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

`\huge\boxed{y=-(1)/(4)x-1\to x+4y=-4}`

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=-(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=(-x)/(4)-(6)/(4)\\\\y=-(1)/(4)x-(3)/(2)\to m_1=-(1)/(4)`

`\text{Lines are to be parallel. Therefore}\ m_2=-(1)/(4).\\\\\text{We initially have the form equation}\ y=-(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=-(1)/(4)(9)+b\\\\-3=-(9)/(4)+b\qquad\text{add}\ (9)/(4)\ \text{to both sides}\\\\-(12)/(4)+(9)/(4)=b\to b=-(3)/(4)`

`\text{Lines are to be parallel. Therefore}\ m_2=-(1)/(4).\\\\\text{We initially have the form equation}\ y=-(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=-(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=-(1)/(4)x-1.\\\\\text{Convert to the standard form}\ Ax+By=C:\\\\y=-(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`