Find the equation of line joining points \[\left( { - 2,3} \right)\] and \[\left( {1,4} \right)\].
Answer
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Hint: Here we are asked to find the equation of the line through the two given points. We will first find the slope of the line through the given points using the formula then we will substitute the slope in the formula of the equation of line through the two given points.
Complete step-by-step answer:
Let,
\[\left( {{x_1},{y_1}} \right) = \left( { - 2,3} \right)\]
\[\left( {{x_2},{y_2}} \right) = \left( {1,4} \right)\]
Now we will find the slope of the line through these points. We know that the slope of the line through the points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by:
\[{\text{slope}}\left( m \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
Now we will put the given two points in this equation to obtain the value of slope. So, we get;
\[ \Rightarrow m = \dfrac{{4 - 3}}{{1 - \left( { - 2} \right)}}\]
On simplification we get,
\[ \Rightarrow m = \dfrac{1}{3}\]
Now we know that the equation of the line passing through points \[\left( {{x_1},{y_1}} \right)\] and having slope \[m\] is;
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Now we will put the values of \[{x_1},{y_1},m\] in the above equation to get the equation of the line. So, we have;
\[ \Rightarrow y - 3 = \dfrac{1}{3}\left( {x - \left( { - 2} \right)} \right)\]
On solving the terms in the bracket, we get;
\[ \Rightarrow y - 3 = \dfrac{1}{3}\left( {x + 2} \right)\]
Now we will shift \[3\] to the LHS. So, we get;
\[ \Rightarrow 3\left( {y - 3} \right) = x + 2\]
Solving further we get;
\[ \Rightarrow 3y - 9 = x + 2\]
Now we will \[9\] to RHS and \[x\] to the LHS. So, we get;
\[ \Rightarrow 3y - x = 11\]
Which is the equation of the desired line.
So, the correct answer is “3y - x = 11”.
Note: One point to note here is that for finding the slope if in the numerator we are writing \[\left( {{y_2} - {y_1}} \right)\] then in the denominator we will have to write \[\left( {{x_2} - {x_1}} \right)\], we cannot write \[\left( {{x_1} - {x_2}} \right)\]. We can also find the lines having the same slope but passing through some other points just by changing the constant term in the above equation of the line and these groups of lines are called parallel lines.
Complete step-by-step answer:
Let,
\[\left( {{x_1},{y_1}} \right) = \left( { - 2,3} \right)\]
\[\left( {{x_2},{y_2}} \right) = \left( {1,4} \right)\]
Now we will find the slope of the line through these points. We know that the slope of the line through the points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by:
\[{\text{slope}}\left( m \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
Now we will put the given two points in this equation to obtain the value of slope. So, we get;
\[ \Rightarrow m = \dfrac{{4 - 3}}{{1 - \left( { - 2} \right)}}\]
On simplification we get,
\[ \Rightarrow m = \dfrac{1}{3}\]
Now we know that the equation of the line passing through points \[\left( {{x_1},{y_1}} \right)\] and having slope \[m\] is;
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Now we will put the values of \[{x_1},{y_1},m\] in the above equation to get the equation of the line. So, we have;
\[ \Rightarrow y - 3 = \dfrac{1}{3}\left( {x - \left( { - 2} \right)} \right)\]
On solving the terms in the bracket, we get;
\[ \Rightarrow y - 3 = \dfrac{1}{3}\left( {x + 2} \right)\]
Now we will shift \[3\] to the LHS. So, we get;
\[ \Rightarrow 3\left( {y - 3} \right) = x + 2\]
Solving further we get;
\[ \Rightarrow 3y - 9 = x + 2\]
Now we will \[9\] to RHS and \[x\] to the LHS. So, we get;
\[ \Rightarrow 3y - x = 11\]
Which is the equation of the desired line.
So, the correct answer is “3y - x = 11”.
Note: One point to note here is that for finding the slope if in the numerator we are writing \[\left( {{y_2} - {y_1}} \right)\] then in the denominator we will have to write \[\left( {{x_2} - {x_1}} \right)\], we cannot write \[\left( {{x_1} - {x_2}} \right)\]. We can also find the lines having the same slope but passing through some other points just by changing the constant term in the above equation of the line and these groups of lines are called parallel lines.
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