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kkurt [141]
3 years ago
13

Write the standard form of the equation of the line passing through (0, 1) and parallel to x = 4.

Mathematics
2 answers:
Olenka [21]3 years ago
8 0
I think a is the right answer
astra-53 [7]3 years ago
3 0

Answer:

Option C

x=0

Step-by-step explanation:

Given is a line x=4

We are to find out a parallel line passing through (0,1)

We know that x=4 is a vertical line parallel to y axis.

Hence any other line parallel to this would be of the form x=a for a suitable a.

Since the required line passes through (0,1)

the point should satisfy x=a

i.e. x=0 is the answer.

This is a vertical line i.e. y axis.

Option C) x=0

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5 0
3 years ago
Note that WXYZ has vertices W(-1, 2), X(-5, 7), y(-1, -2), and Z (3, -7).
svp [43]

a. Slope of WZ = -2.25; Slope of WX = 5

b. WZ = √97; WX = √41

c. WXYZ is not a <em>rectangle, rhombus, nor a square</em>. We can conclude that: <em>D. WXYZ is none of these</em>.

<h3>Slope of a Segment</h3>

Slope = change in y/change in x

Given:

W(-1, 2), X(-5, 7), Y(-1, -2), and Z (3, -7)

a. Slope of WZ and slope of WX:

Slope of WZ = (-7 - 2)/(3 -(-1)) = -2.25

Slope of WX = (7 - 2)/(-1 -(-1)) = 5

b. Use distance formula, d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, to find WZ and WX:

WZ = \sqrt{(3 - (-1))^2 + (-7 - 2)^2}\\\\\mathbf{WZ = \sqrt{97} }

WX = \sqrt{(-5 -(-1))^2 + (7 - 2)^2}\\\\\mathbf{WX = \sqrt{41} }

c. The quadrilateral WXYZ have adjacent sides that are not perpendicular to each other and have different slopes and different lengths, so therefore, WXYZ is not a rectangle, rhombus, nor a square. We can conclude that: <em>D. WXYZ is none of these</em>.

Learn more about slopes on:

brainly.com/question/3493733

5 0
3 years ago
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