How do you write an equation of a line passing through (4, 2), perpendicular to #y=2x+3#?

1 Answer
Apr 18, 2017

Answer:

See the entire solution process below:

Explanation:

The equation in the problem is 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)(2)x + color(blue)(3)#

Therefore the slope of this line is #m = 2#

Let us call the slope of the perpendicular line #m_p#. By definition, the slope of a perpendicular line is:

#m_p = -1/m#

Therefore, for this problem, the slope of the perpendicular line is:

#m_p = -1/2#

We can use the slope-intercept formula, substitute the values from the points for #x# and #y# and substitute the slope we determined and solve for #b#:

#2 = (color(red)(-1/2) * 4) + color(blue)(b)#

#2 = -2 + color(blue)(b)#

#color(red)(2) + 2 = color(red)(2) - 2 + color(blue)(b)#

#4 = 0 + color(blue)(b)#

#4 = color(blue)(b)#

Which means the equation is:

#y = color(red)(-1/2)x + color(blue)(4)#

Or, we could use the point-slope formula to also write and 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 values from the point in the problem gives:

#(y - color(red)(2)) = color(blue)(-1/2)(x - color(red)(4))#