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Find the equation of a line that passes through the point (3,1) and has a gradient of -3. Leave your answer in the form y = m x + c

Sagot :

jacob193

Answer:

[tex]y = (-3)\, x + 10[/tex].

Step-by-step explanation:

The question requires that the equation should be in the slope-intercept form [tex]y = m \, x + c[/tex]. The [tex]m[/tex] and [tex]c[/tex] in this equation are constants. In particular:

  • [tex]m\![/tex] would represent the slope of this line (also known as the gradient of this line,) whereas
  • [tex]c[/tex] would be the [tex]y[/tex]-intercept of this line.

The question states that the slope of this line is [tex](-3)[/tex]. Therefore, [tex]m = -3[/tex]. Thus, the equation for this line should be in the form [tex]y = (-3)\, x + c[/tex].

The value of constant [tex]c[/tex] could be found using the other piece of information: the point [tex](3,\, 1)[/tex] is in this line. In other words, the solution [tex]\text{$x = 3$ and $y = 1$}[/tex] should satisfy the equation [tex]y = (-3)\, x + c[/tex].

Substitute [tex]\text{$x = 3$ and $y = 1$}[/tex] into the equation [tex]y = (-3)\, x + c[/tex] and solve for [tex]c[/tex]:

[tex]1 = (-3) \times 3 + c[/tex].

[tex]1 = -9 + c[/tex].

[tex]c = 10[/tex].

Hence, [tex]y = (-3) \, x + 10[/tex] would be the slope-intercept form of the equation of this line.

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