Question

find the distance off the point (3,-5) from the line 3x-4y-5=0​

Answer 1

Step-by-step explanation:

As we know that distance (d) of a point (h,k) from a line ax+by+c=0 is given as-

d=

a

2

+b

2

∣ah+bk+c∣

Therefore,

Distanvce of the point (3,−5) from the line 3x−7y−4=0 is-

d=

3

2

+(−7)

2

∣3(3)−7(−5)−4∣

⇒d=

9+49

∣9+35−4∣

⇒d=

58

40

⇒d=

29

20

58

units