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