Question

Draw the graphs of the following equations on the same system of co-ordinates. Write the co-ordinates of their points of intersection.

x + 4 = 0, y $-$1 = 0, 2x + 3 = 0, 3y $-$ 15 = 0

Answer 1

x + 4 = 0 ⇒ x = −4
x = −4 is the equation of the line parallel to Y-axis at a distance of 4 units from it to its left.
y − 1 = 0 ⇒ y = 1
y = 1 is the equation of the line parallel to X-axis at a distance of 1 unit and above the X-axis.
2x + 3 = 0 ⇒ x = $-\frac{3}{2}$
x = $-\frac{3}{2}$ is the equation of the line parallel to Y-axis at a distance of $\frac{3}{2}$ units from it to its left.
3y − 15 = 0 ⇒ y = $\frac{15}{3}$ = 5
y = 5 is the equation of the line parallel to X-axis at a distance of 5 units and above the X-axis.


It can be seen from the figure that the co-ordinates of the points of intersection are $\left(-\frac{3}{2},1\right)$, $\left(-\frac{3}{2},5\right)$, (−4, 5) and (−4, 1).