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

Solution:

The given equation of the line is 12 (x + 6) = 5(y – 2)

⇒ 12x + 72 = 5 y - 10

⇒ 12x - 5 y + 82 = 0 ....(1)

On comparing equation (1) with general  equation of a line Ax + By + C = 0 we obtain

A = 12, B = - 5 and C = 82

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point

(x\(_1\), y\(_1\)) is given by d = |Ax\(_1\) + By\(_1\) + c|/√A² + B²

The given point is (x\(_1\), y\(_1\)) = (- 1, 1)

Therefore, the distance of point (- 1, 1) from the given line is

|12(- 1) + (- 5)(1) + 82|/√12² + (- 5)²

= |- 12 - 5 + 82|/√169

= |65|/13

= 5

Hence, the distance of the point (- 1, 1) from the given line is 5 units

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NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 4

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

Summary:

The distance of the point (- 1, 1) from the line 12 (x + 6) = 5(y – 2) is 5 units