Question

Find the values of k for which the length of the perpendicular from the point (4, 1) on the line 3x - 4y + k = 0 is 2 units ?
1. 18
2. - 2
3. - 18
4. None of these
5. 12

Answer 1

Correct Answer - Option 3 : - 18

CONCEPT:

The perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by \(d = \left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\)

CALCULATION:

Given: The length of the perpendicular from the point (4, 1) on the line 3x - 4y + k = 0 is 2 units.

Let P = (4, 1)

Here x₁ = 4, y₁ = 1, a = 3, b = - 4 and d = 2

Now substitute x1 = 4 and y1 = 1 in the equation 3x - 4y + k = 0

⇒ |3⋅ x1 - 4 ⋅ y1 + k| = |12 - 4 + k| = |8 + k|

⇒ \(\sqrt{a^2+b^2} = \sqrt{3^2 + (- 4)^2} = 5\)

As we know that, the perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by \(d = \left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\)

⇒ \(d = \left| {\frac{{3\cdot x_1- 4 \cdot y_1 + k }}{{\sqrt {{3^2} + {(- 4)^2}} }}} \right| = 2\)

⇒ |8 + k| = 10

⇒ k = 2 or - 18

Hence, option C is the correct answer.