Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y - 2) ?

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:

Here, we have to find the distance of the line 12(x + 6) = 5(y -2) from the point (- 1, 1)

The given equation of line can be re-written as: 12x - 5y + 82 = 0

Let P = (-1, 1)

⇒ x1 = - 1 and y1 = 1

Here, a = 12 and b = - 5

Now substitute  x1 = - 1 and y1 = 1 in the equation 12x - 5y + 82 = 0, we get

⇒ |12⋅ x1 - 5 ⋅ y1 + 82| = |- 12 - 5 + 82| = 65

⇒ \(\sqrt{a^2+b^2} = \sqrt{(12)^2 + (- 5)^2} = 13\)

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

⇒ \(d = \left| {\frac{{12\cdot x_1- 5 \cdot y_1 + 82 }}{{\sqrt {{(12)^2} + {(- 5)^2}} }}} \right| = \frac{65}{13} = 5\)

Hence, option C is the correct answer.