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

Asked by Abhisek | 11 months ago |  61

Solution :-

Let us consider (a, 0) be the point on the x-axis that is equidistant from the point (7, 6) and (3, 4).

So,

$$\sqrt{(7 - a)^2 + (6-0)^2}$$

$$\sqrt{(3 - a)^2 + (4-0)^2}$$

$$\sqrt{49 + a^2-14a+36 }$$

$$\sqrt{9 + a^2-6a+16 }$$

$$\sqrt{ a^2-14a+85 }$$

$$\sqrt{ a^2-6a+25 }$$

Now, let us square on both the sides we get,

a2 – 14a + 85 = a2 – 6a + 25

-8a = -60

a = $$\dfrac{60}{8}$$

$$\dfrac{15}{2}$$

The required point is ($$\dfrac{15}{2}$$, 0)

Answered by Pragya Singh | 11 months ago

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