Question

Write an equation that is perpendicular to and 3x+y=3 whose y-intercept 5. Anyone please help me, no link just explain how to do it.

Answer 1

`y = (1)/(3)x + 5`

Explanation:

By definition, two lines are perpendicular if and only if their slopes are negative reciprocals of each other:
`m = - (1)/(m_(2) )`, or equivalently,
`m_(1) * m_(2) = -1`.

Given our linear equation 3x + y = 3 (or y = -3x + 3):

We can find the equation of the line (with a y-intercept of 5) that is perpendicular to y = -3x + 3 by determining the negative reciprocal of its slope, -3, which is
`(1)/(3)`.

To test whether this is correct, we can take first slope,
`m_(1) = -3`, and multiply it with the negative reciprocal slope
`m_(2) = (1)/(3)` :

`m_(1) * m_(2) = -1`

`-3 * (1)/(3) = -1`

Therefore, we came up with the correct slope for the other line, which is
`(1)/(3)`.

Finally, the y-intercept is given by (0, 5). Therefore, the equation of the line that is perpendicular to 3x + y = 3 is:

`y = (1)/(3)x + 5`