At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
\[2y=4\]is this wat u tryin to find
parallel = means the same slope re-write your ew as: 2y=3x-4 or y=3/2 x -2 so, slope your existing line is: m=3/2 so, eq of parallel through the point (3,5): y-5=3/2 *(x-3) y=3/2x -9/2 +5=3/2 x+1/2
That makes sense.... Thank you!
Can you help me with another one?
shoot... I need to go soon
Its like the one you just helped me with only perpendicular to the line 3x-2y=4 and containing points 3,-5
try to do it yourself - I'll help. just remember that perpendicular means that your line will have slope (m2) = -1/m1... so, if re-write your eq: y=3/2 x -2 so, for perp line: m2=-2/3 eq of the line: y-(-5)=-2/3 (x-3) simplify, please
So then y=-2/3x+6/3-5?
y+5=-2/3 *x +2 y=-2/3 *x -3
do you see it?
So the equation is y=-2/3x-3?
yep... to be exact: y=(-2x/3 ) -3
Okay, I guess I just don't get how you suddenly got to x+2 in the end
let's go back to: y-(-5)=-2/3 (x-3) i think that the way it written it could be confusing... let's try differently: y+5= x* (-2/3) -3* (-2/3) y+5=-2x/3 +2 (-3 in numerator & -3 in dominator are canceled out) is it better...?
That does make better sense. Math is like a whole other language for me, lol. Thank you so much for your help
Great! Welcome :)