Question

Write the equation of the line perpendicular to 5x-4y=1 that passes through the point (1,-6) in slope form AND in standard form.

Answer 1

Answer:

$\textsf{Slope form}: \quad y=-\dfrac{4}{5}x-\dfrac{26}{5}$

$\textsf{Standard form}: \quad 4x+5y=-26$

Step-by-step explanation:

$\boxed{\begin{minipage}{6.3 cm}\underline{Slope-intercept form of a linear equation}\\\\ y=mx+b \\\\where:\\ \phantom{ww} \bullet m is the slope. \\ \phantom{ww} \bullet b is the y -intercept.\\\end{minipage}}$

Given equation:

$5x-4y=1$

Rewrite in slope-intercept form:

$\implies 5x-4y+4y=1+4y$

$\implies 5x=1+4y$

$\implies 5x-1=1+4y-1$

$\implies 5x-1=4y$

$\implies \dfrac{5}{4}x-\dfrac{1}{4}=\dfrac{4y}{4}$

$\implies y=\dfrac{5}{4}x-\dfrac{1}{4}$

Therefore, the slope of the line is ⁵/₄.

If two lines are perpendicular to each other, their slopes are negative reciprocals.

Therefore, the slope of the perpendicular line is -⁴/₅.

Substitute the found slope -⁴/₅ and given point (1, -6) into the slope-intercept formula and solve for b:

$\implies -6=-\dfrac{4}{5}(1)+b$

$\implies -6=-\dfrac{4}{5}+b$

$\implies b=-\dfrac{26}{5}$

Therefore, the equation of the perpendicular line in slope form is:

$y=-\dfrac{4}{5}x-\dfrac{26}{5}$

$\boxed{\begin{minipage}{5.5 cm}\underline{Standard form of a linear equation}\\\\ Ax+By=C \\\\where:\\ \phantom{ww} \bullet A, B and C are constants. \\ \phantom{ww} \bullet A must be positive.\\\end{minipage}}$

Multiply both sides of the equation in slope form by 5:

$\implies y \cdot 5=-\dfrac{4}{5}x\cdot 5-\dfrac{26}{5}\cdot 5$

$\implies 5y=-4x-26$

Add 4x to both sides:

$\implies 5y+4x=-4x-26+4x$

$\implies 4x+5y=-26$

Therefore, the equation of the perpendicular line in standard form is:

$4x+5y=-26$