find the length of the perpendicular from the point (3,-2) to the straight line 12x - 5y + 6 = 0?

Open in App

Solution

$D i s \tan c e o f a p o i n t (p, q) f r o m a l i n e a x + b y + c = 0 i s g i v e n b y = \frac{| a p + b q + c |}{\sqrt{(a^{2} + b^{2})}} P u t t i n g h e r e v a l u e s a = 12, b = - 5, c = 6, p = 3, q = - 2 T h u s d i s \tan c e = \frac{| (12 \times 3 + (- 5) \times (- 2) + 6) |}{\sqrt{(12^{2} + (- 5)^{2})}} = \frac{| 36 + 10 + 6 |}{\sqrt{144 + 25}} = \frac{52}{\sqrt{169}} = 52 / 13 = 4 u n i t s$

Suggest Corrections

Similar questions

(- 3, - 4)

12 (x + 6) = 5 (y - 2)

(- 1, 1)

12 (x + 6) = 5 (y - 2)

Find the equation of the line which passes through the point P (4, 2) and perpendicular to the line 5 – 12 – 9 =0.

The perpendicular distance from (1,2) to the straight line 12x+5y=7 is

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

Join BYJU'S Learning Program