Question

find the distance from origin to the line 3x+4y+25=0​

Answer 1

Answer:

The point closest to the given line from the origin, will be the intersection of the perpendicular line passing through the origin.

3x - 4y = 25 can be written as y=3x/4 - 25/4.

The slope of this line is 3/4. So, the slope of the line perpendicular to the given line is - 4/3.

Since this perpendicular line passes through the origin, there will be no intercept on the y or x axes. So the equation of the perpendicular line is y = - 4x/3.

Solve the equations of the two lines given below, to find the point of intersection

y=3x/4 - 25/4 (given line)

y= - 4x/3.

Solving these two equations, gives you the nearest point from origin as (3, - 4).

We can also use derivates of a function to find the minimum value of the distance from origin (x^2 + y^2)^1/2, to get to the same answer.

Answer 2

Answer:

The distance of line 3x + 4y + 25 = 0 from origin is 5 units .

Step-by-step explanation:

• Let, the equation of the line be ax + by + c = 0. Let, the point be P (x , y).
• Distance of point P from line ax + by + c = 0 is

$d = \frac{ax + by + c}{\sqrt{ {a}^{2} + {b}^{2} } }$

• If the point P is origin i.e. P (0,0). Then, distance of line from origin is

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

Given that :

• Equation of line = 3x + 4y + 25 = 0

To find :

• Distance of line from origin.

Solution :

• The distance of line 3x + 4y + 25 = 0 from origin is

$d = \frac{25}{ \sqrt{ {3}^{2} + {4}^{2} } } \\d = \frac{25}{ \sqrt{9 + 16} } \\ d = \frac{25}{5} \\ d= 5 \: units$[/tex]

• Hence, distance of line 3x + 4y + 25 = 0 from origin is 5 units.