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

Answer:

$\large\boxed{a)\ y=\dfrac{5}{3}x+\dfrac{22}{3}}\\\boxed{b)\ 5x-3y=-22}$

Step-by-step explanation:

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

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

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