Question

How do you write an equation of a line that passes through points (-1,3), (2,-3)?

Answer 1

Hint: This question is from the topic of straight line. We will use both the points to find the equation of the line. We will first find the slope or gradient of the equation using the two points. After that, we will use the point-slope form. After using this form, we will solve the further equation and find the equation of line.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find the line which passes through the points (-1,3) and (2,-3)
Using the formula of gradient \[m=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}\], where m is the gradient or slope of line and \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] the points on the straight line.
So, the slope or the gradient of the line whose points are (-1,3) and (2,-3) will be
\[m=\dfrac{3-\left( -3 \right)}{\left( -1 \right)-2}=\dfrac{3+3}{-1-2}=\dfrac{6}{-3}=-2\]
Hence, we get that the slope of the line whose equation we have to find is -2.
Now, we will use the point slope form of a straight line to find the equation of the straight line.
The point slope form of a straight line says that if the slope of a line is given as ‘m’ and a point is also given on the line as \[\left( {{x}_{1}},{{y}_{1}} \right)\], then the equation of line will be
\[y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\]
So, after putting the point \[\left( {{x}_{1}},{{y}_{1}} \right)\] as (-1,3) and putting the value of m as -2 that we have found above, we can write the above equation as
\[\Rightarrow y-3=\left( -2 \right)\left( x-\left( -1 \right) \right)\]
The above equation can also be written as
\[\Rightarrow y-3=\left( -2 \right)\left( x+1 \right)\]
The above equation can also be written as
\[\Rightarrow y-3=-2x-2\]
The above equation can also be written as
\[\Rightarrow 2x+y-1=0\]
Hence, we have found the equation of line that passes through the points (-1,3), (2,-3). The equation we have found is \[2x+y-1=0\].
We can take reference from the following figure.

In the above figure, we can see that the line 2x+y-1=0 is passing through the points (-1,3) and (2,-3).
We can write this equation also in the form as
\[\Rightarrow y=-2x+1\]
\[\Rightarrow y-1=-2x\]
This form of line is slope intercept form, where the line is passing through the y-axis cutting it at y=1 and the slope is -2.

Note:
We should have a better knowledge in the topic of straight line to solve this type of questions easily. We can solve this question using an alternate method also.
For solving this question using an alternate method, we will first see the formula for finding the equation of line using two points on the line.
The formula for the equation is \[\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}\], where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] the points on the straight line.
So, we have given points as (-1,3) and (2,-3) which pass through the line.
Hence, the equation of line is
\[\dfrac{y-3}{x-\left( -1 \right)}=\dfrac{3-\left( -3 \right)}{\left( -1 \right)-2}\]
The above equation can also be written as
\[\Rightarrow \dfrac{y-3}{x+1}=\dfrac{3+3}{-1-2}\]
The above equation can also be written as
\[\Rightarrow \dfrac{y-3}{x+1}=\dfrac{6}{-3}\]
The above equation can also be written as
\[\Rightarrow \dfrac{y-3}{x+1}=-2\]
The above equation can also be written as
\[\Rightarrow y-3=-2\left( x+1 \right)\]
We can write the above equation as
\[\Rightarrow y-3=-2x-2\]
The above equation can also be written as
\[\Rightarrow 2x+y-1=0\]
We have got the same answer, so we can use this formula too to solve this type of question.