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Write an equation for a line perpendicular to y = 3.x – 4 and passing through the point (-6,4) y =

User CJR
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4 votes

Answer:


y=-(1)/(3)x+2

Step-by-step explanation:

First, compare the given equation with the slope-intercept form: y=mx+b


\begin{gathered} y=3x-4 \\ \implies\text{Slope of the line, m = 3} \end{gathered}

Let the slope of the perpendicular line = n.

Two lines are perpendicular if the product of their slopes is -1.


\begin{gathered} 3* n=-1 \\ n=-(1)/(3) \end{gathered}

Thus, we find an equation for a line with a slope of -1/3 and passing through (-6,4).

Using the slope-point form of the equation of a line:


y-y_1=m(x-x_1)

Substitute the point and slope:


\begin{gathered} y-4=-(1)/(3)\lbrack x-(-6)\rbrack \\ y-4=-(1)/(3)\lbrack x+6\rbrack \\ y=-(1)/(3)x-2+4 \\ y=-(1)/(3)x+2 \end{gathered}

The required equation of the line is:


y=-(1)/(3)x+2

User Kakata Kyun
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