Question

Equation of the line perpendicular to the line 3x-y+9=0 and passing through origin is

Answer 1

Answer:

Equation of the line perpendicular to the line $3x-y+9=0$ and passing through origin is $-y-3x=0$

Step-by-step explanation:

The equation of a line that is perpendicular to another line with the equation $ax+by+c=0$ is $by-ax+\alpha =0$

where $\alpha$ is a constant, that can be calculated when passing points are given

From the question, equation of the perpendicular line is $3x-y+9=0$

as compared with the first equation we get $a=3, b=-1$

now the equation of the given line is $-y-3x+\alpha =0$

also, the line is passing through the origin so the point is $(x,y)=(0,0)$

substituting these values

$0+\alpha =0\\\alpha =0$

therefore the equation of the given line is $-y-3x=0$

Answer 2

Step-by-step explanation:

Lines perpendicular to

ax + by + c = 0

are the lines

-bx + ay + d = 0.

So lines perpendicular to 3x - y + 9 = 0 are

x + 3y + d = 0.

Since the point (0,0) is to be on the line, d = O.

So the line is:

x + 3y = 0