How do you write the equation in point slope and slope intercept form given $\left( 1,2 \right)$ and $\left( 2,5 \right)$ ?
Answer
554.7k+ views
Hint: The straight line can be represented in many ways. Two of the ways are point slope form and slope intercept form. Point slope form can be represented as $y-{{y}_{1}}=m(x-{{x}_{1}})$ where ${{x}_{1}},{{y}_{1}}$ are the coordinates and $m$ is the slope. The intercept form the line is $y=mx+c$ .
Complete step-by-step solution:
The points given in the question are $(1,2)$ and $(2,5)$. Firstly we will find the value of the equation in point slope form. Equation in slope form is \[y-{{y}_{1}}=m(x-{{x}_{1}})\] which can further be written as $\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Consider $({{x}_{1}},{{y}_{1}})\equiv (1,2)$ and $({{x}_{2}},{{y}_{2}})\equiv (2,5)$ and put the values respectively in the formula:
$\Rightarrow \dfrac{y-2}{x-1}=\dfrac{5-2}{2-1}$
$\Rightarrow \dfrac{y-2}{x-1}=\dfrac{3}{1}$
$\Rightarrow y-2=3(x-1)$
$\therefore $ The equation in point slope form is $y-2=3(x-1)$.
Now, we shall write the equation in intercept form. The equation in slope-intercept form is $y=mx+c$.
Consider the two points $({{x}_{1}},{{y}_{1}})\equiv (1,2)$ and $({{x}_{2}},{{y}_{2}})\equiv (2,5)$. We will find the value of $m$ and $c$ with the help of two points given. Putting the values one by one we get
\[{{y}_{1}}=m{{x}_{1}}+c\]
\[\Rightarrow 2=m\times 1+c\]
$\Rightarrow c=2-m$
Now putting the value of $c$ in terms of $m$ in the second equation which is
\[{{y}_{2}}=m{{x}_{2}}+c\]
\[{{y}_{2}}=m{{x}_{2}}+2-m\]
Put the values $({{x}_{2}},{{y}_{2}})\equiv (2,5)$
\[\Rightarrow 5=m\times 2+2-m\]
\[\Rightarrow 5=2m+2-m\]
\[\Rightarrow 5-2=2m-m\]
\[\Rightarrow 3=m\]
Therefore the value of
$\begin{align}
& c=2-m \\
& \Rightarrow c=2-(3) \\
\end{align}$
$\Rightarrow c=-1$
$\therefore $ The slope $m$= 3 and $c=-1$ for the equation.
$y=mx+c$
$y=3x-1$
$\therefore $ The equation in slope intercept form is $y=3x-1$.
Note: Do remember there are many forms of equations for straight lines. As per the demand of the question you are supposed to write the equation. Equation being in any form should be the same for the same points. For instance you can see the above equations. In both the forms, the equation remains the same, which is $y=3x-1$.
Complete step-by-step solution:
The points given in the question are $(1,2)$ and $(2,5)$. Firstly we will find the value of the equation in point slope form. Equation in slope form is \[y-{{y}_{1}}=m(x-{{x}_{1}})\] which can further be written as $\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Consider $({{x}_{1}},{{y}_{1}})\equiv (1,2)$ and $({{x}_{2}},{{y}_{2}})\equiv (2,5)$ and put the values respectively in the formula:
$\Rightarrow \dfrac{y-2}{x-1}=\dfrac{5-2}{2-1}$
$\Rightarrow \dfrac{y-2}{x-1}=\dfrac{3}{1}$
$\Rightarrow y-2=3(x-1)$
$\therefore $ The equation in point slope form is $y-2=3(x-1)$.
Now, we shall write the equation in intercept form. The equation in slope-intercept form is $y=mx+c$.
Consider the two points $({{x}_{1}},{{y}_{1}})\equiv (1,2)$ and $({{x}_{2}},{{y}_{2}})\equiv (2,5)$. We will find the value of $m$ and $c$ with the help of two points given. Putting the values one by one we get
\[{{y}_{1}}=m{{x}_{1}}+c\]
\[\Rightarrow 2=m\times 1+c\]
$\Rightarrow c=2-m$
Now putting the value of $c$ in terms of $m$ in the second equation which is
\[{{y}_{2}}=m{{x}_{2}}+c\]
\[{{y}_{2}}=m{{x}_{2}}+2-m\]
Put the values $({{x}_{2}},{{y}_{2}})\equiv (2,5)$
\[\Rightarrow 5=m\times 2+2-m\]
\[\Rightarrow 5=2m+2-m\]
\[\Rightarrow 5-2=2m-m\]
\[\Rightarrow 3=m\]
Therefore the value of
$\begin{align}
& c=2-m \\
& \Rightarrow c=2-(3) \\
\end{align}$
$\Rightarrow c=-1$
$\therefore $ The slope $m$= 3 and $c=-1$ for the equation.
$y=mx+c$
$y=3x-1$
$\therefore $ The equation in slope intercept form is $y=3x-1$.
Note: Do remember there are many forms of equations for straight lines. As per the demand of the question you are supposed to write the equation. Equation being in any form should be the same for the same points. For instance you can see the above equations. In both the forms, the equation remains the same, which is $y=3x-1$.
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