The point on the line 4x + 3y = 5, which is equidistant from (1, 2) an

1.	The point on the line 4x + 3y = 5, which is equidistant from (1, 2) and (3, 4), is(a) (7,-4) (b) (-10,15)(c) (1/7, 8/7)(d) (0, 5/4)
Answer» Correct option: (b) (-10,15) Explanation: Let the point (x₁, y₁) be on the line 4x + 3y = 5. ∴ 4x₁+ 3y₁ = 5 ...(i) Distance of point (x₁, y₁) from (1, 2) is (x₁ –1)² + (y₁ - 2)² Distance of point (x₁, y₁) from (3, 4) is (x₁ - 3)³ + (y₁ - 4)² It is given, point (x₁, y₁) is equidistant from (1, 2) and (3, 4) (x₁ –1)² + (y₁ - 2)² = (x₁ - 3)³ + (y₁ - 4)² => x² + 1 - 2x₁+ y² + 4 - 4y₁ = x² + 9 - 6x₁ + y₁² + 16 - 8y₁ => 4x₁ + 4y₁ = 20 => x₁+ y₁ = 5 ...(ii) From Eqs. (i) and (ii), we get y₁ =15 and x₁ = -10

The point on the line 4x + 3y = 5, which is equidistant from (1, 2) and (3, 4), is(a) (7,-4) (b) (-10,15)(c) (1/7, 8/7)(d) (0, 5/4)

Answer»

Correct option: (b) (-10,15)

Explanation:

Let the point (x₁, y₁) be on the line 4x + 3y = 5.

∴ 4x₁+ 3y₁ = 5 ...(i)

Distance of point (x₁, y₁) from (1, 2) is (x₁ –1)² + (y₁ - 2)²

Distance of point (x₁, y₁) from (3, 4) is (x₁ - 3)³ + (y₁ - 4)²

It is given, point (x₁, y₁) is equidistant from (1, 2) and (3, 4)

(x₁ –1)² + (y₁ - 2)² = (x₁ - 3)³ + (y₁ - 4)²

=> x² + 1 - 2x₁+ y² + 4 - 4y₁ = x² + 9 - 6x₁ + y₁² + 16 - 8y₁

=> 4x₁ + 4y₁ = 20

=> x₁+ y₁ = 5 ...(ii)

From Eqs. (i) and (ii), we get

y₁ =15 and x₁ = -10

Discussion

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