1) The point slope form and slope intercept form of 3x + 4y = -12 are [tex]y+1=\frac{-3}{4}\left(x+\frac{8}{3}\right)[/tex] and [tex]y= \frac{-3}{4}x-3[/tex] respectively
And slope is [tex]\frac{-3}{4}[/tex] and y -intercept is -3
2) The point slope form and slope intercept form of -2x - 3y = 6 are [tex]y+1=-\frac{2}{3}\left(x+\frac{3}{2}\right)[/tex] and [tex]y=-\frac{2}{3} x-2[/tex] respectively
And slope is [tex]\frac{-2}{3}[/tex] 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:
[tex]y-y_{1}=m\left(x-x_{1}\right)[/tex]
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,
[tex]\begin{array}{l}{4 y=-3 x-12} \\\\ {\rightarrow y=\frac{-3}{4}-3}\end{array}[/tex]
Hence the slope intercept form is [tex]y=\frac{-3}{4}-3[/tex]
[tex]\text { where slope is }-\frac{3}{4} \text { and } y-\text { intercept is }-3[/tex]
[tex]\text { Point slope form i.e. } y-y_{1}=m\left(x-x_{1}\right)[/tex]
[tex]\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}[/tex]
2) Solving -2x – 3y = 6
Slope intercept form is given as:
[tex]-2 x-3 y=6 \rightarrow-3 y=2 x+6 \\\\\rightarrow y=-\frac{2}{3} x-2[/tex]
[tex]\text { where slope is }-\frac{2}{3} \text { and } y-\text { intercept is }-2[/tex]
Point slope form is given as:
[tex]\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}[/tex]
Hence the required slope and y-intercept are found
