Ask
your own question, for FREE!
Mathematics
6 Online
OpenStudy (anonymous):
The equation of line GH is y = 4x − 3. Write an equation of a line perpendicular to line GH in slope-intercept form that contains point (2, 3).
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
y = 4x − 3
y = −4x − 5
11 over 4x − 3
y = negative 1 over 4x + 7 over 2
OpenStudy (anonymous):
. y = 4x − 3
y = −4x − 5
11 over 4x − 3
y = negative 1 over 4x + 7 over 2
OpenStudy (anonymous):
@Albany_Goon can you help please
OpenStudy (anonymous):
^_^
OpenStudy (anonymous):
hey :)
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
hey lemme c here
OpenStudy (anonymous):
okay ^.^
OpenStudy (anonymous):
step 1: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
y-(4*x-3)=0
OpenStudy (anonymous):
then Solve
OpenStudy (anonymous):
y-4x+3 = 0
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
then
OpenStudy (anonymous):
im calculating lol
OpenStudy (anonymous):
ok but I recognize that we have here an equation of a straight line. Such an equation is usually written y=mx+b
OpenStudy (anonymous):
wat did you get nikki
OpenStudy (anonymous):
well I did the answers on a grid and on the last one I got a perpendicular line
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
OpenStudy (anonymous):
|dw:1422370136049:dw|
OpenStudy (anonymous):
lol somthimn like dat
OpenStudy (anonymous):
lol so I'm right?
OpenStudy (anonymous):
lol sure but Notice that when x = 0 the value of y is -3/1 so this line "cuts" the y axis at y=-3.0
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
y-intercept = -3/1 = -3.0
OpenStudy (anonymous):
N When y = 0 the value of x is 3/4 Our line therefore "cuts" the x axis at x= 0.75000
x-intercept = 3/4 = 0.75
OpenStudy (anonymous):
Nikki?
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!