Find the distance of the point (3, -5) from the line 3x- 4y -26 = 0
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The shortest distance between the point (3,-5) and the like 3x-4y-26=0 will be the line perpendicular to given line.
length of perpendicular = ax'+by'+c/ |a^2+b^2|
Here ;
a =3
b= -4
c= -26
x' = 3
y'= -5
Length = |(3×3) + (-4×-5) + (-26)}|/ √(3^2+(-4)^2)
= |9 +29 -26| / √9+16
= 12/5
therefore the distance will be 12/5 units.
hope it helped.. ...!☺
length of perpendicular = ax'+by'+c/ |a^2+b^2|
Here ;
a =3
b= -4
c= -26
x' = 3
y'= -5
Length = |(3×3) + (-4×-5) + (-26)}|/ √(3^2+(-4)^2)
= |9 +29 -26| / √9+16
= 12/5
therefore the distance will be 12/5 units.
hope it helped.. ...!☺
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