How do you graph the lines using slope-intercept form $y=-\dfrac{2}{3}x+1?$
Answer
620.4k+ views
Hint: $1.$ This equation is written in slope intercept form.
Which is $y=mx+b$
Where, $m$ is the slope
$b$ is the $y$ intercept.
$2.$ The slope is the constant or number multiplied by the variable $x.$ Which $m$ in this case $-\dfrac{2}{3}$
with the above data we will represent the line graphically.
Complete step by step solution: Here, we know that slope intercept form is.
$y=\dfrac{-2x}{3}+1$
The slope-intercept form of linear equation is $y=mx+b$
Here, $m=\dfrac{-2}{3}$
$b=1$ or $\left( 0,1 \right)$
We can plot this point on the grid as,
Graph $\left\{ \left( \dfrac{x}{2} \right)+\dfrac{\left( y-1 \right)}{2}-0.025 \right\}$
$=0\left[ -10,10,-5,5 \right]$
The slope is, $m=\dfrac{-2}{3}=\dfrac{size}{run}$
In this case,
$y-$intercept: $\left( 0,1 \right)$
The rise is $-2$ and we need to go down $2$ position on the $y-$value and the run is $3$ and we need to go right $3$ position on the $x$ value.
Hence, the second point $B\left( 0+3,1-2 \right)$
$=\left( 3,-1 \right)$
We can now plot this point.
Graph $\left\{ \left( \dfrac{x}{2}+\dfrac{y-1}{2}-0.025 \right)\left( \dfrac{x-3}{2}+\dfrac{y+1}{2}-0.025 \right) \right\}$
$=0\left[ -10,10,-5,5 \right]$
Now, we can also draw a line through the points given.
Graph $\left\{ \left( y+\dfrac{2}{3}x-1 \right)\left( \dfrac{x}{2}+\dfrac{\left( y-1 \right)}{2}-0.025 \right) \right\}$
$\left( \dfrac{x-3}{2}+\dfrac{y-1}{2}-0.025 \right)$
$=0\left[ -10,10,-5,5 \right]$
In this way we can graph the lines using slope-intercept form $y=\dfrac{-2}{3}x+1$
Additional Information:
Slope intercept form is a form of writing an equation of a straight line.
We can also term slope as gradient.
While we write any equation in this form, we usually get information regarding the equation of the line and the value of slope and intercept of the line.
The slope of the line may be either positive, negative or zero.
If the slope of line is positive it means that the slope upwards from left to right and if the slope of line is negative it means that slope downwards from left to right. if the slope of the line is zero it means that slope will be parallel to the horizontal line.
For writing any equations in slope intercept form,
We follow certain patterns,
$\left( 1 \right)$ The slope intercept form of linear equation is $y=mx+b$ Where $m$ is the slope and $b$ is the $y-$intercept.
Step: $\left( 1 \right)$ Graph the equation $y-2x=1$
$\left( 2 \right)$ Rewrite $m$ slope intercept form $y=2x+1$
$\left( 3 \right)$ Identity slope and $y-$intercept $m=2$ and $h=1$
$\left( 4 \right)$ Plot the points $\left( 0,1 \right)$
$\left( 5 \right)$ Second point $\left( 1,3 \right)$ $y=2\times 1+1=3$
Note:
$\left( 1 \right)$ Compare the give equation always by standard slope-intercept form i.e. $y=mx+b$
And identity the value of $m$ and $b$ by the example $y=\dfrac{-2}{3}x+1$
Here,
$m=\dfrac{-2}{3}$ and $b=1$
$\left( 2 \right)$ And find the point of the $x-$axis and $y-$axis.
Which is $y=mx+b$
Where, $m$ is the slope
$b$ is the $y$ intercept.
$2.$ The slope is the constant or number multiplied by the variable $x.$ Which $m$ in this case $-\dfrac{2}{3}$
with the above data we will represent the line graphically.
Complete step by step solution: Here, we know that slope intercept form is.
$y=\dfrac{-2x}{3}+1$
The slope-intercept form of linear equation is $y=mx+b$
Here, $m=\dfrac{-2}{3}$
$b=1$ or $\left( 0,1 \right)$
We can plot this point on the grid as,
Graph $\left\{ \left( \dfrac{x}{2} \right)+\dfrac{\left( y-1 \right)}{2}-0.025 \right\}$
$=0\left[ -10,10,-5,5 \right]$
The slope is, $m=\dfrac{-2}{3}=\dfrac{size}{run}$
In this case,
$y-$intercept: $\left( 0,1 \right)$
The rise is $-2$ and we need to go down $2$ position on the $y-$value and the run is $3$ and we need to go right $3$ position on the $x$ value.
Hence, the second point $B\left( 0+3,1-2 \right)$
$=\left( 3,-1 \right)$
We can now plot this point.
Graph $\left\{ \left( \dfrac{x}{2}+\dfrac{y-1}{2}-0.025 \right)\left( \dfrac{x-3}{2}+\dfrac{y+1}{2}-0.025 \right) \right\}$
$=0\left[ -10,10,-5,5 \right]$
Now, we can also draw a line through the points given.
Graph $\left\{ \left( y+\dfrac{2}{3}x-1 \right)\left( \dfrac{x}{2}+\dfrac{\left( y-1 \right)}{2}-0.025 \right) \right\}$
$\left( \dfrac{x-3}{2}+\dfrac{y-1}{2}-0.025 \right)$
$=0\left[ -10,10,-5,5 \right]$
In this way we can graph the lines using slope-intercept form $y=\dfrac{-2}{3}x+1$
Additional Information:
Slope intercept form is a form of writing an equation of a straight line.
We can also term slope as gradient.
While we write any equation in this form, we usually get information regarding the equation of the line and the value of slope and intercept of the line.
The slope of the line may be either positive, negative or zero.
If the slope of line is positive it means that the slope upwards from left to right and if the slope of line is negative it means that slope downwards from left to right. if the slope of the line is zero it means that slope will be parallel to the horizontal line.
For writing any equations in slope intercept form,
We follow certain patterns,
$\left( 1 \right)$ The slope intercept form of linear equation is $y=mx+b$ Where $m$ is the slope and $b$ is the $y-$intercept.
Step: $\left( 1 \right)$ Graph the equation $y-2x=1$
$\left( 2 \right)$ Rewrite $m$ slope intercept form $y=2x+1$
$\left( 3 \right)$ Identity slope and $y-$intercept $m=2$ and $h=1$
$\left( 4 \right)$ Plot the points $\left( 0,1 \right)$
$\left( 5 \right)$ Second point $\left( 1,3 \right)$ $y=2\times 1+1=3$
Note:
$\left( 1 \right)$ Compare the give equation always by standard slope-intercept form i.e. $y=mx+b$
And identity the value of $m$ and $b$ by the example $y=\dfrac{-2}{3}x+1$
Here,
$m=\dfrac{-2}{3}$ and $b=1$
$\left( 2 \right)$ And find the point of the $x-$axis and $y-$axis.
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