Question

Find the equation of the line passing the poin (-7,2) that is perpendicular to line 4x-3y=10

Answer 1

`y=-(3)/(4)x-(13)/(4)`

Step-by-step explanation

Step 1

find the slope of the given equation:

two lines are perpendicular it the product of the slopes equals -1, so

`\begin{gathered} \text{ line 1 is perpendicular to line 2} \\ L1\parallel L2 \\ \text{if} \\ m_1\cdot m_2=-1 \end{gathered}`

then , let

`\begin{gathered} \text{ Line 1} \\ 4x-3y=10 \end{gathered}`

to know the slope, we need to convert the equation into the form:

`\begin{gathered} y=mx+b \\ \text{where m is the slope} \end{gathered}`

to do that, let's isolate y

so

`\begin{gathered} 4x-3y=10 \\ \text{subtract 4x in both sides} \\ 4x-3y-4x=10-4x \\ -3y=10-4x \\ \text{divide both sides by -3} \\ (-3y)/(-3)=(10-4x)/(-3) \\ y=-(10)/(3)+(4)/(3)x \\ y=(4)/(3)x-(10)/(3) \\ y=(4)/(3)x-(10)/(3)\rightarrow y=\text{ mx+b} \end{gathered}`

hence,

`\text{slope 1=m}_1=(4)/(3)`

Step 2

now, let's find the slope of the line 2

`\begin{gathered} \text{let} \\ m_1=(4)/(3) \\ \end{gathered}`

replacing

`\begin{gathered} m_1\cdot m_2=-1 \\ (4)/(3)\cdot m_2=-1 \\ \text{ to isolate, multiply both sides by 3/4} \\ (4)/(3)\cdot m_2\cdot(3)/(4)=-1\cdot(3)/(4) \\ m_2=-(3)/(4) \end{gathered}`

so, the slope 2 is -3/4

Step 3

finally, let's find the equation of the line

use the expression

`\begin{gathered} y-y_0=m(x-x_0) \\ \text{where} \\ \text{ m is the slope} \\ \text{and (x}_0,y_0)\text{ are the coordinates of a known point} \end{gathered}`

let

`\begin{gathered} m=m_2=-(3)/(4) \\ (x_0,y_0)=(-7,2) \end{gathered}`

replace and isolate y

`\begin{gathered} y-y_0=m(x-x_0) \\ y-2=-(3)/(4)(x-(-7)) \\ y-2=-(3)/(4)(x+7) \\ y-2=-(3)/(4)x-(21)/(4) \\ \text{add 2 in both sides} \\ y-2+2=-(3)/(4)x-(21)/(4)+2 \\ y=-(3)/(4)+(-21+8)/(4) \\ y=-(3)/(4)x-(13)/(4) \\ y=-(3)/(4)x-(13)/(4) \end{gathered}`

therefore, the answer is

`y=-(3)/(4)x-(13)/(4)`

I hope this helps you