Question

Find the distance of the line 2x+3y-5=0 from the origin.​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

In 2x + 3y + 5 = 0,

the constant 5 should be taken to the RHS of the equation, that is 2x + 3y = -5

Dividing both sides with √(a2  + b2) = √(22 + 32) = √13, we get  To make the RHS positive,

we multiply both sides with (−1).

Thus, the normal form is where cos α = -2/√13  and sinα = -3√13

Now, the distance of the line from the orgin is

p=c/root a^2+b^2= -5 by root 13 =5 by root 13

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