Question

The point on the curve y = `sqrt(x - 1)`, where the tangent is perpendicular to the line 2x + y – 5 = 0 is ______.

Options

• (2, –1)

• (10, 3)

• (2, 1)

• (5, –2)

Answer 1

The point on the curve y = `sqrt(x - 1)`, where the tangent is perpendicular to the line 2x + y – 5 = 0 is (2, 1).

Explanation:

Suppose that slope of the curve y = `sqrt(x - 1)` is m₁

∴ m₁ = `dy/dx = d/dxsqrt(x - 1) 1/(2sqrt(x - 1)`

slope of the line 2x + y – 5 = 0 is m₂

∴ m₂ = `dy/dx = d/dx(5 - 2x)` = –2

As, lines are perpendicular if m₁m₂ = –1

So, `1/(2sqrt(x - 1)).(-2)` = –1

`\implies sqrt(x - 1)` = 1

`\implies` x – 1

`\implies` x = 2

After substituting x = 2 in y = `sqrt(x - 1)`, we get y = 1

Hence, the required point is (2, 1).