Find the point slope form and slope intercept form, indicate the slope and y-int:

1 answer:

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1) The point slope form and slope intercept form of 3x + 4y = -12 are $y+1=\frac{-3}{4}\left(x+\frac{8}{3}\right)$ and $y= \frac{-3}{4}x-3$ respectively

And slope is $\frac{-3}{4}$ and y -intercept is -3

2) The point slope form and slope intercept form of -2x - 3y = 6 are $y+1=-\frac{2}{3}\left(x+\frac{3}{2}\right)$ and $y=-\frac{2}{3} x-2$ respectively

And slope is $\frac{-2}{3}$ and y-intercept is -2

Solution:

Given, two equations are 3x + 4y = - 12 ⇒ (1) and – 2x – 3y = 6 ⇒ (2)

We have to find the point slope form and slope intercept form, and have to indicate the slope and y-intercept.

The point slope form is given as:

$y-y_{1}=m\left(x-x_{1}\right)$

Where "m" is the slope of the line

The slope-intercept form is given as:

y = mx + c

Where "c" is the y-intercept

1) Solving 3x + 4y = -12

Given equation is 3x + 4y = -12

On rearranging the terms, we get,

$\begin{array}{l}{4 y=-3 x-12} \\\\ {\rightarrow y=\frac{-3}{4}-3}\end{array}$

Hence the slope intercept form is $y=\frac{-3}{4}-3$

$\text { where slope is }-\frac{3}{4} \text { and } y-\text { intercept is }-3$

$\text { Point slope form i.e. } y-y_{1}=m\left(x-x_{1}\right)$

$\begin{array}{l}{3 x+4 y=-12} \\\\ {\rightarrow 4 y=-3 x-12} \\\\ {\rightarrow y=\frac{-3}{4} x-3} \\\\ {\rightarrow y=\frac{-3}{4} x-2-1} \\\\ {\rightarrow y+1=\frac{-3}{4}\left(x+\frac{8}{3}\right)}\end{array}$

2) Solving -2x – 3y = 6

Slope intercept form is given as:

$-2 x-3 y=6 \rightarrow-3 y=2 x+6 \\\\\rightarrow y=-\frac{2}{3} x-2$

$\text { where slope is }-\frac{2}{3} \text { and } y-\text { intercept is }-2$

Point slope form is given as:

$\begin{array}{l}{-2 x-3 y=6 \rightarrow-3 y=2 x+6} \\\\ {\rightarrow y=-\frac{2}{3} x-2} \\\\ {\rightarrow y=-\frac{2}{3} x-1-1} \\\\ {\quad \rightarrow y+1=-\frac{2}{3}\left(x+\frac{3}{2}\right)}\end{array}$