How do you write an equation in slope intercept form given m=1/3, and contains (-3,-15)?

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How do you write an equation in slope intercept form given m=1/3, and contains (-3,-15)?

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Answer 1 · 2018-04-03T06:44:37

-14 .Check below

Explanation:

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

Where,

$y_{1} = - 15$ $y_1 = -15$
$x_{1} = - 3$ $x_1 = -3$
$m = \frac{1}{3}$ $m =1/3$

Next the equation is :
$y + 15 = \frac{1}{3} (x + 3)$ $y+15=1/3(x+3)$
$3 y + 45 = x + 3$ $3y+45=x+3$
Thus,
$y = \frac{x - 42}{3}$ $y=(x-42)/3$
or u can write
$y = \frac{1}{3} x - 14$ $y=1/3 x - 14$

Now u can use the equation of straight line which is
$Y = m X + c$ $Y=mX+c$

compare the above equation with your result and the y-intercept is -14

Answer 2 · 2018-04-03T06:52:49

$y = \frac{1}{3} x - 14$ $y =1/3x -14$

Explanation:

$m$ $m$ is the slope/gradient of the line

$y = m x + c$ $y =mx + c$ is the general equation of the straight line

So we know $y = \frac{1}{3} x + c$ $y =1/3x +c$ and to find the value of $c$ $c$ we substitute the numbers in the coordinate that the line passes through

$y = \frac{1}{3} x + c$ $y =1/3x + c$ (-3,-15)

$- 15 = \frac{1}{3}$ $-15 =1/3$ $\times$ $xx$ $- 3 + c$ $-3 + c$

$- 15 = - 1 + c$ $-15 =-1 +c$

$c = - 14$ $c = -14$

So the equation is $y = \frac{1}{3} x - 14$ $y =1/3x -14$

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How do you write an equation in slope intercept form given m=1/3, and contains (-3,-15)?

2 Answers

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