How do you write an equation of a line passing through (3, 5), perpendicular to #x - 3y = 9#?

1 Answer
Apr 12, 2017

Answer:

See the entire solution process below:

Explanation:

The equation given in the problem is in standard form. The standard form of a linear equation is: #color(red)(A)x + color(blue)(B)y = color(green)(C)#

Where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

#color(red)(1)x - color(blue)(3)y = color(green)(9)#

The slope of an equation in standard form is: #m = -color(red)(A)/color(blue)(B)#. Therefore, the slope of the line represented by this equation is:

#m = (-color(red)(1))/color(blue)(-3) = 1/3#

Let's call the slope of a line perpendicular to this line #m_p#.

#m_p = -1/m#

Then the slope perpendicular to the line from the equation is:

#m_p = -3/1 = -3#

Now, use the point-slope formula to find an equation for the line. The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #color(red)(((x_1, y_1)))# is a point the line passes through.

Substituting the slope we calculated and the point gives:

#(y - color(red)(5)) = color(blue)(-3)(x - color(red)(3))#

We can also solve for #y# to put the equation in slope-intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.

#y - color(red)(5) = (color(blue)(-3) xx x) - (color(blue)(-3) xx color(red)(3))#

#y - color(red)(5) = -3x - (-9)#

#y - color(red)(5) = -3x + 9#

#y - color(red)(5) + 5 = -3x + 9 + 5#

#y - 0 = -3x + 14#

#y = color(red)(-3)x + color(blue)(14)#