Question

Write an equation of the line that is perpendicular to the line whose equation is 2y=3x+12 and passes through the origin

Answer 1

The equation of the line that is perpendicular to 2y = 3x + 12 and passes through the origin is y = (-2/3)x.

Step-by-step explanation:

We are looking for the equation of a line that is perpendicular to the given line 2y = 3x + 12 and also passes through the origin (0,0). First, we will find the slope of the given line by rewriting its equation in slope-intercept form, which is y = mx + b, where m is the slope.

The given equation is 2y = 3x + 12. Dividing by 2, we get y = (3/2)x + 6. Here, the slope of the given line is 3/2.

Since perpendicular lines have slopes that are negative reciprocals of each other, we take the negative reciprocal of 3/2, which is -2/3. The equation of the line we are looking for will have this slope.

Because this new line has to pass through the origin, its y-intercept is 0. Thus, the equation of our new line is y = (-2/3)x.

Answer 2

the slope of the line 2y=3x+12, is 3/2, because when you divide both sides of the equation, you get y=(3/2)x+6
the perpendicular line has a slope -2/3
so the equation is y=(-2/3)x. (because the line passes thought the origin, the y intercept is 0)
Answer is in bold.