Question

The equation of a line is $3x+4y-7=0$. Find the equation of a line perpendicular to the given line and passing through the intersection of the lines $x-y+2=0$and $3x+y-10=0$.

Answer 1

Step1-Finding the slope of the given line:
Given equation of line is $3x+4y-7=0$

$\Rightarrow 4y=-3x+7$
$\Rightarrow y=\frac{-3}{4}x+\frac{7}{4}$
On comparing with general equation of a line $y=mx+c$, we get:
$m=\frac{-3}{4}$(where ‘$m$’ is the slope of the line)

Hence, slope of the given line is $\frac{-3}{4}$

Step 2: Finding the intersection point of the lines $x-y+2=0$ and $3x+y-10=0$:

The given equations are: $x-y+2=0$ and $3x+y-10=0$

Adding both the equations, we get:
$x-y+3x+y=-2+10$

$\Rightarrow 4x=8$

$\Rightarrow x=2$
Substituting $x=2$ in $3x+y=10$

$3\times 2+y=10$
$y=4$
Thus, point of intersection is $\left(2,4\right)$

Step 3: Finding the slope of the required line:
Given that the line $3x+4y-7=0$ and the required line are perpendicular to each other.
$\Rightarrow m_1.m_2=-1$ (where $m_2$ is the slope of the required line)
$\Rightarrow \frac{-3}{4}\times m_2=-1$
$\Rightarrow m=\frac{4}{3}$

Step 4: Finding the equation of the required line:
We know that, the slope of the required line is $\frac{4}{3}$and it passes through the point $\left(2,4\right)$.
Equation of a line in slope-point form is given as:
$y-y_1=m\left(x-x_1\right)$
$\Rightarrow y-4=\frac{4}{3}\left(x-2\right)$
$\Rightarrow 3y-12=4x-8$
$\Rightarrow 4x-3y+4=0$

Hence, equation of the required line is $4x-3y+4=0$.