Question

Which equation of a line below is perpendicular to –7x – 8y = 12 and contains P(–3, 1).

Answer 1

`\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \parallel\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)\\\\----------------------\\\\\text{The slope-intercept form:}\ y=mx+b.\\\\\text{We have}\ -7x-8y=12.\ \text{Convert to the slope-intercept form:}\\\\-7x-8y=12\qquad\text{add 7x to both sides}\\\\-8y=7x+12\qquad\text{divide both sides by (-8)}\\\\y=-(7)/(8)x-(12)/(8)\to m_1=-(7)/(8)\\\\\text{Therefore}\\\\m_2=-(1)/(-(7)/(8))=(8)/(7)`

`\text{We have the equation of a line:}\\\\y=(8)/(7)x+b\\\\\text{Put the coordinates of the given point P(-3, 1) to the equation:}\\\\1=(8)/(7)(-3)+b\\\\1=(-24)/(7)+b\\\\(7)/(7)=-(24)/(7)+b\qquad\text{add}\ (24)/(7)\ \text{to both sides}\\\\(31)/(7)=b\\\\Answer:\ \boxed{y=(8)/(7)x+(31)/(7)}`