NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 8

NCERT Class X Chapter 7: Coordinate Geometry

Question:

Find the values of \( y \) for which the distance between the points \( P(2, -3) \) and \( Q(10, y) \) is 10 units.

Given:

Points: \( P(2, -3) \), \( Q(10, y) \)
Distance \( PQ = 10 \) units

To Find:

The values of \( y \) for which the distance between \( P \) and \( Q \) is 10 units.

Formula:

Distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Solution:

Step 1: Assign the coordinates to the points.

\[ P(x_1, y_1) = (2, -3), \quad Q(x_2, y_2) = (10, y) \]

Step 2: Write the distance formula for \( PQ \) and substitute the values.

\[ PQ = \sqrt{(10 - 2)^2 + (y - (-3))^2} \] \[ 10 = \sqrt{8^2 + (y + 3)^2} \]

Step 3: Simplify inside the square root.

\[ 10 = \sqrt{64 + (y + 3)^2} \]

Step 4: Square both sides to remove the square root.

\[ (10)^2 = 64 + (y + 3)^2 \] \[ 100 = 64 + (y + 3)^2 \]

Step 5: Rearrange to solve for \( (y + 3)^2 \).

\[ (y + 3)^2 = 100 - 64 = 36 \]

Step 6: Take square roots on both sides.

\[ y + 3 = \pm 6 \]

Step 7: Solve for \( y \) for both cases.

Case 1: \[ y + 3 = 6 \implies y = 6 - 3 = 3 \] Case 2: \[ y + 3 = -6 \implies y = -6 - 3 = -9 \]

Result:

The values of \( y \) are 3 and -9.

Result:

The distance between the points (2, 3) and (4, 1) is \(2\sqrt{2}\).

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