Question

Write an equation for a line perpendicular to y=3x+1 and passing through the point (6,2)

Answer 1

Answer:

$y = -\frac{1}{3}x + 4$

Step-by-step explanation:

Required

Equation of line

passes through $(6,2)$

In an equation of the form $y =mx + b$; the slope is $m$

So, by comparison;

The slope of $y = 3x + 1$ is: $m =3$

From the question, we understand that the required equation is perpendicular to $y = 3x + 1$

This means that its slope is:

$m_2 =-\frac{1}{m}$

So, we have:

$m_2 =-\frac{1}{3}$

The line equation is:

$y = m_2(x - x_1) + y_1$

Where:

$(x_1,y_1) = (6,2)$

So, we have:

$y = -\frac{1}{3}(x - 6) + 2$

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

$y = -\frac{1}{3}x + 4$