Question

Find an equation of a line through (−5, 4) and parallel to 3x − 6y = 3.

Answer 1

Solution:

The equation of a line, in slope-intercept form, is expressed as

$\begin{gathered} y=mx+c\text{ ----- equation 1} \\ where \\ m\Rightarrow slope\text{ of the line} \\ c\Rightarrow y-intercept\text{ of the line} \end{gathered}$

Suppose lines A and B are parallel to each other, their slopes become equal.

This implies that

$\begin{gathered} m_A=m_B\text{ ----- equation 2} \\ where \\ m_A\Rightarrow slope\text{ of line A} \\ m_B\Rightarrow slope\text{ of line B} \end{gathered}$

Given that a line passes through (-5, 4) and is parallel to

$3x-6y=3$

Step 1: Let the equation of line A be

$3x-6y=3$

Step 2: Express the equation of line A in the slope-intercept form as expressed in equation 1.

Thus,

$\begin{gathered} 3x-6y=3 \\ add\text{ -3x to both sides of the equation,} \\ -3x+3x-6y=-3x+3 \\ \Rightarrow-6y=-3x+3 \\ divide\text{ both sides by the coefficient of y, which is -6} \\ -\frac{6y}{-6}=\frac{-3x+3}{-6} \\ \Rightarrow y=\frac{1}{2}x-\frac{1}{2}\text{ ---- equation 3} \end{gathered}$

Thus, the equation of line A is expressed as

$y=\frac{1}{2}x-\frac{1}{2}$

Step 3: Determine the slope of line A.

Comparing equations 1 and 3, we have

$m=\frac{1}{2}$

Thus, the slope of line A is

$m_A=\frac{1}{2}$

Step 4: Determine the slope of line B.

Recall that lines A and B are parallel,

$\begin{gathered} m_A=m_B \\ \Rightarrow m_B=\frac{1}{2} \end{gathered}$

Step 5: Express the equation of line B.

The equation of a line that passes through a point is expressed as

$\begin{gathered} y-y_1=m(x-x_1)\text{ ----equation 4} \\ where \\ m\Rightarrow slope\text{ of the line} \\ (x_1,y_1)\text{ }\Rightarrow coordinate\text{ of the point through which the line passes} \end{gathered}$

Given that line B passes through the point (-5,4), this implies that

$\begin{gathered} x_1=-5 \\ y_1=4 \end{gathered}$

Thus, the equation of the line is expressed as

$\begin{gathered} y-4=m_B(x-(-5)) \\ \Rightarrow y-4=m_B(x+5) \\ add\text{ 4 to both sides of the equation,} \\ y-4+4=m_B(x+5)+4 \\ \Rightarrow y=m_B(x+5)+4 \\ but\text{ m}_B=\frac{1}{2} \\ thus, \\ y=\frac{1}{2}(x+5)+4 \\ open\text{ parentheses,} \\ y=\frac{1}{2}x+\frac{5}{2}+4 \\ \Rightarrow y=\frac{1}{2}x+\frac{13}{2}\text{ ----- equation 5} \end{gathered}$

Hence, the equation of the line is expressed as

$y=\frac{1}{2}x+\frac{13}{2}$

The third option is the correct answer.