
Find the equation of the line joining the points \[\left( -1,3 \right)\] and \[\left( 4,-2 \right)\]?
Answer
521.4k+ views
Hint: In this problem, we have to find the equation of a line with given two points. We have two points in this problem, so we can use the two points form to find the equation of the line. We can also find the equation using the slope intercept form, by finding the slope using two points and substituting any one of the points in the equation to get a line equation.
Complete step by step solution:
We know that the two points form is,
\[\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
We know that the given two points are,
\[\left( {{x}_{1}},{{y}_{1}} \right)\]=\[\left( -1,3 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]=\[\left( 4,-2 \right)\].
We can substitute the above points in the formula, we get
\[\Rightarrow \dfrac{y-3}{-2-3}=\dfrac{x+1}{4+1}\]
Now we can simplify the above step,
\[\Rightarrow \dfrac{y-3}{-5}=\dfrac{x+1}{5}\]
Now we can cross multiply the above step, we get
\[\begin{align}
& \Rightarrow 5\left( y-3 \right)=-5\left( x+1 \right) \\
& \Rightarrow 5y-15=-5x-5 \\
& \Rightarrow 5x+5y-10=0 \\
\end{align}\]
We can now divide the above step by 5 on both sides, we get
\[\Rightarrow x+y-2=0\]
Therefore, the required equation is \[x+y-2=0\].
Note: We can also find the equation by using slope intercept form.
We know that the slope intercept form of the line is,
\[y=mx+c\] ……. (1)
Where, m is the slope and c is the y-intercept.
We have two points, \[\left( {{x}_{1}},{{y}_{1}} \right)\]=\[\left( -1,3 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]=\[\left( 4,-2 \right)\].
Slope, m = \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{-2-3}{4+1}=\dfrac{-5}{5}=-1\] .
We can substitute in equation (1), we get
\[y=-1x+c\]
We can now substitute the x and y value from any of the two points to get the y-intercept.
We can take the point \[\left( -1,3 \right)\]
\[\begin{align}
& \Rightarrow 3=-1\left( -1 \right)+c \\
& \Rightarrow c=2 \\
\end{align}\]
The y-intercept is -1 and the slope is 2.
Therefore, the required equation is \[y=-x+2\].
Complete step by step solution:
We know that the two points form is,
\[\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
We know that the given two points are,
\[\left( {{x}_{1}},{{y}_{1}} \right)\]=\[\left( -1,3 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]=\[\left( 4,-2 \right)\].
We can substitute the above points in the formula, we get
\[\Rightarrow \dfrac{y-3}{-2-3}=\dfrac{x+1}{4+1}\]
Now we can simplify the above step,
\[\Rightarrow \dfrac{y-3}{-5}=\dfrac{x+1}{5}\]
Now we can cross multiply the above step, we get
\[\begin{align}
& \Rightarrow 5\left( y-3 \right)=-5\left( x+1 \right) \\
& \Rightarrow 5y-15=-5x-5 \\
& \Rightarrow 5x+5y-10=0 \\
\end{align}\]
We can now divide the above step by 5 on both sides, we get
\[\Rightarrow x+y-2=0\]
Therefore, the required equation is \[x+y-2=0\].
Note: We can also find the equation by using slope intercept form.
We know that the slope intercept form of the line is,
\[y=mx+c\] ……. (1)
Where, m is the slope and c is the y-intercept.
We have two points, \[\left( {{x}_{1}},{{y}_{1}} \right)\]=\[\left( -1,3 \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]=\[\left( 4,-2 \right)\].
Slope, m = \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{-2-3}{4+1}=\dfrac{-5}{5}=-1\] .
We can substitute in equation (1), we get
\[y=-1x+c\]
We can now substitute the x and y value from any of the two points to get the y-intercept.
We can take the point \[\left( -1,3 \right)\]
\[\begin{align}
& \Rightarrow 3=-1\left( -1 \right)+c \\
& \Rightarrow c=2 \\
\end{align}\]
The y-intercept is -1 and the slope is 2.
Therefore, the required equation is \[y=-x+2\].
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