Question

The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) using a point and the slope. Which point did Harold use?

Answer 1

Answer:

Point used by Harold was:

(7, 0)

Step-by-step explanation:

Given that

Equation of linear function used by Harold:

$y = 3(x - 7)$

We know that linear equation in point slope form can be represented as:

$y - y_1 = m(x - x_1)$

Where $(x_1,y_1)$ are the coordinates of a given point.

$m$ is the slope of line.

Formula for Slope, m is given as:

$m = \dfrac{y_2-y_1}{x_2-x_1}$

Where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points on the line.

If slope and a point with coordinates $(x_1,y_1)$ is know, the equation of a line can be represented in linear form as:

$y - y_1 = m(x - x_1)$ ....... (1)

Now, the given equation is:

$y = 3(x - 7)$

Re-writing the equation with a slight modification:

$y-0 = 3(x - 7)$

Now, comparing the above equation with equation (1):

We get that:

$x_1=7\\y_1=0$

So, the point used by Harold is (7, 0).