Question

Perpendicular distance from the origin to the line 3x+4y+10 = 0.

Answer 1

Answer:

$x1 = 0 \: \: \: \: y1 = 0 \: (from \: origin) \\ perpendicular \: distance\: is \: \\ d = \frac{ |ax1 + by1 + c| }{ \sqrt{ { a}^{2} + {b}^{2} } } \\ = \frac{ |3(0) + 4(0) + 10| }{ \sqrt{ {3}^{2} + {4}^{2 } } } \\ = \frac{10}{ \sqrt{25} } \\ = \frac{10}{5} \\ = 2 \\ (d)$

Answer 2

Answer:

The distance of line 3x + 4y + 10 = 0 from origin is 2 units .

Step-by-step explanation:

• The perpendicular distance of a point (x1, y1) from the line ax + by + c = 0 is

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

• If the point P(x1, y1) is origin, then x1 = 0 and y1 = 0.
• The perpendicular distance of line from origin is

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

Given that :

• Equation of line = 3x + 4y + 10 = 0

Solution :

• Perpendicular distance of line 3x + 4y + 10 = 0 from origin is given by :
• Here, a = 3 ; b = 4 and c = 10

$d = | \frac{10}{ \sqrt{ {3}^{2} + {4}^{2} } } | \\ d= | \frac{10}{ \sqrt{9 + 16} } | \\d = | \frac{10}{ \sqrt{25} } | \\ d = | \frac{10}{5} | \\d = 2 \: units$

• Hence, the distance of line 3x + 4y + 10 = 0 from origin is 2 units .