Question

The equation of a line passing through the point of intersection of the lines $x+5y+7=0,3x+2y-5=0$and perpendicular to the line $7x+2y-5=0$, is given by:

• A. $2x-7y-20=0$
• B. $2x+7y-20=0$
• C. $-2x+7y-20=0$
• D. $2x+7y+20=0$

Answer 1

The correct option is A

$2x-7y-20=0$

Determine the equation of the line :

Let the required line be $AB.$

Given that: $AB$ passes through the point of intersection of $x+5y+7=0\&3x+2y-5=0$

On solving the given two equations simultaneously, we have, $x=3,y=-2$

therefore, The point of intersection is $(3,-2)$.

$AB.$ is perpendicular to $7x+2y-5=0$

Slope of $7x+2y-5=0$ is:

$\begin{array}{rcl}& \Rightarrow & 2y=-7x+5\\ & \Rightarrow & y=-\frac{7}{2}x+\frac{5}{2}\\ & & \\ \therefore \text{slope}& =& -\frac{7}{2}\end{array}$

Slope of $AB=-\frac{1}{-{\displaystyle \frac{7}{2}}}=\frac{2}{7}$ [Since, $m_1\times m_2=-1$]

thus, the equation of $AB$ which passes through $(3,-2)$ and slope $-\frac{7}{2}$ will be:

$$\begin{gathered} y-(-2)=\frac{2}{7}(x-3) \\ \Rightarrow 7(y+2)=2(x-3) \\ \Rightarrow 2x-7y-20=0 \end{gathered}$$

Hence, the correct answer is Option (A).