Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find a Point on the X-axis, Which is Equidistant from the Points (7, 6) and (3, 4). - Mathematics

Sum

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

#### Solution

Let C(x, 0) be a point on the x-axis, which is equidistant from the points A(7, 6) and B(3, 4).
$\therefore$ AC = BC
$\Rightarrow A C^2 = B C^2$

$\Rightarrow \left( 7 - x \right)^2 + \left( 6 - 0 \right)^2 = \left( 3 - x \right)^2 + \left( 4 - 0 \right)^2$
$\Rightarrow 49 + x^2 - 14x + 36 = 9 + x^2 - 6x + 16$
$\Rightarrow 85 - 14x = 25 - 6x$
$\Rightarrow 60 = 8x$
$\Rightarrow \frac{15}{2} = x$
Thus, the point on the x-axis, which is equidistant from the points (7, 6) and (3, 4) is $\left( \frac{15}{2}, 0 \right)$

Concept: Brief Review of Cartesian System of Rectanglar Co-ordinates
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 22 Brief review of cartesian system of rectangular co-ordinates
Exercise 22.1 | Q 8 | Page 13