Find the coordinates of a point on the line x + y = 5 which is nequidistant from (6, 4) and (5, 2) .

CBSE Class 10 Maths PYQ · Coordinate Geometry · Distance Formula · 2 Marks · March 2026 · Standard

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352 Marks · March 2026 · Standard
Find the coordinates of a point on the line $x + y = 5$ which is
nequidistant from $(6, 4)$ and $(5, 2)$.
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Let the required point be $(x, y)$ which is equidistant from $(6, 4)$ and $(5, 2)$
$\therefore (6 - x)^2 + (4 - y)^2 = (5 - x)^2 + (2 - y)^2$
$\Rightarrow 2x + 4y = 23$ --- (i) (1 Mark)
Since point $(x, y)$ also lies on the line $x + y = 5$
$\therefore x + y = 5$ --- (ii) (1/2 Mark)
Solving (i) and (ii), we get $x = -\frac{3}{2}$ and $y = \frac{13}{2}$ (1/2 Mark)
So, the coordinates of the required point are $(-\frac{3}{2}, \frac{13}{2})$
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