Question

How do you find the intercepts of $4y=2x+6$?

Answer 1

Change of form of the given equation will give the y intercept and x-intercept of the line $4y=2x+6$. We change it to the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$ to find the x intercept, and y intercept of the line as p and q respectively. Then we place the line on the graph based on that.

Complete step by step answer:
We are taking the general equation of line to understand the intercept form of the line $4y=2x+6$.
The given equation $4y=2x+6$ is of the form $ax+by=c$. Here a, b, c are the constants.
Now we have to find the y intercept, and x-intercept of the same line $4y=2x+6$.
For this we convert the given equation into the form of $\dfrac{x}{p}+\dfrac{y}{q}=1$. From the form we get that the x intercept, and y intercept of the line will be p and q respectively.
The intersecting points for the line with the axes will be $\left( p,0 \right)$ and $\left( 0,q \right)$.
The given equation is $4y=2x+6$.
$\begin{aligned} & 4y=2x+6 \\\ & \Rightarrow 2x-4y=-6 \\\ \end{aligned}$
Dividing both sides with -6 we get
$\begin{aligned} & 2x-4y=-6 \\\ & \Rightarrow \dfrac{2x}{-6}+\dfrac{-4y}{-6}=1 \\\ & \Rightarrow \dfrac{x}{-3}+\dfrac{y}{{}^{3}/{}_{2}}=1 \\\ \end{aligned}$
The intersecting points for the line $4y=2x+6$ with the axes will be $\left( -3,0 \right)$ and $\left( 0,\dfrac{3}{2} \right)$.

Therefore, the x intercept, and y intercept of the line $2x+y=5$ is 4 and 8 respectively.

Note: A line parallel to the X-axis does not intersect the X-axis at any finite distance and hence we cannot get any finite x-intercept of such a line. Same goes for lines parallel to the Y-axis. In case of slope of a line the range of the slope is 0 to $\infty$.