Question

find the distance of origin from the line 3x+4y+15=0​

Answer 1

Step-by-step explanation:

Use intercepts form for this type of questions

$3x + 4y + 15 = 0$

$3x + 4y = - 15$

$\frac{3x}{ - 15} + \frac{4y}{ - 15} = 1$

$\frac{x}{ \frac{ - 15}{3} } + \frac{y}{ \frac{ - 15}{4} } = 1$

HERE X-INTERCEPT IS (-5)

Y-INTERCEPT IS (-15/4)

THE DISTANCE FROM ORIGIN IS

$distance = \frac{ |c| }{ \sqrt{ {a}^{2} + {b}^{2} } } = \frac{15}{5} = 3$