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jolli1 [7]
3 years ago
10

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) us

ing a point and the slope. Which point did Harold use?
Mathematics
1 answer:
babunello [35]3 years ago
6 0

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 <em>(7, 0)</em>.

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