Question

Find distance of point ( 3,-5) from line 3x-4y-26=0

Answer 1

Answer:

3/5

Step-by-step explanation:

Line is 3x - 4y - 26 = 0.

Comparing with Ax + By + C = 0, we get

A = 3, B = -4, C = -26.

Point is (3,-5).

x₁ = 3, y₁ = -5

The distance of the point P(x₁,y₁) to ax + by + c = 0 is

d = |Ax₁ + By₁ + C/√A² + B²|

  = |3(3) + (-4)(-5) + (-26)/√(3)² + (-4)²

  = |9 + 20 - 26|/√25|

  = 3/5

Hence, the distance = 3/5

Hope it helps you

#Bebrainly

Answer 2

$hello \: frnd$


line is :

→ 3x-4y-26=0.....( 1 )

comparing with Ax+ßy+c=0

A=3
ß=(-4)
c=(-26)

point is (3,-5)

x1=3,y1=(-5)



↪the distance of point P (x1,y1) to Ax+say+c=0 is⤵⤵


$d = |\frac{ax1 + by1 + c}{ \sqrt[]{a { }^{2} + b {}^{2} } } |$

$d = \frac{9 + 20 - 26}{25}$

$d = \frac{3}{5}$


distance is
$\frac{3}{5}$


$hope \: it \: helps$