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Question: Find the equation of line joining the points \(( - 1,3)\) and \((4, - 2)\)....

Find the equation of line joining the points (−1,3)( - 1,3) and (4,−2)(4, - 2).

Explanation

Solution

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 (x1,y1)({x_1},{y_1}) and (x2,y2)({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)( - 1,3) and (4,−2)(4, - 2) , we can consider them as(x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) for further equation purpose.

Now, for the equation for which the line passing through two points is
y2  − y1  = m(x2  − x1){{\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 x1,y1,x2andy2{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
 m=(y2  − y1)(x2  − x1)  {\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.

 m=(−2  − 3)(4  +1)   m=−55 m=−1  {\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.