Identify an equation in point-slope form for the line perpendicular toy=-2x + 8 that passes through

Identify an equation in point-slope form for the line perpendicular to y=-2x + 8 that passes through (-3,9).
O A. y - 9 =-2(x+3)
O B. y+3 (x-9)
O C. y-9 - (x+3)
O D. y + 9 = 2(x-3)

Related Posts

This Post Has 6 Comments

  1. If the line is perpendicular, then as a rule the gradient of the perpendicular line will be the negative inverse of the gradient of the line given.
    Therefore the gradient of the perpendicular line would be 1/2 
    so now just substitute in the equation of a line (y=mx+c) where y and x are given by -3 and 9, and your new m=1/2 
    Hope this helps 

  2. y - 9 = 1/2(x + 3)

    Step-by-step explanation:

    For the equation y = -2x + 8, your slope, m, is -2. To find the slope for a line perpendicular to the original line, we do the opposite reciprocal of this number. The slope of the perpendicular line is 1/2.

    Point-slope form is: y - y1 = m(x - x1) where x1 and y1 are the x- and y-coordinates of an ordered pair.

    With this new slope for the perpendicular line, and the point we are given, we just plug in the info.

    Your equation is: y - 9 = 1/2(x + 3)

  3. [tex]y - 9 = \frac{1}{2}(x +3)[/tex]

    Step-by-step explanation:

    Given

    Function; [tex]y = -2x + 8[/tex]

    Required

    Find an equation perpendicular to the given function if it passes through (-3,9)

    First, we need to determine the slope of:  [tex]y = -2x + 8[/tex]

    The slope intercept of an equation is in form;

    [tex]y = mx + b[/tex]

    Where m represent the slope

    Comparing  [tex]y = m_1x + b[/tex] to [tex]y = -2x + 8[/tex];

    We'll have that

    [tex]m_1 = -2[/tex]

    Going from there; we need to calculate the slope of the parallel line

    The condition for parallel line is;

    [tex]m_1 * m_2 = -1[/tex]

    Substitute [tex]m_1 = -2[/tex]

    [tex](-2) * m_2 = -1[/tex]

    Divide both sides by -2

    [tex]m_2 =\frac{ -1}{-2}[/tex]

    [tex]m_2 =\frac{1}{2}[/tex]

    The point slope form of a line is;

    [tex]y - y_1 = m_2(x - x_1)[/tex]

    Where [tex](x_1,y_1) = (-3,9)[/tex] and [tex]m_2 =\frac{1}{2}[/tex]

    [tex]y - y_1 = m_2(x - x_1)[/tex]becomes

    [tex]y - 9 = \frac{1}{2}(x - (-3))[/tex]

    Open the inner bracket

    [tex]y - 9 = \frac{1}{2}(x +3)[/tex]

    Hence, the point slope form of the perpendicular line is:

    [tex]y - 9 = \frac{1}{2}(x +3)[/tex]

  4. For this case we have the following linear equation:
     [tex]y = -2x + 8
[/tex]
     As the lines are perpendicular, then the slope of the line is:
     [tex]m' = \frac{-1}{m}[/tex]
     Where,
     m: is the slope of the original line
     Substituting values we have:
     [tex]m' = \frac{-1}{-2}[/tex]
     Rewriting:
     [tex]m' = \frac{1}{2}[/tex]
     Then, the equation in point-slope form is given by:
     [tex]y-yo = m '(x-xo)
[/tex]
     Where,
     (xo, yo): ordered pair that goes through the line.
     Substituting values we have:
     [tex]y-9 = \frac{1}{2}(x+3)
[/tex]
     
     an equation in point-slope form for the line perpendicular to y = –2x + 8 that passes through (–3, 9) is:
     [tex]y-9 = \frac{1}{2}(x+3)[/tex]

Leave a Reply

Your email address will not be published. Required fields are marked *