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

1 Answer
Aug 15, 2016

Answer:

#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"#