
Find the equation of line joining the points $( - 1,3)$ and $(4, - 2)$.
Answer
550.5k+ views
Hint: Find slope and substitute in the normal equation for the line through the two points.
First, we start by finding the slope of the line with the help of a formula and once we have that, we can now, consider the given two points as $({x_1},{y_1})$ and $({x_2},{y_2})$. Then substitute in the equation for the line joining through two points. Then, there we go we have the required equation.
Complete step by step solution:
We are given two points which are $( - 1,3)$ and $(4, - 2)$ , we can consider them as$({x_1},{y_1})$ and $({x_2},{y_2})$ for further equation purpose.
Now, for the equation for which the line passing through two points is
\[{{\mathbf{y}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{y}}_{\mathbf{1}}}\; = {\text{ }}{\mathbf{m}}({{\mathbf{x}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{x}}_{\mathbf{1}}})\]
Where m is the slope and ${x_1},{y_1},{x_2}and{y_2}$ are the coordinates of the two points.
Since we don’t have the slope, we have to find it.
So, the formula for slope is
\[{\text{ }}{\mathbf{m}} = \dfrac{{({{\mathbf{y}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{y}}_{\mathbf{1}}})}}{{({{\mathbf{x}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{x}}_{\mathbf{1}}})}}\;\]
Now, we substitute the coordinates into the formula and get the slope.
\[
{\text{ }}{\mathbf{m}} = \dfrac{{( - 2\;-{\text{ 3}})}}{{(4\; + 1)}}\; \\
m = \dfrac{{ - 5}}{5} \\
m = - 1 \\
\]
Now, we have obtained the slope as well.
Since, we need an equation, the above formula for an equation with two points cannot be used as an equation cannot be derived from it.
So, we use this formula to get the required equation.
This formula is called point-slope formula
We can use one of the given two points and
\[
{\text{ }}{\mathbf{m}} = \dfrac{{( - 2\;-{\text{ 3}})}}{{(4\; + 1)}}\; \\
m = \dfrac{{ - 5}}{5} \\
m = - 1 \\
\]substitute in the above formula.
Then we get,
$
y - 3 = - 1(x + 1) \\
y - 3 = - x - 1 \\
x + y - 3 + 1 = 0 \\
x + y - 2 = 0 \\
$
And substituting the other point, we get another equation which is
$
y + 2 = - 1(x - 4) \\
y + 2 = - x + 4 \\
x + y + 2 - 4 = 0 \\
x + y - 2 = 0 \\
$
Note: In this problem, we have to be careful while substituting the values in the formula of slope and the equation, as we might miss ${x_1}$ and write ${x_2}$ in its place which can give us a wrong answer. Hence, substituting is the critical place in the problem.
First, we start by finding the slope of the line with the help of a formula and once we have that, we can now, consider the given two points as $({x_1},{y_1})$ and $({x_2},{y_2})$. Then substitute in the equation for the line joining through two points. Then, there we go we have the required equation.
Complete step by step solution:
We are given two points which are $( - 1,3)$ and $(4, - 2)$ , we can consider them as$({x_1},{y_1})$ and $({x_2},{y_2})$ for further equation purpose.
Now, for the equation for which the line passing through two points is
\[{{\mathbf{y}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{y}}_{\mathbf{1}}}\; = {\text{ }}{\mathbf{m}}({{\mathbf{x}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{x}}_{\mathbf{1}}})\]
Where m is the slope and ${x_1},{y_1},{x_2}and{y_2}$ are the coordinates of the two points.
Since we don’t have the slope, we have to find it.
So, the formula for slope is
\[{\text{ }}{\mathbf{m}} = \dfrac{{({{\mathbf{y}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{y}}_{\mathbf{1}}})}}{{({{\mathbf{x}}_{\mathbf{2}}}\;-{\text{ }}{{\mathbf{x}}_{\mathbf{1}}})}}\;\]
Now, we substitute the coordinates into the formula and get the slope.
\[
{\text{ }}{\mathbf{m}} = \dfrac{{( - 2\;-{\text{ 3}})}}{{(4\; + 1)}}\; \\
m = \dfrac{{ - 5}}{5} \\
m = - 1 \\
\]
Now, we have obtained the slope as well.
Since, we need an equation, the above formula for an equation with two points cannot be used as an equation cannot be derived from it.
So, we use this formula to get the required equation.
This formula is called point-slope formula
We can use one of the given two points and
\[
{\text{ }}{\mathbf{m}} = \dfrac{{( - 2\;-{\text{ 3}})}}{{(4\; + 1)}}\; \\
m = \dfrac{{ - 5}}{5} \\
m = - 1 \\
\]substitute in the above formula.
Then we get,
$
y - 3 = - 1(x + 1) \\
y - 3 = - x - 1 \\
x + y - 3 + 1 = 0 \\
x + y - 2 = 0 \\
$
And substituting the other point, we get another equation which is
$
y + 2 = - 1(x - 4) \\
y + 2 = - x + 4 \\
x + y + 2 - 4 = 0 \\
x + y - 2 = 0 \\
$
Note: In this problem, we have to be careful while substituting the values in the formula of slope and the equation, as we might miss ${x_1}$ and write ${x_2}$ in its place which can give us a wrong answer. Hence, substituting is the critical place in the problem.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

