How do you write an equation of a line passing through (10, 5), perpendicular to #5x+4y=8#?

1 Answer
Apr 11, 2017

Answer:

See the entire solution process.

Explanation:

The line 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)(5)x + color(blue)(4)y = color(green)(8)#

The slope of an equation in standard form is: #m = -color(red)(A)/color(blue)(B)#. Substituting the values from the equation in the problem gives:

#m = -color(red)(5)/color(blue)(4)#

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

#m_p = -1/m = 4/5#

We can now use the point-slope formula to find the equation of 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)(5)) = color(blue)(4/5)(x - color(red)(10))#

We can convert this to the slope-intercept form by solving for #y#. 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)(4/5) xx x) - (color(blue)(4/5) xx color(red)(10))#

#y - color(red)(5) = 4/5x - 40/5#

#y - color(red)(5) = 4/5x - 8#

#y - color(red)(5) + 5 = 4/5x - 8 + 5#

#y - 0 = 4/5x - 3#

#y = color(red)(4/5)x - color(blue)(3)#