Question

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

Answer 1

Answer:

$y=-\frac{1}{3}x+2$

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.

$\begin{gathered} 3\times n=-1 \\ n=-\frac{1}{3} \end{gathered}$

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:

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

The required equation of the line is:

$y=-\frac{1}{3}x+2$