Question

Find the equation of a line that is perpendicular to line $g$ that contains $\left(\text{P,Q}\right)$. Coordinate plane with line $g$ that passes through the points $(-2,6)$and $(-3,2)$.

• A. $4x+y=Q+4P$
• B. $x-4y=-4Q+P$
• C. $-4x+y=Q-4P$
• D. $x+4y=4Q+P$

Answer 1

The correct option is D

$x+4y=4Q+P$

Explanation for the correct option.

Option (D):

Find the slope of line $g$ using the formula $\text{slope}=\frac{y_2-y_1}{x_2-x_1}$.

Let, $m_1$be the slope of $g$.

Since, the line $g$ passes through the points $(-2,6)$and $(-3,2)$.

Therefore, the slope, $\begin{array}{rcl}{m}_1& =& \frac{2-6}{-3-\left(-2\right)}\end{array}$

$\begin{array}{rcl}& =& \frac{-4}{-1}\\ & =& 4\end{array}$

It is known that the product of the slope of perpendicular lines is $-1$.

Let, $m_2$ be the slope of the line perpendicular to $g$.

$\begin{array}{rcl}{m}_1\times m_2& =& -1\\ & \Rightarrow & 4\times m_2=-1\\ & \Rightarrow & m_2=-\frac{1}{4}\end{array}$

The slope of the line, perpendicular to line $g$, is $-\frac{1}{4}$and the point $\left(P,Q\right)$is on the line.

The slope point form is $\left(y-y_1\right)=m\left(x-x_1\right)$. Here $m$ is the slope and $\left(x_1,y_1\right)$is the point on the line.

The equation of the line perpendicular to the line $g$ is as follows.

$\left(y-Q\right)=-\frac{1}{4}\left(x-P\right)$

Simplify the equation:

$\begin{array}{rcl}\left(y-Q\right)& =& -\frac{1}{4}\left(x-P\right)\\ & \Rightarrow & 4y-4Q=-x+P\\ & \Rightarrow & x+4y=4Q+P\end{array}$

Equation of line having point $\left(P,Q\right)$ on it and perpendicular to the line $g$ is $x+4y=4Q+P$.

Therefore, the correct option is option (D).