How do you write an equation with slope of 5/3 and contains the point (-6, -2)?

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How do you write an equation with slope of 5/3 and contains the point (-6, -2)?

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Answer 1 · 2015-09-24T03:05:18

$y = \frac{5}{3} x + 8$ $y = 5/3x+8$

Explanation:

To do this, we use a linear equation called point slope form. This is basically another way of writing a linear equation, like $y = m x + b$ $y = mx+b$ . Point slope form is as follows: $y - y_{1} = m (x - x_{1})$ $y-y_1 = m(x-x_1)$ . I won't go into the specifics of what this equation is or how it's derived, but I encourage you to do so. In this equation, $y_{1}$ $y_1$ and $x_{1}$ $x_1$ are points on the line $y$ $y$ and $m$ $m$ is the slope.

Here, we already have the elements: points on the line, and the slope. To solve, we just substitute these values into the equation and simplify:
$y - (- 2) = (\frac{5}{3}) (x - (- 6))$ $y-(-2) = (5/3)(x-(-6))$ ; $x_{1} = - 6$ $x_1 = -6$ , $y_{1} = - 2$ $y_1 = -2$ , $m = \frac{5}{3}$ $m = 5/3$
$y + 2 = \frac{5}{3} (x + 6)$ $y+2 = 5/3(x+6)$
$y + 2 = \frac{5}{3} x + 10$ $y+2 = 5/3x+10$
$y = \frac{5}{3} x + 10 - 2$ $y = 5/3x+10-2$
$y = \frac{5}{3} x + 8$ $y = 5/3x+8$
And there you have it - the equation of the line with slope 5/3 and passing through the point (-6,-2).

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How do you write an equation with slope of 5/3 and contains the point (-6, -2)?

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