Question

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

Answer 1

Answer:

$y = (-3)\, x + 10$.

Step-by-step explanation:

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

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

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

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

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

$1 = (-3) \times 3 + c$.

$1 = -9 + c$.

$c = 10$.

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