Question

Find the equation of a line, which has the $y$-intercept $4$; and is parallel to the line $2x-3y-7=0$. Hence, find the co-ordinates of the point, where it cuts $x-axis$.

Answer 1

Step 1: Finding the slope of the given line:

The equation of the given line is $2x-3y-7=0$
$\Rightarrow 2x-7=3y$
$\Rightarrow y=\frac{2}{3}x-\frac{7}{3}$
Comparing with $y=mx+c$,

we get,
$m=\frac{2}{3}$ (where ‘m’ is the slope of the given line)

Step 2: Finding the equation of the required line using Slope-Intercept form:

Since the lines are parallel. Therefore, the slope of the given line is equal to the slope of the required line
$y$-intercept of the given line $\left(c\right)=4$
Equation of the line in Slope-Intercept form is given as:
$y=mx+c$
Substituting known values in the above equation, we get:
$y=\frac{2}{3}x+4$

Step 3: Finding the coordinates of the point where the required line cuts $x-axis$:
The required line will cut the $x-axis$ at $y=0$
Thus, substituting $y=0$in the equation $y=\frac{2}{3}x+4$, we get:
$0=\frac{2}{3}x+4$
$\Rightarrow \frac{2}{3}x=-4$
$\Rightarrow x=-6$

Hence, the equation of the required line is $y=\frac{2}{3}x+4$and it cuts the X-axis at $\left(-6,0\right)$.