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:

Harold used the point (7,0)

Step-by-step explanation:

Formula used:

Slope-point form of straight line is

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

Given equation is

y = 3 (x - 7)

But it can be written as

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

comparing this equation with

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

$(x_1,y_1)=(7,0)\:and\:m=3$

Therefore, Harold used the point (7, 0)

Answer 2

Harold used the point (7, 0).

Step-by-step explanation:

Given : The equation of a linear function in point-slope form $y - y_1= m(x-x_1)$. Harold correctly wrote the equation y = 3(x – 7) using a point and the slope.

To find : Which point did Harold use?

Solution :

Slope-point form of straight line $y - y_1= m(x-x_1)$.

Re-write the equation $y = 3 (x - 7)$ as,

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

Comparing this equation with slope form, we get

$x_1=7, y_1=0$

Therefore, Harold used the point (7, 0).

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