How do you find the x and y intercept given (4, -2), (5, -4)?

1 Answer
Jun 11, 2017

See a solution process below:

Explanation:

First, we need to determine the equation for the line going through the two points in the problem. The first thing we need to do for this is determine the slope of the line. The slope can be found by using the formula: #m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))#

Where #m# is the slope and (#color(blue)(x_1, y_1)#) and (#color(red)(x_2, y_2)#) are the two points on the line.

Substituting the values from the points in the problem gives:

#m = (color(red)(-4) - color(blue)(-2))/(color(red)(5) - color(blue)(4)) = (color(red)(-4) + color(blue)(2))/(color(red)(5) - color(blue)(4)) =-2/1 = -2#

Now, we can 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 values from the first point in the problem gives:

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

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

x#-intercept:

To find the #x#-intercept set #y# to #0# and solve for #x#:

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

Becomes:

#(0 + color(red)(2)) = (color(blue)(-2) xx x) + (color(blue)(-2) xx -color(red)(4))#

#2 = -2x + 8#

#2 - color(red)(8) = -2x + 8 - color(red)(8)#

#-6 = -2x + 0#

#-6 = -2x#

#(-6)/color(red)(-2) = (-2x)/color(red)(-2)#

#3 = (color(red)(cancel(color(black)(-2)))x)/cancel(color(red)(-2))#

#x = 3#

The #x#-intercept is: #color(red)(3)# or #(color(red)(3), color(red)(0))#

y#-intercept:

To find the #y#-intercept set #x# to #0# and solve for #y#:

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

Becomes:

#y + color(red)(2) = color(blue)(-2)(0 - color(red)(4))#

#y + color(red)(2) = color(blue)(-2) xx -color(red)(4)#

#y + color(red)(2) = 8#

#y + color(red)(2) - 2 = 8 - 2#

#y + 0 = 6#

#y = 6#

The #y#-intercept is: #color(red)(6)# or #(color(red)(0), color(red)(6))#