Q6. - Algebra 1

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 $\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{)}$?

 

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

 

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

 

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

 

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

Step-by-Step Solution

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 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 4 x − 2 y = 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.