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

Answer:

$y = \frac{1}{3}x + 5$

Step-by-step explanation:

By definition, two lines are perpendicular if and only if their slopes are negative reciprocals of each other:  $m = - \frac{1}{m_{2} }$, 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 $\frac{1}{3}$.

To test whether this is correct, we can take first slope,  $m_{1} = -3$,  and multiply it with the negative reciprocal slope $m_{2} = \frac{1}{3}$ :

$m_{1} * m_{2} = -1$

$-3 * \frac{1}{3} = -1$

Therefore, we came up with the correct slope for the other line, which is  $\frac{1}{3}$.

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

$y = \frac{1}{3}x + 5$