Question

How do you find slope, point slope form, slope intercept form, standard form, domain and range a line joining the points (6,0) and (10,3)?

Answer 1

In this question, we are given coordinates of two points on a line and we need to find slope, point slope form, slope intercept form, standard form, domain and range for the lines. We will use following formula for each part,
(I) Slope: Slope (m) of a line having two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
(II) Point slope form: Point slope form of the line having any one point $\left( {{x}_{1}},{{y}_{1}} \right)$ and slope m is given by $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$.
(III) Slope intercept form: The slope intercept form of a b is the y intercept value. For finding slope intercept form we just need to rearrange the point slope form and solve for y.
(IV) Standard form: For standard form, we will convert the slope intercept form in the form as Ax + By = C where A, B, C are some constant.
(V) Domain and range: The value of x which can be put in the equation gives us the domain and value of y which will get after putting x will give us the range.

Complete step by step solution:
Here we are given two points as (6,0) and (10,3). We need to find slope, point slope form, slope intercept form, standard form, domain and range for line passing through these points. Let us find all things required one by one,
(I) Slope: We know that the slope of a line passing through the point $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here we have (6,0) and (10,3) so the slope becomes $m=\dfrac{3-0}{10-6}$.
Solving we get $m=\dfrac{3}{4}$.
Hence the slope of the line is $\dfrac{3}{4}$.
(II) Point slope form: For any point $\left( {{x}_{1}},{{y}_{1}} \right)$ passing through line having slope m, the equation of line in point slope form is given as $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$.
Slope is found as $m=\dfrac{3}{4}$.
Taking $\left( {{x}_{1}},{{y}_{1}} \right)$ as (6,0) we get $\left( y-0 \right)=\dfrac{3}{4}\left( x-6 \right)$ which is the required point slope form.
(III) Slope intercept form: The slope intercept form for a linear equation is given as y = mx+b where m is the slope and b is the slope intercept. Let us change the point slope form in the form as y = mx+b.
Point slope form is given as $\left( y-0 \right)=\dfrac{3}{4}\left( x-6 \right)\Rightarrow y=\dfrac{3}{4}\left( x-6 \right)$.
Multiplying $\dfrac{3}{4}$ inside the bracket we get $y=\dfrac{3}{4}x-\dfrac{3}{4}\times 6$.
Cancelling the factor 2 from second term we get $y=\dfrac{3}{4}x-\dfrac{3\times 3}{2}\Rightarrow y=\dfrac{3}{4}x-\dfrac{9}{2}$.
Which is in the form as y = mx+b where $m=\dfrac{3}{4}\text{ and }c=-\dfrac{9}{2}$.
So the required slope intercept form is $y=\dfrac{3}{4}x-\dfrac{9}{2}$.
(IV) Standard form: The standard form of a linear equation is Ax + By = C where A, B, C are integers. Let us try to convert the slope intercept form in the standard form.
We have slope intercept form as $y=\dfrac{3}{4}x-\dfrac{9}{2}$.
As we can see, we want variable term on one side and constant term on the other side. So rearranging the equation we get $\dfrac{3}{4}x-y=\dfrac{9}{2}$.
Comparing with Ax + By = C we see $A=\dfrac{3}{4}\text{ and }C=\dfrac{9}{2}$ which are not integers, so let us simplify the equation. Let us multiply both sides of the equation by 4 we get $4\left( \dfrac{3}{4}x-y \right)=4\left( \dfrac{9}{2} \right)$.
Multiplying 4 with individual terms on both sides we get $4\times \dfrac{3}{4}x-4y=4\times \dfrac{9}{2}$.
Solving and cancelling common factors from all terms we get $3x-4y=2\times 9\Rightarrow 3x-4y=18$.
We know that, -a can be written as +(-a) so, writing 4y in the same way we get $3x+\left( -4y \right)=18$.
This equation is of the form Ax + By = C where A = 3, B = -4 and C = 18.
Therefore, $3x+\left( -4y \right)=18$ is the required standard form of the equation.
(V) Domain and range: As we can see from the standard form that there are no restrictions on the value of x, therefore the domain can take any values from real numbers. Hence, domain $\left( -\infty ,\infty \right)$.
From slope intercept form, we can see that as x increases, y increases and as x decreases, y decreases. So range is also the value from all real numbers. Hence range $\left( -\infty ,\infty \right)$.

Note: Students should keep in mind all the forms for a basic linear equation. Note that, we can calculate point slope form using the other point (10,3) as well. The standard form and slope intercept form will become the same as found using point (6,0).