Question

The equation of the straight line is perpendicular to $5x-2y=7$ and passing through the point of intersection of the lines $2x+3y=1$ and $3x+4y=6$, is

• A. $2x+5y+17=0$
• B. $2x+5y-17=0$
• C. $2x-5y+17=0$
• D. $2x–5y=17$

Answer 1

The correct option is A

$2x+5y+17=0$

Step 1: Solve the given equations of intersecting lines

Given lines: $2x+3y=1...\left(\mathrm{i}\right)$ ,

$3x+4y=6...\left(\mathrm{ii}\right)$

By multiplying eq.$\left(\mathrm{i}\right)$ by $3$ and eq.$\left(\mathrm{ii}\right)$ by $2$ we get,

$6x+9y=3...\left(\mathrm{iii}\right)$

$6x+8y=12...\left(\mathrm{iv}\right)$

By subtracting eq.$\left(\mathrm{iii}\right)$ from eq.$\left(\mathrm{iv}\right)$ we get,

$y=-9$

By substituting the value of $y$ in eq.$\left(\mathrm{i}\right)$ we get,

$$\begin{gathered} 2x+3(-9)=1 \\ \Rightarrow 2x-27=1 \\ \Rightarrow 2x=28 \\ \Rightarrow x=14 \end{gathered}$$

So the point of intersection of these lines is $(x,y)=(14,-9)$.

Step 2: Find the required equation of line

the required equation of the straight line is perpendicular to $5x-2y=7$

$\Rightarrow y=\frac{5}{2}x–\frac{7}{2}$

By comparing with the equation of line in slope-intercept form $y=mx+c$ we get,

The slope of the given line $m_1=\frac{5}{2}$

We know that if the lines are perpendicular to each other then the product of their slopes is $-1$

So the slope of the required perpendicular line is

$m=\frac{-2}{5}\left[\because \frac{5}{2}\times \frac{-2}{5}=-1\right]$

We also know that the equation of line in point slope form is given by: $y-y_1=m\left(x-x_1\right)$

So we have $(x_1,y_1)=(14,-9)$ and $m=\frac{-2}{5}$

By substituting the values we get,

$$\begin{gathered} y+9=\frac{-2}{5}\left(x-14\right) \\ \Rightarrow 5y+45=-2x+28 \\ \Rightarrow 5y+2x+45-28=0 \\ \Rightarrow 2x+5y+17=0 \end{gathered}$$

Hence, the correct answer is option $\left(a\right)$.