Question

Find the perpendicular distance from the point (-3,4) to the straight line 5x-12y=2​

Answer 1

The correct answer is 5.

Given: The point =  (-3,4).

Equation of line = 5x-12y=2​.

To Find: The perpendicular distance of point from the line.

Solution:

Perpendicular distance(d) of point (p, q) from ax + by + c = 0.

d = $\frac{|pa+qb+c|}{\sqrt{a^{2} +b^{2} } }$

For point (-3, 4) and line 5x-12y=2.

d = $\frac{|(-3)5+4(-12)-2|}{\sqrt{5^{2} +(-12)^{2} } }$

d = $\frac{|-15-48-2|}{\sqrt{25 +144 } }$

d = $\frac{|65|}{\sqrt{169 } }$

d = $\frac{65}{13}$

d = 5

Hence, the distance of point (-3, 4) from 5x-12y=2​ is 5.

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Answer 2

The answer  is $\frac{31}{13 }} }$ or 2.3846

Given: 5x - 12y = 2 is straight line and the point (-3,4)

To find: The perpendicular distance between given straight line to the point

Solution: 5x - 12y = 2 is straight line

⇒ 5x - 12y - 2 = 0

The formula to find the perpendicular distance between a line and point is given by  

Perpendicular Distance =  $|\frac{ax+by+c}{\sqrt{a^{2}+b^{2} } } } |$

here the point (x, y) = (-3, 4)

⇒ Perpendicular Distance  =  $|\frac{5(-3)+12(4)-2}{\sqrt{5^{2}+12^{2} } } } |$  

⇒  $|\frac{-15+48 -2}{\sqrt{25+144 } } } |$

⇒ $|\frac{31}{\sqrt{169 } } } |$        

⇒ $\frac{31}{13 }} }$= 2.3846

The perpendicular distance is $\frac{31}{13 }} }$ or 2.3846

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