What is the equation of the line passing through #(96,72)# and #(19,4)#?

3 Answers
May 7, 2018

The slope is 0.88311688312.

Explanation:

#(Y_2 - Y_1)/ (X_2 - X_1)# = #m#, the slope

Label your ordered pairs.

(96, 72) #(X_1, Y_1)#
(19, 4) #(X_2, Y_2)#

Plug-in your variables.

#(4 - 72)/(19 - 96)# = #m#

-68/-77 = #m#

Two negatives make a positive, so:

0.88311688312 = #m#

May 7, 2018

#y = = 68/77x - 984/77#

Explanation:

Recall;

#y = mx + c#

#m = (y_2 - y_1)/(x_2 - x_1)#

#y_2 = 4#

#y_1 = 72#

#x_2 = 19#

#x_1 = 96#

Inputing the values..

#m = (4 - 72)/(19 - 96)#

#m = (-68)/-77#

#m= 68/77#

The new equation is;

#(y - y_1) = m(x - x_1)#

Inputing their values..

#y - 72 = 68/77(x - 96)#

#y - 72 = (68x - 6528)/77#

Cross multiplying..

#77(y - 72) = 68x - 6528#

#77y - 5544 = 68x - 6528#

Collecting like terms..

#77y = 68x - 6528 + 5544#

#77y = 68x - 984#

Dividing through by #77#

#y = = 68/77x - 984/77#

May 7, 2018

Point-slope form: #y-4=68/77(x-19)#

Slope-intercept form: #y=68/77x-984/77#

Standard form: #68x-77y=984#

Explanation:

First determine the slope using the slope formula and the two points.

#m=(y_2-y_1)/(x_2-x_1)#,

where #m# is the slope, and #(x_1,y_1)# is one point and #(x_2,y_2)# is the other point.

I'm going to use #(19,4)# as #(x_1,y_1)# and #(96,72)# as #(x_2,y_2)#.

#m=(72-4)/(96-19)#

#m=68/77#

Now use the slope and one of the points to write the equation in point-slope form:

#y-y_1=m(x-x_1)#,

where:

#m# is the slope and #(x_1,y_1)# is one of the points.

I'm going to use #(19,4)# for the point.

#y-4=68/77(x-19)# #larr# point-slope form

Solve the point-slope form for #y# to get the slope-intercept form:

#y=mx+b#,

where:

#m# is the slope and #b# is the y-intercept.

#y-4=68/77(x-19)#

Add #4# to both sides of the equation.

#y=68/77(x-19) +4#

Expand.

#y=68/77x-1292/77 + 4#

Multiply #4# by #77/77# to get an equivalent fraction with #77# as the denominator.

#y=68/77x-1292/77+4xx77/77#

#y=68/77x-1292/77+308/77#

#y=68/77x-984/77# #larr# slope-intercept form

You can convert the slope-intercept form to the standard form:

#Ax+By=C#

#y=68/77x-984/77#

Multiply both sides by #77#.

#77y=68x-984#

Subtract #68x# from both sides.

#-68x+77y=-984#

Multiply both sides by #-1#. This will reverse the signs, but the equation represents the same line.

#68x-77y=984# #larr# standard form

graph{68x-77y=984 [-10, 10, -5, 5]}