How do you write an equation for the horizontal line passing through the point $\left( -1,-8 \right)$?
Answer
558.3k+ views
Hint: From the question given a horizontal line passing through the point $\left( -1,-8 \right)$, we have to find the equation of that horizontal line. As we know that the slope of any horizontal line is 0. As we know that if any line has a slope “m” and it is passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$, then the line equation is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. By this we will get the required line equation.
Complete step-by-step solution:
From the question given a horizontal line passing through the point,
$\Rightarrow \left( -1,-8 \right)$
As mentioned in the question that the line is a horizontal line.
By this we can conclude that the slope of the given line is 0, because any horizontal line makes angle ${{0}^{\circ }}$with the x-axis, as we know that formula for the slope “m” is
$\Rightarrow m=\tan \theta =\tan {{0}^{\circ }}=0$
From this we can conclude that slope of the given line is 0, that is
$\Rightarrow m=0$
As we know that if any line has a slope “m” and it is passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$, then the line equation is
$\Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
By comparing here, we will get,
$\Rightarrow m=0$
$\Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( -1,-8 \right)$
By substituting the above values in their respective positions, we will get the equation of line,
$\Rightarrow y-\left( -8 \right)=0\left( x-\left( -1 \right) \right)$
By simplifying further, we will get,
$\Rightarrow y=-8$
Therefore, this is the required line equation which is horizontal and passes through the point $\left( -1,-8 \right)$.
Note: Students should know the basic formulas of coordinate geometry, students can also do this sum directly as we know that any horizontal line equation is $y=k$ where k is any constant, in the above question the line passing through the point is $\left( -1,-8 \right)$, therefore the equation of the line is $y=-8$. Students should also know that the equation of any vertical line is $x=k$ where k is any constant.
Complete step-by-step solution:
From the question given a horizontal line passing through the point,
$\Rightarrow \left( -1,-8 \right)$
As mentioned in the question that the line is a horizontal line.
By this we can conclude that the slope of the given line is 0, because any horizontal line makes angle ${{0}^{\circ }}$with the x-axis, as we know that formula for the slope “m” is
$\Rightarrow m=\tan \theta =\tan {{0}^{\circ }}=0$
From this we can conclude that slope of the given line is 0, that is
$\Rightarrow m=0$
As we know that if any line has a slope “m” and it is passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$, then the line equation is
$\Rightarrow y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
By comparing here, we will get,
$\Rightarrow m=0$
$\Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( -1,-8 \right)$
By substituting the above values in their respective positions, we will get the equation of line,
$\Rightarrow y-\left( -8 \right)=0\left( x-\left( -1 \right) \right)$
By simplifying further, we will get,
$\Rightarrow y=-8$
Therefore, this is the required line equation which is horizontal and passes through the point $\left( -1,-8 \right)$.
Note: Students should know the basic formulas of coordinate geometry, students can also do this sum directly as we know that any horizontal line equation is $y=k$ where k is any constant, in the above question the line passing through the point is $\left( -1,-8 \right)$, therefore the equation of the line is $y=-8$. Students should also know that the equation of any vertical line is $x=k$ where k is any constant.
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