Question

Which is an equation of the line perpendicular to y+5X=7and passes through (10,-4)[A] y = 1/5x +7[B] y = 5x + 25/4[C] y = 1/5x - 6[D] y = 5x + 7

Answer 1

Given the equation:

`y+5x=7`

we can find its slope if we write it in the y=mx+b form:

`\begin{gathered} y+5x=7 \\ \Rightarrow y=-5x+7 \end{gathered}`

Now, we know as a general rule, that the slope of the perpendicular of the line that has slope m, is -1/m, more clearly:

`\begin{gathered} \text{if m is the slope of the line} \\ \Rightarrow m_p=-(1)/(m)\text{ is the slope of the perpendicular line} \end{gathered}`

So, in this case we have:

`\begin{gathered} m=-5 \\ \Rightarrow m_p=-(1)/(m)=-(1)/(-5)=(1)/(5) \\ m_p=(1)/(5) \end{gathered}`

now we use the slope-point formula to find the equation of the perpendicular line:

`\begin{gathered} (x_0,y_0)=(10,-4) \\ m_p=(1)/(5)_{} \\ y-y_0=m(x-x_0)_{} \\ \Rightarrow y-(-4)=(1)/(5)(x-10) \\ \Rightarrow y+4=(1)/(5)x-(10)/(5) \\ \Rightarrow y=(1)/(5)x-2-4=(1)/(5)x-6 \\ y=(1)/(5)x-6 \end{gathered}`

therefore, the line perpendicular to y+5x=7 that passes through (10,-4) is y=1/5x-6