Mathematics — Standard (041) — CBSE Class 10 board question

Let the point be $(x, y)$ on line $x + y = 5$, so $y = 5 - x$.

Since it is equidistant from A(6, 4) and B(5, 2):

$(x-6)^2 + (y-4)^2 = (x-5)^2 + (y-2)^2$

Substituting $y = 5 - x$:

$(x-6)^2 + (1-x)^2 = (x-5)^2 + (3-x)^2$

$x^2 - 12x + 36 + x^2 - 2x + 1 = x^2 - 10x + 25 + x^2 - 6x + 9$

$-14x + 37 = -16x + 34$

$2x = -3 \Rightarrow x = -\dfrac{3}{2}$, $\quad y = 5 - \left(-\dfrac{3}{2}\right) = \dfrac{13}{2}$

The required point is $\left(-\dfrac{3}{2},\ \dfrac{13}{2}\right)$.

Source: Distance Formula, Chapter 7

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• Let the point be $(x, y)$ and immediately use $x + y = 5$ to reduce to one variable.
• Set the two distance-squared expressions equal (avoids square roots).
• Examiners award marks for: correct setup of distance equation (1 mark) + correct coordinates (1 mark).
• Always verify: $-\frac{3}{2} + \frac{13}{2} = \frac{10}{2} = 5$ ✓ (lies on the line).