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? When Harold wrote his equation, the point he used was (7, 3). When Harold wrote his equation, the point he used was (0, 7). When Harold wrote his equation, the point he used was (7, 0). When Harold wrote his equation, the point he used was (3, 7).

Answer 1

When Harold wrote his equation, the point he used was (7, 0).

EXPLANATION

The point-slope form is given as

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

The equation Harold wrote correctly is:

`y = 3(x - 7)`

This is the same as:

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

Comparing to point-slope form, we have

`x_1=7 \: \: and \: \: y_1=0`

Hence the point is (7,0)

When Harold wrote his equation, the point he used was (7, 0).

Answer 2

When Harold wrote his equation, the point he used was (7, 0) ⇒ the third answer

Explanation:

* Lets look to the equation to find the correct answer

- He used the point-slope form is y – y1 = m(x – x1), where m is the

slope of the line , (x1 , y1) are the coordinates of the point which

the line passes through it and (x , y) are the coordinates of any

general point on the line

- Lets solve the problem

- Harold correctly wrote the equation y = 3(x – 7)

∵ y - y1 = m (x - x1)

∵ y = 3 (x - 7)

- By comparing between the two equations

∴ y1 = 0

∴ m = 3

∴ x1 = 7

- He used the point (x1 , y1)

∴ Harold used the point (7 , 0) to write the equation

∴ The answer is when Harold wrote his equation, the point he used

was (7, 0)