Question

What is the equation of the line that is perpendicular to the line 3x+y= -8 and passes through the point (2, 2)?

Answer 1

`x-3y+4=0`

Explanation:

`\mathrm{Let\ }m_1\ \mathrm{be\ the\ slope\ of\ the\ line}\ 3x+y=-8.\\\mathrm{Let\ }m_2\ \mathrm{be\ the\ slope\ of\ the\ line\ perpendicular\ to\ }3x+y=-8.\\\mathrm{From\ the\ condition\ of\ perpendicular\ lines,}\\m_1.m_2=-1\\\mathrm{or,\ }(-3)m_2=-1\\\mathrm{or,\ }m_2=(1)/(3)`

`\mathrm{Equation\ of\ the\ line\ having\ slope\ (1)/(3)\ and\ passing\ through\ (2,2)\ is:}\\\mathrm{y-2=(1)/(3)(x-2)}\\\\\mathrm{or,\ }3y-6=x-2\\\\\mathrm{or,\ }x-3y+4=0\mathrm{\ is\ the\ required\ equation.}`

Answer 2

Info required for the question

If one line is perpendicular to another one, then their slopes are opposite reciprocals.

To find the opposite reciprocal of a number, we change its sign, and flop it over, like this:

(Let's say we're looking for the opposite reciprocal of 4).

So first, I change the sign:

-4

Then, I flop it over:

-1/4

_________________

Now, we should be able to find the slope of the line which is perpendicular to the given line, i.e., 3x + y =-8.

First, I'll write its equation in slope-intercept:

y = -8 - 3x

y = -3x - 8

Now, the slope is the number before x, i.e., -3.

The opposite reciprocal of -3 is:

(changing the sign) -3 ==> 3

(flopping it over) 3 ==> 1/3

Now, we have all the information that is required for writing the equation in point-slope form. The format of point-slope form is
`\sf{y-y_1=m(x-x_1)}`.

Where:

m = slope

y₁ = y-coordinate of the point

x₁ = x-coordinate of the point

Here:

m = 1/3

(x₁, y₁) = (2,2)

Plug in the data:

`\large\begin{gathered}\sf{y-2=(1)/(3)(x-2)}\\\sf{y-2=(1)/(3)x-(2)/(1)}\\\sf{y-2=(1)/(3)x-((1)/(3)*(2)/(1))\\\sf{y-2=(1)/(3)x-(2)/(3)}\\\sf{y=(1)/(3)x-(2)/(3)+(2)/(1)}\\\sf{y=(1)/(3)x-(2*2)/(3*2)+(2*6)/(1*6)\\\sf{y=(1)/(3)x-(4)/(6)+(12)/(6)\\\sf{y=(1)/(3)x+(8)/(6)\\\sf{y=(1)/(3)x+(4)/(3)\\\\\bigstar\end{gathered}`

Hence, the equation is y = 1/3x + 4/3.