Question

Write an equation (a) in slope intercept form and (b) in standard form for the line passing through (1,9) and perpendicular to 3x+5y=1.

Answer 1

`\large\boxed{a)\ y=(5)/(3)x+(22)/(3)}\\\boxed{b)\ 5x-3y=-22}`

Explanation:

`\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\m-slope\\b-y-intercept\\\\\text{Let}\ k:y=m_1x+b_1,\ l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)\\\\l\ \parallel\ k\iff m_1=m_2\\\\=========================\\\\\text{We have the equation of a line in the standard form.}\\\text{Convert it to the slope-intercept form.}\\\\3x+5y=1\qquad\text{subtract}\ 3x\ \text{from both sides}\\\\5y=-3x+1\qquad\text{divide both sides bvy 5}\\\\y=-(3)/(5)x+(1)/(5)\to m_1=-(3)/(5)`

`a)\\\\y=m_2x+b\\\\m_1=-(3)/(5)\to m_2=-(1)/(-(3)/(5))=(5)/(3)\\\\\text{Put the value of slope and the coordinates of the given point (1, 9)}\\\text{to the equation of a line:}\\\\9=(5)/(3)(1)+b\\\\9=(5)/(3)+b\qquad\text{subtract}\ (5)/(3)\ \text{from both sides}\\\\(27)/(3)-(5)/(3)=b\\\\(22)/(3)=b\\\\\text{Finally:}\\\\y=(5)/(3)x+(22)/(3)`

`b)\\\\\text{The standard form of an equation of a line:}\\\\Ax+By=C\\\\\text{Convert the equation}\ y=(5)/(3)x+(22)/(3)\ \text{to the standard form:}\\\\y=(5)/(3)x+(22)/(3)\qquad\text{multiply both sides by 3}\\\\3y=5x+22\qquad\text{subtract}\ 5x\ \text{from both sides}\\\\-5x+3y=22\qquad\text{change the signs}\\\\5x-3y=-22`