find the equation of the line through the origin and perpendiclar to the line with equation x+3y=5?

Answer 1

Bring the equation to the form y = mx + c, where m is the slope,
y = -1/3(x) + 5/3
slope of the line = -1/3

slope of the perpendicular = -1/(slope of the line) = 3

y - 0 = 3(x - 0)
y = 3x ---------> equation of the perpendicular.

Answer 2

Starting with the general format of y = mx + b, you know the line is through the origin so b=0 and you're left with y = mx. Now it's just a matter of finding the slope. Remember that the slope of line perpendicular to a given line is the negative reciprocal of the slope. Since x + 3y = 5 means y = (-1/3)x + (5/3), the slope of this line is -1/3, so the slope we're looking for is 3. This gives us y = 3x.

Answer 3

first of all you need to make the line in the form of
y=mx+c
3y=5-x
y=(5/3)-(x/3)
the slope of the line is -1/3
therefore the perpendicular line will have a slope of 3
y=3x

Answer 4

x + 3y = 5

3y = -x + 5

y = -x/3 + 5/3

The slope is -1/3

The slope of a perpendicular line is -1/(-1/3) = 3

y=3x