Question

Find the equation of straight lines which are perpendicular to the line 12x + 5y = 17 and at a distance of 2 units from point v(-4,1 )

Answer 1

5x - 12y + 58 = 0, 5x - 12y + 6 = 0

Step-by-step explanation:

The given straight line is

12x + 5y = 17 ..... (1)

Let the equation of the straight line perpendicular to (1) be

5x - 12y + k = 0 ..... (2)

∴ the distance of the straight line (2) from the point V (- 4, 1) is

= {| 5 (- 4) - 12 (1) + k |}/√{5² + (- 12)²} units

= (| - 20 - 12 + k |)/13 units

= (| k - 32 |)/13 units

By the given condition,

(| k - 32 |)/13 = 2

or, | k - 32 | = 26

or, k - 32 = ± 26

or, k = 58, 6

Therefore the required straight lines are

5x - 12y + 58 = 0 and

5x - 12y + 6 = 0.

Answer 2

The equations of straight lines which are perpendicular to the line

12x + 5y = 17 are  5x - 12y + 58 = 0, 5x - 12y + 6 = 0

• Given line

                12x + 5y = 17  ---------(1)

• Equation of line perpendicular to (1) is given by

                5x - 12y + c = 0 -----(2)

• Distance of V(-4, 1) to (2) = 2

               $\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2} }} =2$

                 a = 5, b = -12, c = c

                 $x_{1}=-4,y_{1}=1$

                $\frac{|5(-4)-12(1)+c|}{\sqrt{5^{2}+(-12)^{2} }} =2$

                $\frac{|-32+c|}{\sqrt{169} }} =2$

                 $\frac{|-32+c|}{13 }} =2$

                 |-32 + c| = 26

                 -32 + c = ± 26

Case (i):                   Case (ii):

-32 + c = 26          -32 + c = -26

c = 26 + 32            c = - 26 + 32

c = 58                     c = 6

• Substituting 'c' in (2), we get

            5x - 12y + 58 = 0, 5x - 12y + 6 = 0