Question

How do you graph the line $5x-2=0$? $$$$

Answer 1

We recall the definition of $x$ and $y-$ coordinate of the point. We use the fact that that the locus of all points equidistant from a line will be a line parallel to the original line and deduce that $5x-2=0\Rightarrow x=\dfrac{2}{5}$ is a line parallel to $y-$axis at a distance $\dfrac{2}{5}$ from $y-$axis passing through point $\left( \dfrac{2}{5},0 \right)$.$$$$

Complete answer:
We know that all points in plane are represented as the ordered pair $\left( a,b \right)$ where $\left| a \right|$ is the distance from $y-$axis (called as abscissa or $x-$coordinate) and $\left| b \right|$ is the distance of the point from the $y-$axis (called as ordinate or $y-$coordinate). We are given the line $5x-2=0\Rightarrow x=\dfrac{2}{5}$ in the question. Here $x=\dfrac{2}{5}$ means all the points on the line $x=\dfrac{5}{2}$the $x-$coordinate of the points will remain same irrespective of the $y-$ordinate which means $x=\dfrac{5}{2}$ is the locus of the points of the type $\left( \dfrac{5}{2},b \right)$ where $b\in R$.
We know that locus of all points equidistant from a line will be a line parallel to the original line. Since $x-$coordinate which is also the distance $\left( \left| \dfrac{5}{2} \right|=2 \right)$ from $y-$axis is constant, all the points of the type $\left( \dfrac{5}{2},b \right)$ from $y-$axis will be equidistant from the points of the type $\left( 0,b \right)$. So the distance between $y-$axis and $5x-2=0$ is constant and hence $5x-2=0$ is line parallel to $y-$axis . So we the line $5x-2=0$ will also pass through $\left( \dfrac{2}{5},b \right)$ for any $b$ it will pass through $\left( \dfrac{2}{5},0 \right)$where it will cut $x-$axis. We can take any other $b$ say $b=1$ to have $\left( \dfrac{2}{5},1 \right)$ and join them using a line to graph the line. $$$$

Note: We know that the general equation of line is $ax+by+c=0$ and the line parallel to it is given by $ax+by+k=0,k\ne c$. Since the equation of the $y-$axis is $x=0$ line parallel to it will be $x=k,k\ne 0$. We can alternatively graph by finding the slope of the line $ax+by+c=0$ as $\dfrac{-a}{b}$ . Here we have from $0\cdot y+5\cdot x-2=0$ slope $\dfrac{5}{0}$ which is undefined and we know that a line with undefined slope is perpendicular to $x-$axis and we get point $\left( \dfrac{2}{5},0 \right)$ to draw the perpendicular line.