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

(Q 25B – 30/3/3 – CBSE 2026 Question Paper)

Ans:

Let the coordinates of the this point be (x, y)

Step 1: Since this point lies on equation, it will satisfy the line equation

∴ x + y = 5 …….. (i)

Step 2: Since its equidistant from points (6, 4) and (5, 2);

∴ Distance between (x, y) and (6, 4) = Distance between (x, y) and (5, 2)

Since the distance between (x₁, y₁) and (x₂, y₂) is given by:

D =

∴

By squaring on both sides, we get:

(6 – x) ² + (4 – y) ² = ( 5 – x) ² + ( 2 – y) ² 

By applying albebraic identity of (a – b) ² = a ² + b ² – 2 a b, we get:

(36 + x ² – 12 x + 16 + y ² – 8 y) = (25 + x ² – 10 x + 4 + y ² – 4 y)

(36 +  – 12 x + 16 + – 8 y) = (25 + – 10 x + 4 + – 4 y)

∴ 52 – 12 x – 8 y = 29 – 10 x – 4 y

∴ (29 – 10 x – 4 y) – (52 – 12 x – 8 y) = 0

∴ 29 – 10 x – 4 y – 52 + 12 x + 8y = 0

∴ 2 x + 4 y = 23 …… (ii)

Step 3: Let’s multiply equation (i) by 2 and subtract equation (ii):

∴ 2 x + 4 y – 2 (x + y) = 23 – 2 (5)

∴ 2 x + 4 y – 2 x – 2 y = 23 – 10

∴ + 4 y – – 2 y = 13

∴ 2 y = 13

∴ y =

Step 4: By substituting value of y in equation (i), we get:

x + y = 5

∴ x = 5 – y = 5 –

∴ x =

Therefore, the coordinates of the point are .

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