Question

Which of the following is an equation of the line perpendicular to$\mathbf{4}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{6}$ and passing through$\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{)}$?

$\mathit{F}$$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}\mathit{x}\mathbf{+}\mathbf{3}$

$\mathit{G}$$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}\mathit{x}\mathbf{-}\mathbf{1}$

$\mathit{H}$$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{-}\mathbf{4}$

$\mathit{J}$$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{-}\mathbf{2}$

Short answer

The equation of the line perpendicular to$4x-2y=6$ and passing through$(4,-4)$ is

$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{-}\mathbf{2}$.

Option $\mathit{J}$is correct.

Step-by-step solution

Step 1. State the concept of equation of line in ‘Slope intercept form’.

Suppose ‘$m$’ is the slope of the a and ‘$c$’ is the y-intercept of a line , then the equation of the line in slope intercept form is given as,

$y=mx+c\dots \left(1\right)$.

Step 2. State the concept of equation of line in ‘Slope-point form’.

Suppose a line has a slope ‘$m$’ and passes through the point, then the equation of the line in slope point is given as,

$(y-y_1)=m(x-x_1)\dots \left(2\right)$

Step 3. State the concept of ‘perpendicularity of two lines’.

If two lines are pendicular to each other, then the product of their slopes is equal to ‘$-1$’.

That is, suppose $m_1$is the slope of line one and $m_2$is the slope of line two,

Then$m_1\left(m_2\right)=-1\dots \left(3\right)$

Step 3. Calculate the equation of the line perpendicular to 4x−2y=6 and passing  through (4,−4).

First write the equation $4x-2y=6$in slope intercept form and find its slope.

$\begin{array}{l}4x-2y=6\\ 4x-6=2y\\ 2y=4x-6\\ \frac{2y}{2}=\frac{4x-6}{2}\\ y=\frac{4x}{2}-\frac{6}{2}\\ y=2x-3\end{array}$

On comparing $y=2x-3$with $y=mx+c$

The slope of the line $y=2x-3$is $m=2$.

Let $m_1$be the slope of the line perpendicular to the line $y=2x-3$.

$\begin{array}{l}m\left(m_1\right)=-1\left[\text{Using\hspace{0.17em}\hspace{0.17em}}3\right]\\ 2\left(m_1\right)=-1\\ m_1=-\frac{1}{2}\end{array}$

The line passes through the point $(4,-4)$and has a slope $m_1=-\frac{1}{2}$.

Substitute $(4,-4)$as $(x_1,y_1)$and $-\frac{1}{2}$as m in (2) and find the required equation.

Therefore, by slope point form

$\begin{array}{l}(y-(-4))=-\frac{1}{2}(x-4)\\ y+4=\left(-\frac{1}{2}\right)\left(x\right)-\left(-\frac{1}{2}\right)\left(4\right)\\ y+4=-\frac{1}{2}x+\frac{4}{2}\\ y+4=-\frac{1}{2}x+2\\ y=-\frac{1}{2}x+2-4\\ y=-\frac{1}{2}x-2\end{array}$

Therefore, the required equation $y=-\frac{1}{2}x-2$.

Option $J$is correct.