Given points C(-2,-5) and D(-5,-11), write an equation, in standard form, for a line that has a x-intercept of

one-third and is also parallel to line CD





i have a test tomorow and i want to know how to do this

Given points C(-2,-5) and D(-5,-11), write an equation, in standard form, for a line that has a x-intercept of
Slope m=(-11-(-5))/(-5-(-2))=2





For now use slope-intercept y=mx+b to find the equation of the line. We have m and the x-intercept is when y=0. Since the x-intercept is 1/3 then plug in the numbers for x and y and solve for b





0=2(1/3)+b


b=-2/3


y=2x-2/3





convert to Standard form by subtracting 2x from both sides then multiply both sides by -3 to get rid of the fraction.





6x-3y=2
Reply:Rise over run...





Rise of -6 over a run of -3...therefore your line has a slope of 2.





Now, you should know an equation for a line on a graph is:





y = mx + b





A parallel line will have the same slope as the line you have been given. If m (slope) = 2 and your x-intercept (b) = 1/3 then: y = (2)x + 1/3
Reply:first you have to find the slope of the line through the two given points





M=(y-y.)/(x-x.)


M=(-11-(-5))/(-5-(-2))


M=(-6/-3)


M=2





Then you put this into standard form





y=Mx + b


y=2x + (1/3)
Reply:um isnt it -5y+-11 and then you make a table?
Reply:What you do is you take the Y1-Y2/ X2-X1 to find the slope then you plug it in to the Y=mx+b then it becomes a y=-2x+b.


Then plug in the points in X and Y -11=-2(-5)+b subtract both sides by 10 the it is -21=b y=-2x-21 add 2x both sides then it becomes 2x+y=-21