Question

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

Answer 1

`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`