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3) Find the equation of the line:
a) with a gradient of 2 and cutting the y-axis at 7
b) with a gradient of -2 and passing through the point (2;4)
c) passing through the points (2; 3) and (-1; 2)
d) parallel to the x-axis cutting the y-axis at 5



Answer :

reinhard10158

Step-by-step explanation:

this is very much doing the exact same things as the previous question, just with a little bit different numbers.

remember, gradient = slope.

the slope is always the factor of x in the slope-intercept form

y = ax + b

our in the point-slope form

y - y1 = a(x - x1)

"a" is the slope, b is the y-intercept (the y- value when x = 0).

(x1, y1) is a point on the line.

the slope is the ratio (y coordinate change / x coordinate change) when going from one point on the line to another.

a)

y = 2x + 7

b)

y - 4 = -2(x - 2) = -2x + 4

y = -2x + 8

c)

going from (2, 3) to (-1, 2)

x changes by -3 (from 2 to -1)

y charges by -1 (from 3 to 2)

the slope is -1/-3 = 1/3

we use one of the points, e.g. (2, 3)

y - 3 = (1/3)×(x - 2) = x/3 - 2/3

y = x/3 - 2/3 + 3 = x/3 - 2/3 + 9/3 = x/3 + 7/3

d)

y = 5

this is a horizontal line (parallel to the x-axis) and represents every point on the grid, for which y = 5.

the slope is 0/x = 0, as y never changes at all.

the y- intercept is 5, of course.

semsee45

Answer:

[tex]\textsf{a) \quad $y=2x+7$}[/tex]

[tex]\textsf{b) \quad $y=-2x+8$}[/tex]

[tex]\textsf{c) \quad $y=\dfrac{1}{3}x+\dfrac{7}{3}$}[/tex]

[tex]\textsf{d) \quad $y=5$}[/tex]

Step-by-step explanation:

Part (a)

Slope-intercept form of a linear equation:

[tex]y=mx+b[/tex]

where:

  • m is the slope.
  • b is the y-intercept.

Given values:

  • Slope = 2
  • y-intercept = 7

Substitute the given values into the formula to create the equation of the line:

[tex]\implies y=2x+7[/tex]

---------------------------------------------------------------------------

Part (b)

Point-slope form of a linear equation:  

[tex]y-y_1=m(x-x_1)[/tex]

where:

  • m is the slope.
  • (x₁, y₁) is a point on the line.

Given:

  • Slope = -2
  • (x₁, y₁) = (2, 4)

Substitute the given values into the formula to create the equation of the line:

[tex]\implies y-4=-2(x-2)[/tex]

[tex]\implies y-4=-2x+4[/tex]

[tex]\implies y=-2x+8[/tex]

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Part (c)

Slope formula:

[tex]\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

where (x₁, y₁) and (x₂, y₂) are points on the line.

Given points:

  • (x₁, y₁) = (2, 3)
  • (x₂, y₂) = (-1, 2)

Substitute the points into the slope formula to calculate the slope of the line:

[tex]\implies m=\dfrac{2-3}{-1-2}=\dfrac{-1}{-3}=\dfrac{1}{3}[/tex]

Substitute the found slope and one of the points into the point-slope formula to create the equation of the line:

[tex]\implies y-y_1=m(x-x_1)[/tex]

[tex]\implies y-3=\dfrac{1}{3}(x-2)[/tex]

[tex]\implies y-3=\dfrac{1}{3}x-\dfrac{2}{3}[/tex]

[tex]\implies y=\dfrac{1}{3}x+\dfrac{7}{3}[/tex]

---------------------------------------------------------------------------

Part (d)

Slope-intercept form of a linear equation:

[tex]y=mx+b[/tex]

where:

  • m is the slope.
  • b is the y-intercept.

If the line is parallel to the x-axis, its slope is zero.

If the line intersects the y-axis at y = 5, then its y-intercept is 5.

Therefore:

  • m = 0
  • b = 5

Substitute the given values into the formula to create the equation of the line:

[tex]\implies y=0x + 5[/tex]

[tex]\implies y=5[/tex]

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