Question

How do you write the equation of a line in point slope form and slope intercept form given point (6, -3) and has a slope of $\dfrac{1}{2}$?

Answer 1

Assume the given point (6, -3) as $\left( {{x}_{1}},{{y}_{1}} \right)$ and the given slope as ‘m’. Now, first consider the point slope form and write the equation as: - $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$, substitute the values to get the answer. Now, consider the general equation of a line in slope intercept form given as $y=mx+c$, where ‘c’ denotes the intercept on y – axis. Solve for the value of ‘c’ by using the points and slope given and get the required equation.

Complete answer:
Here, we have been given that a line passes through the point (6, -3) and its slope is $\dfrac{1}{2}$. We have been asked to write the equation of the line in point slope form and in slope intercept form.
(1). Let us consider the point form first.
Now, if a line passes through a point $\left( {{x}_{1}},{{y}_{1}} \right)$ and has slope ‘m’ then its equation in point slope form is given as: - $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$. So, considering the given point (6, -3) as $\left( {{x}_{1}},{{y}_{1}} \right)$ and the slope $\dfrac{1}{2}$ as ‘m’ we have the equation of the line: -

& \Rightarrow \left[ y-\left( -3 \right) \right]=\dfrac{1}{2}\left( x-6 \right) \\\ & \Rightarrow \left( y+3 \right)=\dfrac{1}{2}\left( x-6 \right) \\\ \end{aligned}$$ Hence, the above relation represents the equation of line in point slope form. (2). Now, let us consider the slope – intercept form. We know that the general equation of a straight line in slope – intercept form is given as: - $$y=mx+c$$, where ‘c’ is the intercept on y – axis. Now, let us assume $$y=mx+c$$ as the equation of our line. Since, the value of slope (m) is $$\dfrac{1}{2}$$, so we have, $$\Rightarrow y=\dfrac{1}{2}x+c$$ Since, the above line passes through the point (6, -3), therefore this point will satisfy the above equation. So, substituting x and y coordinate in the above relation, we get, $$\begin{aligned} & \Rightarrow -3=\dfrac{1}{2}\times 6+c \\\ & \Rightarrow -3=3+c \\\ & \Rightarrow c=-6 \\\ \end{aligned}$$ So, the equation of the line becomes, $$\begin{aligned} & \Rightarrow y=\dfrac{x}{2}+\left( -6 \right) \\\ & \Rightarrow y=\dfrac{1}{2}x-6 \\\ \end{aligned}$$ Hence, the above relation represents the equation of line in slope – intercept form. **Note:** One may note that we can easily derive any form of a straight line if we know one form of an equation. Here, we can derive the slope – intercept form from the obtained point slope form by simplifying the relation by removing the bracket. Note that we have another form of a straight line known as standard form, represented as $$\dfrac{x}{a}+\dfrac{y}{b}=1$$, where ‘a’ and ‘b’ are intercepts on the x and y – axis respectively. You must remember all the forms of a straight line including parametric form and polar form.