How do you write an equation of a line with point (2,-3), slope 2/3?

1 Answer
Nov 3, 2017

See a solution process below:

Explanation:

We can use the point-slope formula for writing the equation for the line in the problem. The point-slope form of a linear equation is: #(y - color(blue)(y_1)) = color(red)(m)(x - color(blue)(x_1))#

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

Substituting the slope and values from the point in the problem gives:

#(y - color(blue)(-3)) = color(red)(2/3)(x - color(blue)(2))#

#(y + color(blue)(3)) = color(red)(2/3)(x - color(blue)(2))#

We can solve this equation for #y# to put the equation in slope-intercept format. 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(blue)(3) = (color(red)(2/3) xx x) - (color(red)(2/3) xx color(blue)(2))#

#y + color(blue)(3) = color(red)(2/3)x - 4/3#

#y + color(blue)(3) - 3 = color(red)(2/3)x - 4/3 - 3#

#y + 0 = color(red)(2/3)x - 4/3 - (3/3 xx 3)#

#y = color(red)(2/3)x - 4/3 - 9/3#

#y = color(red)(2/3)x - color(blue)(13/3)#