Question

find the equation of the line that is parallel to the given line and passes through the given point
$x+4y=-6 ; (8, -3)$

Answer 1

Answer:

$\huge\boxed{y=-\dfrac{1}{4}x-1\to x+4y=-4}$

Step-by-step explanation:

$\text{Let}\ k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\l\ ||\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\==========================\\\\\text{We have}\ x+4y=-6.\\\text{Convert to the slope-intercept form:}\\\\x+4y=-6\qquad\text{subtract}\ x\ \text{from both sides}\\\\4y=-x-6\qquad\text{divide both sides by 4}\\\\y=\dfrac{-x}{4}-\dfrac{6}{4}\\\\y=-\dfrac{1}{4}x-\dfrac{3}{2}\to m_1=-\dfrac{1}{4}$

$\text{Lines are to be parallel. Therefore}\ m_2=-\dfrac{1}{4}.\\\\\text{We initially have the form equation}\ y=-\dfrac{1}{4}x+b.\\\\\text{The line passes through the point}\ (9,\ -3).\\\\\text{Substitute the coordinates of the point to the equation of a line:}\\\\x=9,\ y=-3\\\\-3=-\dfrac{1}{4}(9)+b\\\\-3=-\dfrac{9}{4}+b\qquad\text{add}\ \dfrac{9}{4}\ \text{to both sides}\\\\-\dfrac{12}{4}+\dfrac{9}{4}=b\to b=-\dfrac{3}{4}$

$\text{Lines are to be parallel. Therefore}\ m_2=-\dfrac{1}{4}.\\\\\text{We initially have the form equation}\ y=-\dfrac{1}{4}x+b.\\\\\text{The line passes through the point}\ (8,\ -3).\\\\\text{Substitute the coordinates of the point to the equation of a line:}$

$x=8,\ y=-3\\\\-3=-\dfrac{1}{4}(8)+b\\\\-3=-2+b\qquad\text{add 2 to both sides}\\\\-1=b\to b=-1$

$\text{Therefore the equation is:}\ y=-\dfrac{1}{4}x-1.\\\\\text{Convert to the standard form}\ Ax+By=C:\\\\y=-\dfrac{1}{4}x-1\qquad\text{multiply both sides by 4}\\\\4y=-x-4\qquad\text{add}\ x\ \text{to both sides}\\\\x+4y=-4$