
Find the equation of a line with slope is $ - 1$ and cut off an intercept of $4$ units in the negative direction of the $y$-axis.
Answer
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Hint: To solve this question use the point-intercept formula to find the equation of the line. The point $(x,y)$ on the line with slope $m$ and $y$ -intercept $c$ lies on the line if and only if \[y = mx + c\;\].
$\therefore $ substitute $ - 1$ as m and $ - 4$ as $c$ into the equation \[y = mx + c\;\] that is the equation of line.
Take $c$ as $-4$ as the intercept is made on the negative side of the $y$-axis.
Complete answer:
The slope of the line is given as $ - 1$ and the intercept on the negative $y$ -axis is 4. Since the intercept is made on the negative side of the $y$ -axis therefore, the value of $c$ is $ - 4$.
\[ \Rightarrow \]$m = - 1$
\[ \Rightarrow \]$c = - 4$
The point-intercept formula to find the equation of the line is \[y = mx + c\;\], where $m$ is the slope and $c$ is the $y$-intercept.
Substitute $m = - 1$ and $c = - 4$ into the equation of the line,
\[ \Rightarrow \]\[y = mx + c\;\]
\[ \Rightarrow y = - 1x + ( - 4)\]
\[ \Rightarrow y = - x - 4\]
After rearrange the equation we get,
\[ \Rightarrow \]\[y + x + 4 = 0\]
The equation of a line with slope is $ - 1$ and cutting off an intercept of $4$ units in the negative direction of $y$-axis is \[y + x + 4 = 0\].
Note:
If the intercept on the positive side of the $y$-axis then the value of $c$will be positive and If the intercept on the negative side of the $y$-axis then the value of $c$will be negative. In this question take intercept $c = - 4$.
Another method:
The line with the slope m cuts the y-axis from the negative direction at a distance c from the origin.
Here, $c$ represents the y-intercept of the line. That means \[x\] -coordinate of the point is \[0\] and the \[y\]-coordinate is $ - c$ because the intercept on the negative side of the $y$-axis .The coordinates of the point where the line meet the y-axis is \[(0, - c)\].
$\therefore $ The point is $(0, - 4)$ and slope is $ - 1$ .
Therefore, by point-slope form with slope $m$and point \[\left( {{x_{1,}}{y_1}} \right)\] to find the equation of line is,
$y - {y_1} = m(x - {x_1})$
Substitute ${x_1} = 0$,${y_1} = - 4$ and $m = - 1$ into the equation,
$ \Rightarrow y - ( - 4) = ( - 1)(x - 0)$
$ \Rightarrow y + 4 = - 1x$
$ \Rightarrow y + 4 + x = 0$
The equation of a line with slope is $ - 1$ and cutting off an intercept of $4$ units in the negative direction of $y$-axis is \[y + x + 4 = 0\].
$\therefore $ substitute $ - 1$ as m and $ - 4$ as $c$ into the equation \[y = mx + c\;\] that is the equation of line.
Take $c$ as $-4$ as the intercept is made on the negative side of the $y$-axis.
Complete answer:
The slope of the line is given as $ - 1$ and the intercept on the negative $y$ -axis is 4. Since the intercept is made on the negative side of the $y$ -axis therefore, the value of $c$ is $ - 4$.
\[ \Rightarrow \]$m = - 1$
\[ \Rightarrow \]$c = - 4$
The point-intercept formula to find the equation of the line is \[y = mx + c\;\], where $m$ is the slope and $c$ is the $y$-intercept.
Substitute $m = - 1$ and $c = - 4$ into the equation of the line,
\[ \Rightarrow \]\[y = mx + c\;\]
\[ \Rightarrow y = - 1x + ( - 4)\]
\[ \Rightarrow y = - x - 4\]
After rearrange the equation we get,
\[ \Rightarrow \]\[y + x + 4 = 0\]
The equation of a line with slope is $ - 1$ and cutting off an intercept of $4$ units in the negative direction of $y$-axis is \[y + x + 4 = 0\].
Note:
If the intercept on the positive side of the $y$-axis then the value of $c$will be positive and If the intercept on the negative side of the $y$-axis then the value of $c$will be negative. In this question take intercept $c = - 4$.
Another method:
The line with the slope m cuts the y-axis from the negative direction at a distance c from the origin.
Here, $c$ represents the y-intercept of the line. That means \[x\] -coordinate of the point is \[0\] and the \[y\]-coordinate is $ - c$ because the intercept on the negative side of the $y$-axis .The coordinates of the point where the line meet the y-axis is \[(0, - c)\].
$\therefore $ The point is $(0, - 4)$ and slope is $ - 1$ .
Therefore, by point-slope form with slope $m$and point \[\left( {{x_{1,}}{y_1}} \right)\] to find the equation of line is,
$y - {y_1} = m(x - {x_1})$
Substitute ${x_1} = 0$,${y_1} = - 4$ and $m = - 1$ into the equation,
$ \Rightarrow y - ( - 4) = ( - 1)(x - 0)$
$ \Rightarrow y + 4 = - 1x$
$ \Rightarrow y + 4 + x = 0$
The equation of a line with slope is $ - 1$ and cutting off an intercept of $4$ units in the negative direction of $y$-axis is \[y + x + 4 = 0\].
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