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Find the equation of the line:
a) with a gradient of ½ and a y-intercept of -2
b) with a gradient of 3 and passing through the point (-2:2)
c) passing through the points (1; -2) and (-2; 1)
d) parallel to the y-axis cutting the x-axis at -1



Answer :

reinhard10158

Step-by-step explanation:

gradient means slope.

the slope is anyways the factor a of x on the slope-intercept form

y = ax + b

and in the point-slope form

y - y1 = a(x - x1)

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

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

a)

y = (1/2)×x - 2

b)

y - 2 = 3(x - -2) = 3(x + 2) = 3x + 6

y = 3x + 8

c)

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

so, let's go from (1, -2) to (-2, 1)

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

y charges by +3 (from -2 to 1)

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

so, we pick one point, e.g. (1, -2), and the line is

y - -2 = -(x - 1) = -x + 1

y + 2 = -x + 1

y = -x - 1

d)

x = -1

this is a vertical line (parallel to the y-axis) representing all points on the grid for which x = -1.

the slope is undefined, as it is y/0 (because x never changes at all). and there is no y-intercept, as x can never be 0, and the line is parallel to the y-axis, so, there cannot be any intersection point.

semsee45

Answer:

[tex]\textsf{a) \quad $y=\dfrac{1}{2}x-2$}[/tex]

[tex]\textsf{b) \quad $y=3x+8$}[/tex]

[tex]\textsf{c) \quad $y=-x-1$}[/tex]

[tex]\textsf{d) \quad $x=-1$}[/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 = ¹/₂
  • y-intercept = -2

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

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

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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 = 3
  • (x₁, y₁) = (-2, 2)

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

[tex]\implies y-2=3(x-(-2))[/tex]

[tex]\implies y-2=3(x+2)[/tex]

[tex]\implies y-2=3x+6[/tex]

[tex]\implies y=3x+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) = (1, -2)
  • (x₂, y₂) = (-2, 1)

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

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

[tex]\implies y+2=-x+1[/tex]

[tex]\implies y=-x-1[/tex]

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Part (d)

If the line is parallel to the y-axis, it is a vertical line.

The equation of a vertical line is x = c.

Therefore, as the line intersects the x-axis at -1, the equation of the line is:

[tex]\implies x=-1[/tex]

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