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denis-greek [22]
3 years ago
11

PLEASE HELP! urgent! Write the equation of a line in a slope-intercept form and standard form

Mathematics
1 answer:
m_a_m_a [10]3 years ago
7 0

Answer:

y = \frac{3}{4}x -3 --- Slope intercept

4y - 3x =- 12 --- Standard form

Step-by-step explanation:

Given

The attached linear graph

Required

The equation

Select any two points on the line

(x_1,y_1) = (0,-3)

(x_2,y_2) = (4,0)

Next, calculate the slope (m)

m = \frac{y_2 - y_1}{x_2 - x_1}

So, we have:

m = \frac{0 --3}{4 - 0}

m = \frac{3}{4}

The slope intercept is then calculated using:

y = m(x - x_1) + y_1

So, we have:

y = \frac{3}{4}(x - 0) -3

y = \frac{3}{4}(x) -3

y = \frac{3}{4}x -3

Multiply through bu 4

4y = 3x - 12

Subtract 3x from both sides

4y - 3x =- 12 --- Standard form

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slader (a) Find parametric equations for the line through (4, 1, 8) that is perpendicular to the plane x − y + 4z = 2. (Use the
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Answer:

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

xy - plane    (x, y, z) = (2, -1, 0)

yz - plane    (x, y, z) = (0, 5, -8)

xz - plane     (x, y, z) = (5, 0, 12)

Step-by-step explanation:

The given point (x, y ,z) = (4, 1, 8)

The plane x -y + 4z = 2

Normal vector (n) = < 1, -1, 4 >

The equation of line through point (4, 1, 8) and the plane is:

(x(t), y(t), z(t)) = (4, 1, 8) + t(1, -1, 4)

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

Any point on the line P(x, y, z) = ( 4 + t, 1 - t, 8 + 4t)

xy-Pane ⇒ z = 0

8 + 4t = 0

4t = - 8

t = -8/4

t = -2

∴

(x, y, z) = (4 - 2, 1 - 2, 8 + 4(-2))

(x, y, z) = (2, -1, 0)

yz-plane ⇒ x = 0

4 + t = 0

t = -4

∴

(x, y, z) = (4 + (-4) , 1-(-4), 8 + 4(-4)

(x, y, z) = (0, 5, -8)

xz-plane ⇒ y = 0

1 - t = 0

-t = -1

t = 1

∴

(x, y, z) = ( 4 + 1, 1 - 1, 8 + 4(1) )

(x, y, z) = (5, 0, 12)

6 0
3 years ago
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