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

`\textsf{Slope form}: \quad y=-(4)/(5)x-(26)/(5)`

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

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 (5)/(4)x-(1)/(4)=(4y)/(4)`

`\implies y=(5)/(4)x-(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=-(4)/(5)(1)+b`

`\implies -6=-(4)/(5)+b`

`\implies b=-(26)/(5)`

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

`y=-(4)/(5)x-(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=-(4)/(5)x\cdot 5-(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`