How do you write an equation of a line with point (2,5), slope -2?

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Answer 1 · 2018-05-19T04:27:32

both answers:

$y - 5 = - 2 (x - 2)$ $y−5=-2(x−2)$

and

$y = - 2 x + 9$ $y=-2x +9$

are correct.

Point slope form of a line is the standard equation:

$y - y_{1} = m (x - x_{1})$ $y−y_1=m(x−x_1)$

where $x_{1}$ $x_1$ and $y_{1}$ $y_1$ are a point the line intersects and $m$ $m$ in the slope, so your line:

$y - 5 = - 2 (x - 2)$ $y−5=-2(x−2)$ is your line.

you can convert it to slope intercept form:

$y = - 2 x + 9$ $y=-2x +9$

so the slope $m = - 2$ $m =-2$ and the y-intercept is 9

graph{y=-2x +9 [-17.04, 22.96, -5.36, 14.64]}

Answer 2 · 2018-05-19T04:29:52

$y = - 2 x + 9$ $y=-2x+9$

We must assume that the line is a straight line.

The equation of a straight line in slope/point form is:

$(y - y_{1}) = m (x - x_{1})$ $(y-y_1) = m(x-x_1)$

Where the line has a slope $m$ $m$ and passes through the point $(x_{1}, y_{1})$ $(x_1,y_1)$

Here we have a line of slope $- 2$ $-2$ passing through the point $(2, 5)$ $(2,5)$

$:. (y-5) = -2(x-2)$

$y = -2x+4+5$

$y=-2x+9$
is the equation of the line in slope/intercept form.