On a coordinate plane, a line goes through (negative 3, negative 4) and (3, 0). What are the necessary criteria for a line to be perpendicular to the given line and have the same y-intercept? The slope is Negative three-halves and contains the point (0, 2). The slope is Negative two-thirds and contains the point (0, −2). The slope is Three-halves and contains the point (0, 2). The slope is Negative three-halves and contains the point (0, −2).

Therefore, we conclude that correct answer is: The slope is and contains the point .

Step-by-step explanation:

From Analytical Geometry we remember that slopes of two lines that are perpendicular to each other observe the following relationship:

(Eq. 1)

Where:

- Slope of the original line, dimensionless.

- Slope of the perpendicular line, dimensionless.

At first we must determine the slope of original line by definition of secant line:

(Eq. 2)

Where:

, - Initial and final independent variables, dimensionless.

, - Initial and final dependent variables, dimensionless.

If we know that and , the slope of original line is:

In addition, the standard form of original line is represented by this formula:

(Eq. 3)

Where is the y-Intercept of original line, dimensionless.

If we get that , , , then y-Intercept of the line is:

From (Eq. 1) we get that slope of perpendicular line is:

The slope of perpendicular line is .

And we procced to use the equation of perpendicular line in standard form is:

(Eq. 4)

If we know that , and , the value of is:

Therefore, we conclude that correct answer is: The slope is and contains the point .

The slope is Negative three-halves and contains the point (0, −2).

Step-by-step explanation:

Did it on edge 2020

step-by-step explanation:

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answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°