Find a point on the x-axis, which is equidistant from the point (7, 6)

1.	Find a point on the x-axis, which is equidistant from the point (7, 6) and (3, 4).
Answer» Key points to solve the problem: Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) As, the point is on x-axis so y-coordinate is 0. Let the coordinate be (x,0) Given distance of (x,0) from (7,6) and (3,4) is same. ∴ using distance formula we have: \(\sqrt{(x-7)^2 + (0-6)^2}\) = \(\sqrt{(x - 3)^2 + (0 - 4)^2}\) Squaring both sides, we have: (x - 7)² + (0 - 6)² = (x -3)² + (0 - 4)² \(\Rightarrow\) x² + 49 - 14x + 36 = x² + 9 - 6x + 16 \(\Rightarrow\) 8x = 60 \(\Rightarrow\) x = \(\frac{60}{8} = \frac{15}{2} = 7.5\) ∴ point on x-axis is (7.5,0)

Find a point on the x-axis, which is equidistant from the point (7, 6) and (3, 4).

Answer»

Key points to solve the problem:

Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

As, the point is on x-axis so y-coordinate is 0.

Let the coordinate be (x,0)

Given distance of (x,0) from (7,6) and (3,4) is same.

∴ using distance formula we have:

\(\sqrt{(x-7)^2 + (0-6)^2}\) = \(\sqrt{(x - 3)^2 + (0 - 4)^2}\)

Squaring both sides,

we have:

(x - 7)² + (0 - 6)² = (x -3)² + (0 - 4)²

\(\Rightarrow\) x² + 49 - 14x + 36 = x² + 9 - 6x + 16

\(\Rightarrow\) 8x = 60 \(\Rightarrow\) x = \(\frac{60}{8} = \frac{15}{2} = 7.5\)

∴ point on x-axis is (7.5,0)

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