Choose the slope-intercept equation of the line that passes through the point shown and is parallel to the line shown. line through the points 0 comma negative 3 and 9 comma 0 with a point 6 comma 4 not on the line y = 3x - 14 y = one thirdx + 2 y = -3x + 22 y = -one thirdx + 8
@hartnn Could you help?
could you find the slope of line , since 2 points are given ??
\(\large m=\dfrac{y_2-y_1}{x_2-x_1}=...?\)
Let me try to out :)
take your time :)
I ended up getting negative one
you too, x1,y1 = 0,-3 x2,y2 = 9,0 right ?
Yes :)
then you should get m = 1/3 isn't it ?
Wait, I got 1/2
m= (0-(-3))/(9-0) = 3/9 =1/3
Oh, lol I feel dumb.
My eyes just looked all over the page.
So, what is the next step, of there is one, I am kinda confused.
ok, now you have a slope ,m and take any one point say, x1,y1= 9,0 put those in y-y1 = m(x-x1)
what u get ?
After I got 1/3, I don't know the next step:/
x1= 9,y1=0 in y-y1 = m(x-x1) gives y-0 = (1/3) (x-9)
I am confused.
where ?
Y-0 = (1/3) (x-9) how did you get that? And where do I go from there
i just plugged in known values in this general formula! y-y1 = m(x-x1)
Ohh, I see now, where do I go from there?
y-0 = (1/3) (x-9) y= one third x -9/3 y= one third x-3 thats it
y = 3x - 14 y = one thirdx + 2 y = -3x + 22 y = -one thirdx + 8
Those are my options
and last step is to consider point x1,y1=6,4 because parallel lines have same slope, m= 1/3 so y-y1 = m(x-x1) y-4= (1/3) (x-6) simplify this and you will get the answer :)
Would I simplify it by cross canceling?
y-4 = 1/3 x - 1/3 (6) y-4 = 1/3 x -2 y= 1/3 x -2+4 y=1/3x+2 like this way ...
Ohh, I might need help with more problems, maybe, but thank you! I think I can get it.
ok, welcome ^_^
Can you help me with slope intercept equation?
y= mx+c is the slope intercept equation.
Choose the slope-intercept equation of the line that passes through the point shown and is perpendicular to the line shown. line through the points 0 comma 6 and negative 2 comma 0 with a point 6 comma 2 not on the line y = one thirdx y = -one thirdx + 4 y = 3x + 20 y = -3x - 16
the slope of that line again comes out to be 1/3 the same way. but in this case we need equation of PERPENDICULAR line so, its slope will be negative reciprocal so, m = -1/(1/3) = -3 now just plug in given point x1,y1=6,2 in y-y1 = m(x-x1) and simplify like y- 2 = (-3)(x-6) can u simplify this ?
Not really, I am so sorry.
y-2 = -3x+18 y=-3x+20
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