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What are the necessary criteria for a line to be perpendicular to the given line and have the same y-intercept?

A. The slope is [tex]\(-\frac{3}{2}\)[/tex] and contains the point [tex]\((0, 2)\)[/tex].
B. The slope is [tex]\(-\frac{2}{3}\)[/tex] and contains the point [tex]\((0, -2)\)[/tex].
C. The slope is [tex]\(\frac{3}{2}\)[/tex] and contains the point [tex]\((0, 2)\)[/tex].
D. The slope is [tex]\(-\frac{3}{2}\)[/tex] and contains the point [tex]\((0, -2)\)[/tex].

Sagot :

To determine the necessary criteria for a line to be perpendicular to a given line and have the same [tex]\( y \)[/tex]-intercept, we need to follow these steps:

1. Understand the criteria for perpendicularity:
- Two lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. Therefore, if we have a line with slope [tex]\( m \)[/tex], the slope of the line that is perpendicular to it will be the negative reciprocal of [tex]\( m \)[/tex].

2. Understand the role of [tex]\( y \)[/tex]-intercept:
- The [tex]\( y \)[/tex]-intercept is the point where the line crosses the [tex]\( y \)[/tex]-axis. For two lines to intersect at a given point on the [tex]\( y \)[/tex]-axis, they must have the same [tex]\( y \)[/tex]-intercept.

Let's analyze the given question with this understanding:

### Slope [tex]\(-\frac{3}{2}\)[/tex] and contains the point [tex]\((0, 2)\)[/tex]:

- Given slope: [tex]\( -\frac{3}{2} \)[/tex]
- To find the perpendicular slope, we take the negative reciprocal: [tex]\( \frac{2}{3} \)[/tex]
- Given point [tex]\((0, 2)\)[/tex] provides the [tex]\( y \)[/tex]-intercept directly, which is 2.

Thus, the line with slope [tex]\( \frac{2}{3} \)[/tex] passing through [tex]\((0, 2)\)[/tex] is described by [tex]\( y = \frac{2}{3}x + 2 \)[/tex].

### Slope [tex]\(-\frac{2}{3}\)[/tex] and contains the point [tex]\((0, -2)\)[/tex]:

- Given slope: [tex]\( -\frac{2}{3} \)[/tex]
- The slope of the perpendicular line is the negative reciprocal: [tex]\( \frac{3}{2} \)[/tex]
- Given point [tex]\((0, -2)\)[/tex] provides the [tex]\( y \)[/tex]-intercept directly, which is -2.

The line with slope [tex]\( \frac{3}{2} \)[/tex] passing through [tex]\((0, -2)\)[/tex] is described by [tex]\( y = \frac{3}{2}x - 2 \)[/tex].

### Slope [tex]\(\frac{3}{2}\)[/tex] and contains the point [tex]\((0, 2)\)[/tex]:

- Given slope: [tex]\( \frac{3}{2} \)[/tex]
- The slope of the perpendicular line is the negative reciprocal: [tex]\( -\frac{2}{3} \)[/tex]
- Given point [tex]\((0, 2)\)[/tex] provides the [tex]\( y \)[/tex]-intercept directly, which is 2.

The line with slope [tex]\( -\frac{2}{3} \)[/tex] passing through [tex]\((0, 2)\)[/tex] is described by [tex]\( y = -\frac{2}{3}x + 2 \)[/tex].

### Slope [tex]\(-\frac{3}{2}\)[/tex] and contains the point [tex]\((0, -2)\)[/tex]:

- Given slope: [tex]\( -\frac{3}{2} \)[/tex]
- The slope of the perpendicular line is the negative reciprocal: [tex]\( \frac{2}{3} \)[/tex]
- Given point [tex]\((0, -2)\)[/tex] provides the [tex]\( y \)[/tex]-intercept directly, which is -2.

The line with slope [tex]\( \frac{2}{3} \)[/tex] passing through [tex]\((0, -2)\)[/tex] is described by [tex]\( y = \frac{2}{3}x - 2 \)[/tex].

### Conclusion

The criteria for a line to be perpendicular to another and have the same [tex]\( y \)[/tex]-intercept are:

- The slope of the perpendicular line must be the negative reciprocal of the given line's slope.
- The [tex]\( y \)[/tex]-intercept (as given in the point) must remain the same for both lines.

The solution for the given initial slope of [tex]\(-\frac{3}{2}\)[/tex] and having the same [tex]\( y \)[/tex]-intercept (0) is that the perpendicular line's slope must be [tex]\( 0.6666666666666666 \)[/tex] and the [tex]\( y \)[/tex]-intercept must remain 0. This means the perpendicular line’s equation would be [tex]\( y = 0.66666...x + 0 \)[/tex].

Thus, only options with a corresponding slope of [tex]\(0.6666666666666666\)[/tex] and y-intercept [tex]\(0\)[/tex] satisfy these conditions.