Question

Find the equation of the line passing through the point of intersection of$7x+6y=71$ and $5x-8y=-23$ and perpendicular to the line$4x-2y=1.$

Answer 1

Step 1- Finding the intersection point :

As the line passes through the point of intersection of $7x+6y=71$ and $5x-8y=-23$

We need to Solve the equations to get the point of intersection,

$7x+6y=71$ -(I)

$5x-8y=-23$ -(II)

Multiply eq.(I) by $5$ and Eq.(II) by $7$ and then subtracting we get

$35x+30y=355$ -(III)

$35x-56y=-161$ -(IV)

Subtract eq.(III) and eq.(IV) we get,

$86y=516$

$\Rightarrow y=6$

Put value of $y$ in eq.(I) we get,

$$\begin{gathered} \Rightarrow 7x+6\left(6\right)=71 \\ \Rightarrow 7x=71-36 \\ \Rightarrow 7x=35 \\ \Rightarrow x=5 \end{gathered}$$

Therefore, the point of intersection is$(5,6)$

Step2- Finding the slope :

The given line is $4x-2y=1$

On rearranging we get ,

$$\begin{gathered} -2y=-4x+1 \\ \Rightarrow y=2x-\frac{1}{2} \end{gathered}$$

Now, on comparing with the general equation of a line $y=mx+c$, which on comparing gives the slope as $2$

Given that the required line is perpendicular to line$4x-2y=1$ having slope as $2$

Thus , the multiplication of their slopes will be $-1$

$$\begin{gathered} \Rightarrow m\times 2=-1 \\ \Rightarrow m=\frac{-1}{2} \end{gathered}$$

Therefore , the slope of line will be $\frac{-1}{2}$

Step3- Find the equation of line:

The general equation of a Line is given by $\left(\mathrm{y}-{\mathrm{y}}_1\right)=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)$

Here ${\text{x}}_1=5,{\mathrm{y}}_1=6,\mathrm{m}=\frac{-1}{2}$

Equation of a Line:

$$\begin{gathered} \text{⇒ (y-6) =}\frac{-1}{2}(\mathrm{x}-5) \\ \Rightarrow 2(\mathrm{y}-6)=-(\mathrm{x}-5) \\ \Rightarrow 2\mathrm{y}-12=-\mathrm{x}+5 \\ \Rightarrow \mathrm{x}+2\mathrm{y}-17=0 \end{gathered}$$

Therefore, equation of a line is $x+2y-17=0$