How do you write an equation in point slope form given slope 2/3, (5,6)?

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How do you write an equation in point slope form given slope 2/3, (5,6)?

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Answer 1 · 2016-09-03T02:06:13

The linear equation in point slope form is $y - 6 = \frac{2}{3} (x - 5)$ $y-6=2/3(x-5)$ .

Explanation:

The general equation for a line in point slope form is $y - y_{1} = m (x - x_{1})$ $y-y_1=m(x-x_1)$ , where $m = slope = \frac{2}{3}$ $m="slope"=2/3"$ , $x_{1} = 5$ $x_1=5$ , and $y_{1} = 6$ $y_1=6$ .

To get the point slope form for the variables given, plug the known values into the equation.

$y - 6 = \frac{2}{3} (x - 5)$ $y-6=2/3(x-5)$

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Now, if you wish, you can determine the slope intercept form for easier graphing, by solving the point slope form for $y$ $y$ . This will give the x and y interceptss, where the x-intercept is the value of $x$ $x$ when $y = 0$ $y=0$ , and the y-intercept is the value of $y$ $y$ when $x = 0$ $x=0$ . Once you have the x and y intercepts, you only need to plot two points to graph the line.

The general equation for the slope intercept form is $y = m x + b$ $y=mx+b$ , where $m$ $m$ is the slope, $(\frac{2}{3})$ $(2/3)$ , and $b$ $b$ is the y-intercept.

Start with the point slope form and solve for $y$ $y$ .

$y - 6 = \frac{2}{3} (x - 5)$ $y-6=2/3(x-5)$

Add $6$ $6$ to both sides.

$y = \frac{2}{3} (x - 5) + 6$ $y=2/3(x-5)+6$

Simplify.

$y = \frac{2}{3} x - \frac{10}{3} + 6$ $y=2/3x-10/3+6$

Multiply $6$ $6$ by $\frac{3}{3}$ $3/3$ to get $\frac{18}{3}$ $18/3$ .

Simplify.

$y = \frac{2}{3} x - \frac{10}{3} + \frac{18}{3}$ $y=2/3x-10/3+18/3$

Simplify.
The slope intercept form is $y = \frac{2}{3} x + \frac{8}{3}$ $y=2/3x+8/3$ , where the slope, $m$ $m$ , is $\frac{2}{3}$ $2/3$ , and the y-intercept is $\frac{8}{3}$ $8/3$ and the point is $(0, \frac{8}{3})$ $(0,8/3)$ .

To find the x-intercept , make $y = 0$ $y=0$ and solve for $x$ $x$ .

$0 = \frac{2}{3} x + \frac{8}{3}$ $0=2/3x+8/3$

$- \frac{2}{3} x = \frac{8}{3}$ $-2/3x=8/3$

$- 6 x = 24$ $-6x=24$

$x = \frac{24}{- 6}$ $x=24/(-6)$

$x = - 4$ $x=-4$

The x-intercept is $- 4$ $-4$ and the point is $(- 4, 0)$ $(-4,0)$ .

graph{y=2/3x+8/3 [-10, 10, -5, 5]}

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How do you write an equation in point slope form given slope 2/3, (5,6)?

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