
How do you find an equation in standard form for the line containing $\left( -1,1 \right)$, $\left( -2,1 \right)$?
Answer
548.7k+ views
Hint: For this problem we need to calculate the equation of the line which contains the given points. In algebra we have the formula for the line passing through the two points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ as $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$. Now we will compare the given points with $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and substitute the values of ${{x}_{1}}$, ${{x}_{2}}$, ${{y}_{1}}$, ${{y}_{2}}$ in the formula and simplify the equation to get the required result.
Complete step by step solution:
Given points, $\left( -1,1 \right)$, $\left( -2,1 \right)$.
Comparing the above points with $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$, then we will get
${{x}_{1}}=-1$, ${{x}_{2}}=-2$, ${{y}_{1}}=1$, ${{y}_{2}}=1$
In algebra we have the formula for the equation of the line which passes through the points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ as
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$
Substituting the values ${{x}_{1}}$, ${{x}_{2}}$, ${{y}_{1}}$, ${{y}_{2}}$ in the above formula, then we will get
$\Rightarrow y-1=\dfrac{1-1}{-2-\left( -1 \right)}\left( x-\left( -1 \right) \right)$
When we multiply a negative sign with the negative sign, then we will get a positive sign as a result. Using this rule in the above equation and simplifying it, then we will get
$\Rightarrow y-1=\dfrac{0}{-2+1}\left( x+1 \right)$
We know that anything multiplied or divided by zero is always a zero. Applying this rule in the above equation, then we will have
$\Rightarrow y-1=0$
Hence the equation of the line that passes through the given points $\left( -1,1 \right)$, $\left( -2,1 \right)$ is $y-1=0$.
Note:
We can also mark the given points in the graph paper and join them together to get the required equation. When we mark the points in the graph and joined them then we will get
Complete step by step solution:
Given points, $\left( -1,1 \right)$, $\left( -2,1 \right)$.
Comparing the above points with $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$, then we will get
${{x}_{1}}=-1$, ${{x}_{2}}=-2$, ${{y}_{1}}=1$, ${{y}_{2}}=1$
In algebra we have the formula for the equation of the line which passes through the points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ as
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$
Substituting the values ${{x}_{1}}$, ${{x}_{2}}$, ${{y}_{1}}$, ${{y}_{2}}$ in the above formula, then we will get
$\Rightarrow y-1=\dfrac{1-1}{-2-\left( -1 \right)}\left( x-\left( -1 \right) \right)$
When we multiply a negative sign with the negative sign, then we will get a positive sign as a result. Using this rule in the above equation and simplifying it, then we will get
$\Rightarrow y-1=\dfrac{0}{-2+1}\left( x+1 \right)$
We know that anything multiplied or divided by zero is always a zero. Applying this rule in the above equation, then we will have
$\Rightarrow y-1=0$
Hence the equation of the line that passes through the given points $\left( -1,1 \right)$, $\left( -2,1 \right)$ is $y-1=0$.
Note:
We can also mark the given points in the graph paper and join them together to get the required equation. When we mark the points in the graph and joined them then we will get
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