Can u help me with this ?

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Answer 1 · 2018-06-27T12:13:00

Have a look:

You can use the general relationship for a line passing through $(x_{0}, y_{0})$ $(x_0,y_0)$ and with slope $m$ $m$ :

$y - y_{0} = m (x - x_{0})$ $y-y_0=m(x-x_0)$

Substitute your values and rearrange to get your equation in the form:

$y = m x + c$ $y=mx+c$

Answer 2 · 2018-06-27T12:15:16

See below

You have to memorize that the equation point-slope has the general form

$y - y_{0} = m (x - x_{0})$ $y-y_0=m(x-x_0)$ where m is slope and $(x_{0}, y_{0})$ $(x_0,y_0)$ is a line passing point.

In our case

$y - (- 3) = - \frac{1}{4} (x - 8)$ $y-(-3)=-1/4(x-8)$

$y + 3 = - \frac{1}{4} (x - 8)$ $y+3=-1/4(x-8)$

we can stop here, but usualy we re-arrange terms in form:

$y = - \frac{1}{4} x + 2 - 3 = - \frac{1}{4} x - 1$ $y=-1/4x+2-3=-1/4x-1$ and finally

$y = - \frac{1}{4} x - 1$ $y=-1/4x-1$

Answer 3 · 2018-06-27T12:19:50

$(y - (- 3)) = (- \frac{1}{4}) \cdot (x - 8)$ $color(violet)((y - (-3)) = (-1/4) * (x - 8)$ is the required point-slope equation form.

Point-Slope form of equation is $y - y_{1} = m \cdot (x - x_{1})$ $y - y_1 = m * (x - x_1)$

Given : $m = - (\frac{1}{4}), (x_{1}, y_{1}) = (8, - 3)$ $m = -(1/4), (x_1,y_1) = (8, -3)$

$(y - (- 3)) = (- \frac{1}{4}) \cdot (x - 8)$ $color(violet)((y - (-3)) = (-1/4) * (x - 8)$ is the required point-slope equation form.

$4 y + 12 = - x + 8$ $4y + 12 = -x + 8$

$x + 4 y + 4 = 0$ $x + 4y + 4 = 0$