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# Write an equation with standard form with given $\left( {2,2} \right)$ and $\left( {6,3} \right)$ .

Last updated date: 29th Feb 2024
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Hint: In the given question, we have been given the coordinates of two points. There is a line joining the two points. We have to find the standard form equation of that line joining the two points. For doing this first we are going to represent the line in slope intercept form and then then turn that slope intercept form to standard form.

The slope intercept form of an equation is $y = mx + b$
The given points of the line are $\left( {2,2} \right)$ and $\left( {6,3} \right)$ .
The formula to be used in this question is given below:
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
First we are going to calculate the slope of the line.
$m = \dfrac{{3 - 2}}{{6 - 2}} \\ \Rightarrow m = \dfrac{1}{4} \;$
So, we can write the slope intercept form of the equation as
$y = \left( {\dfrac{1}{4}} \right)x + b$
To find “b”, we can plug in any point. We are given $\left( {2,2} \right)$ and $\left( {6,3} \right)$ .
Hence we have,
$2 = \dfrac{1}{4} \times 2 + b \\ \Rightarrow 2 = \dfrac{2}{4} + b \\ \Rightarrow b = \dfrac{3}{2} \;$
Or we can have,
$3 = \dfrac{1}{4} \times 6 + b \\ \Rightarrow 3 = \dfrac{3}{2} + b \\ \Rightarrow b = \dfrac{3}{2} \;$
Hence, either way we will get “b” as $\dfrac{3}{2}$ .
Now writing out what we have,
$y = \dfrac{1}{4}x + \dfrac{3}{2}$
Converting this to standard form we have,
$- \dfrac{1}{4}x + y = \dfrac{3}{2} \\ \Rightarrow - 4\left( {\dfrac{{ - 1}}{4}x + y = \dfrac{3}{2}} \right) \\ \Rightarrow x - 4y = - 6 \;$
Hence, this is the required equation.
So, the correct answer is “x - 4y = - 6”.

Note: So, for solving such questions we should first write what has been given to us. Then we should write what we have to find. Since, in this question we need to find the standard equation of a line with given endpoints. To solve that we should first represent it in slope intercept form and then we can turn it into standard form. It is important to know the formula of the concept being used.