Question

Find the equation of a straight line passing through the intersection of $2x+5y-4=0$with X-axis and parallel to the line $3x-7y+8=0$.

Answer 1

Step 1: Finding the point of intersection of the line $2x+5y-4=0$with $\mathit{x}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$:
The given line will intersect with $x-axis$ at $y=0$
Substituting $y=0$in the equation, we get:
$2x-4=0$
$\Rightarrow x=2$
Thus, the point of intersection will be $\left(2,0\right)$.

Step 2: Finding the slope required line:
The equation of a line parallel to the required line is:
$3x-7y+8=0$

$\Rightarrow 7y=3x+8$
$\Rightarrow y=\frac{3}{7}x+\frac{8}{7}$

Comparing with $y=mx+c$, we get:
$m=\frac{3}{7}$ (where ‘m’ is the slope)
Since the required line is parallel to the above line, their slopes are equal.
Therefore, the slope of the required line $=m=\frac{3}{7}$

Step 3: Finding the equation of the required line using Slope-Point form:
We know that, slope of the required line is $\frac{3}{7}$and it passes through the point $\left(2,0\right)$
Let $x_1=2$and $y_1=0$
The equation of a line in slope-point form is given as:

$y-y_1=m\left(x-x_1\right)$

$\Rightarrow y-0=\frac{3}{7}\left(x-2\right)$

$\Rightarrow 7y=3x-6$

$\Rightarrow 3x-7y-6=0$

Therefore, the equation of the required line is $3x-7y-6=0$.