How do you write the equation in point slope form given ( 1/2 , -1 ) , ( -2/3 , 6 )?

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How do you write the equation in point slope form given ( 1/2 , -1 ) , ( -2/3 , 6 )?

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Answer 1 · 2017-06-30T15:21:22

$y + 1 = - 6 (x - \frac{1}{2})$ $y+1=-6(x-1/2)$

Explanation:

First, you need to know what the point-slope form of a line is. That is usually expressed as

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

This form of the equation of a line is the final goal. We need to find both the slope, $m$ $m$ , and the point the line passes through at $(x_{1}, y_{1})$ $(x_1,y_1)$ .

To find the slope, $m$ $m$ , you need to have the slope formula.

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m=(y_2-y_1)/(x_2-x_1)$

Plugging in the points you were given, we get

$m = \frac{6 - (- 1)}{(- \frac{2}{3}) - \frac{1}{2}}$ $m=(6-(-1))/((-2/3)-1/2)$

$m = \frac{6 + 1}{- \frac{4}{6} - \frac{3}{6}}$ $m=(6+1)/(-4/6-3/6)$

$m = \frac{7}{- \frac{7}{6}}$ $m=(7)/(-7/6)$

$m = 7 \times (- \frac{6}{7}) = - 6$ $m=7xx(-6/7)=-6$

The first point you were given was $(\frac{1}{2}, - 1)$ $(1/2,-1)$ , so you can plug this in for $(x_{1}, y_{1})$ $(x_1,y_1)$ in the point-slope equation above.

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

$y - (- 1) = - 6 (x - \frac{1}{2})$ $y-(-1)=-6(x-1/2)$

$y + 1 = - 6 (x - \frac{1}{2})$ $y+1=-6(x-1/2)$

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How do you write the equation in point slope form given ( 1/2 , -1 ) , ( -2/3 , 6 )?

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