NCERT Class X Chapter 7: Coordinate Geometry Example 5

NCERT Class X Chapter 7: Coordinate Geometry Example 5

Question:

Find a point on the y-axis which is equidistant from the points A(6, 5) and B(–4, 3).

Given:

  • Point \(A(6, 5)\)
  • Point \(B(-4, 3)\)

To Find:

A point on the y-axis which is equidistant from \(A\) and \(B\).

Formula:

Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Solution:

Step 1: Let the required point on the y-axis be \(P(0, y)\).

Step 2: Since \(P\) is equidistant from \(A\) and \(B\), we have:

$$ PA = PB $$ $$ \sqrt{(6 - 0)^2 + (5 - y)^2} = \sqrt{(-4 - 0)^2 + (3 - y)^2} $$

Step 3: Squaring both sides to remove the square roots:

$$ (6)^2 + (5 - y)^2 = (-4)^2 + (3 - y)^2 $$ $$ 36 + (5 - y)^2 = 16 + (3 - y)^2 $$

Step 4: Expand both sides:

$$ 36 + [25 - 10y + y^2] = 16 + [9 - 6y + y^2] $$ $$ (36 + 25 - 10y + y^2) = (16 + 9 - 6y + y^2) $$ $$ (61 - 10y + y^2) = (25 - 6y + y^2) $$

Step 5: Subtract \(y^2\) from both sides and solve for \(y\):

$$ 61 - 10y = 25 - 6y $$ $$ 61 - 25 = 10y - 6y $$ $$ 36 = 4y $$ $$ y = \frac{36}{4} = 9 $$

Step 6: Therefore, the required point is \((0, 9)\).

Result:

The point on the y-axis which is equidistant from \(A(6, 5)\) and \(B(-4, 3)\) is \((0, 9)\).

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