Question

How do you write an equation of a line passing through (-6, 1), perpendicular to `y = –3x + 1`?

Answer 1

`y=1/3x+3`

Explanation:

Given 2 lines with gradients `m_1" and " m_2`. If the lines are perpendicular, then.

`color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(m_1xxm_2=-1)color(white)(a/a)|)))........ (A)`

The equation of a line in `color(blue)"slope-intercept form"` is

`color(red)(|bar(ul(color(white)(a/a)color(black)(y=mx+b)color(white)(a/a)|)))`
where m represents the gradient and b, the y-intercept.

The equation here is in this form `rArrm=-3`

Using (A) The gradient of the line perpendicular to this is

`m_("perp")=(-1)/(-3)=1/3`

The equation of a line in `color(blue)"point-slope form"` is

`color(red)(|bar(ul(color(white)(a/a)color(black)(y-y_1=m(x-x_1))color(white)(a/a)|`
where `(x_1,y_1)" is a point on the line"`

here `m=1/3" and " (x_1,y_1)=(-6,1)`

substitute into point-slope form of the equation.

`y-1=1/3(x+6)rArry-1=1/3x+2`

`rArry=1/3x+3" is the equation of the line"`