The value of (2α + β) is:

  3. Find the coordinates of the midpoint of the segment joining the points (-4, 7) and (2, 3).

  4. Find the slope of the line given by the equation 4x + 6y = 9.

  5. Reflection of point (-2, -6) on the Y-axis is:

  6. Reflection of the point (2, 3) on the X-axis is:


Important Questions from Co-ordinate Geometry

  1. In which ratio the point (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3)?

  2. The area (in sq. units) of the triangle formed by the graphs of 8x + 3y = 24, 2x + 8 = y and the x-axis is:

  3. In which quadrant both abscissa and ordinate are negative?

  4. Find the slope of the line joining the points (3, -4) and (5, 2).

  5. Find the value of K for which equation x – Ky = 2, 3x + 2y = 5 has unique solution.

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