How do you write an equation of a line given (-5, -8) and is perpendicular to 10x – 6y = -11?

1 Answer
Jul 22, 2017

Answer:

See a solution process below:

Explanation:

The equation in the problem is in Standard Linear 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

The slope of an equation in standard form is: #m = -color(red)(A)/color(blue)(B)#

Given: #color(red)(10)x - color(blue)(6)y = color(green)(-11)#

Then the slope is: #m = color(red)(-10)/color(blue)(-6) = 5/3#

The formula for the slope of a perpendicular line, #m_p# is:

#m_p = -1/m#

Substituting gives: #m_p = (-1)/(5/3) = -3/5#

Now we can use the point-slope formula to write and equation for the line to solve the problem. 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)(-8)) = color(blue)(-3/5)(x - color(red)(-5))#

#(y + color(red)(8)) = color(blue)(-3/5)(x + color(red)(5))#

If necessary we can solve this equation for #y# to put it 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)(8) = (color(blue)(-3/5) * x) + (color(blue)(-3/5) * color(red)(5))#

#y + color(red)(8) = -3/5x - 3#

#y + color(red)(8) - 8 = -3/5x - 3 - 8#

#y + 0 = -3/5x - 11#

#y = color(red)(-3/5)x - color(blue)(11)#