Question #bcad8

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Question #bcad8

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Answer 1 · 2018-01-02T19:16:14

$y = \frac{1}{10} x - \frac{13}{5}$ $y=1/10x-13/5$

Explanation:

First find the gradient by doing change in y/change in x such as...

m(gradient)= $\frac{- 3 - (- 4)}{- 4 - (- 14)}$ $(-3-(-4))/(-4-(-14))$
= $\frac{1}{10}$ $1/10$

equation is y= $\frac{1}{10} x + b$ $1/10x+b$

to find b you will need to substitute one of the points to the equation and solve for b
such as...

(-4,-3)
(x,y)

$= - 3 = \frac{1}{10} (- 4) + b$ $=-3=1/10(-4)+b$
$= - 3 = - \frac{2}{5} + b$ $=-3=-2/5+b$
$= - \frac{13}{5} = b$ $=-13/5=b$

equation is $y = \frac{1}{10} x - \frac{13}{5}$ $y=1/10x-13/5$

you can also use $y - y 1 = m (x - x 1)$ $y-y1=m(x-x1)$ (point-slope formula).

Answer 2 · 2018-01-02T19:23:07

$y = \frac{1}{10} x - \frac{13}{5}$ $y = \frac{1}{10}x-\frac{13}{5}$

Explanation:

Given two points $(x_{1}, y_{1})$ $(x_1, y_1)$ and $(x_{2}, y_{2})$ $(x_2, y_2)$ you have to use this formula here:
$\frac{y - y_{1}}{y_{2} - y_{1}} = \frac{x - x_{1}}{x_{2} - x_{1}}$ $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

Let's substitute:
$\frac{y - (- 4)}{- 3 - (- 4)} = \frac{x - (- 14)}{- 4 - (- 14)}$ $\frac{y-(-4)}{-3-(-4)}=\frac{x-(-14)}{-4-(-14)}$

$\frac{y + 4}{- 3 + 4} = \frac{x + 14}{- 4 + 14}$ $\frac{y+4}{-3+4}=\frac{x+14}{-4+14}$

$\frac{y + 4}{1} = \frac{x + 14}{10}$ $\frac{y+4}{1}=\frac{x+14}{10}$

$y + 4 = \frac{x + 14}{10}$ $y+4=\frac{x+14}{10}$

$10 (y + 4) = x + 14$ $10(y+4)=x+14$

$10 y + 40 = x + 14$ $10y+40=x+14$

$10 y = x + 14 - 40$ $10y=x+14-40$

$10 y = x - 26$ $10y=x-26$

$y = \frac{1}{10} x - \frac{26}{10}$ $y=\frac{1}{10}x-\frac{26}{10}$

The final result:
$y = \frac{1}{10} x - \frac{13}{5}$ $y=\frac{1}{10}x-\frac{13}{5}$

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Question #bcad8

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