On a coordinate plane, line M N goes through (negative 4, 0) and (4, 2). Point P is at (2, negative 4). Which point on the x-axis lies on the line that passes through point P and is perpendicular to line MN? (0, 1) (0, 4) (1, 0) (4, 0)

Answer:

C, (1,0)

Explanation:

A perpendicular line is when 2 lines cross each other with right angles, which are 90⁰ angles. Two lines are only perpendicular when the lines slopes are negative reciprocals to each other.

To find the slope of the first graph use, (y2-y1)/(x2-x1)

This is, 2-0/4+4

2/8

1/4 is the slope for the first line.

The negative reciprocal of this is -4.

We are also looking for a point on the x axis, so this eliminates answer choice A, and B, because they aren't even on the x axis.

When you test out answer choice c it will show that,

-4-0/2-1=-4/1 which is just -4.

These points on line p will pass each other when you add +4, y=-4x+4. So because of both of these lines negative reciprocating slopes, it shows that it will be answer C.

The point on the x-axis that is perpendicular to MN and passes through point P is: C. (1, 0).

The slope values of lines that are perpendicular to each other are negative reciprocals.

Slope of MN = rise / run = [tex]\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{4 - (-4)} = \frac{2}{8} = \frac{1}{4}[/tex]

Thus, the slope between the point on line MN and point P, making both lines perpendicular would be negative reciprocal of 1/4, which is -4.

Let (x1, y1) represent the point we are looking for.

(2, -4) = (x2, y2)

Slope = -4

Therefore:

[tex]\frac{-4 - y_1}{2 - x_1} = -4[/tex]

-4 - y1 = -4

-y1 = -4 + 4

-y1 = 0

y1 = 0

2 - x1 = 1

-x1 = 1 - 2

-x1 = -1

x1 = 1

The coordinate of the point is: C. (1, 0).

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