Question

The point-slope form of the equation of a line that passes through points (8,4) and (0, 2) is y-

Answer 1

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

While the point-slope equation of a line is given by:

$y-y_ {0} = m (x-x_ {0})$

Where:

m: It's the slope

$(x_ {0}, y_ {0}):$It is a point through which the line passes

In this case we have a line through:

(8,4) and (0,2)

Therefore, its slope is:

$m = \frac {2-4} {0-8} = \frac {-2} {- 8} = \frac {1} {4}$

Its point-slope equation is:

$y-4 = \frac {1} {4} (x-8)$

Then, we manipulate the expression to find the equation of the slope-intersection form:

$y-4 = \frac {1} {4} x- \frac {8} {4}\\y-4 = \frac {1} {4} x-2\\y = \frac {1} {4} x-2 + 4\\y = \frac {1} {4} x + 2$

Therefore, the cut-off point with the y-axis is $b = 2$

ANswer:

$y = \frac {1} {4} x + 2$