Find the slope of a line parallel to the line through the given points. P(-2, 4), Q(6, -2)
First we would find the slope of these points.... \(\Huge{\frac{y2-y1}{x2-x1}}\) We input the coordinates... \(\Huge{\frac{-2 - 4}{6-(-2)}}\) Simplify....what would the slope be?
3/4?
oops sorry i thought you were looking for the perpendicular slope :P
so its wrrong?
If you were to find a slope parallel to yours it would be exactly the same so the parallel slope would also be \(\LARGE{-\frac{3}{4}}\)
Find the slope of the line that contains the following points. A(5, 6), B(10, 8)
so for this one wuld it be 5/2 or 2/5?
For this one use the slope formula.... \(\Huge{\frac{y2-y1}{x2-x1}}\) We input the coords... \(\Huge{\frac{8-6}{10-5} \implies \frac{2}{5}}\)
Find the slope of a line perpendicular to a line through the given points. E(5, 7), F(3, 1)
what about this one ?
First find the slope of the points... \(\Huge{\frac{1-7}{3-5}}\) What would this simplify down too?
so it would be -3?
Close it would actually be \(\LARGE{3}\) because a negative divided by a negative is a positive.... Now to find the perpendicular slope we would flip the whole number into a fraction and change the sign as well.... \(\Huge{3 \implies - \frac{1}{3}}\)
Line M passes through points (4,2) and (-7,1). What is the slope of a line N that is parallel to line M?
i also need help with this one
First find the slope of the points.... \(\Huge{\frac{1-2}{-7 - 4}}\) What is the slope?
-1/11 ?
they are both negative so does that turn positive ?
Correct ^^ they do turn to a positive so it will be \(\LARGE{-\frac{1}{11}}\)
oops positive
not negative
\(\LARGE{\frac{1}{11}}\)*
Given the points A(0, 0), B(e, f), C(0, e) and D(f, 0), determine if line segments AB and CD are parallel, perpendicular or neither.
ignore that last post
that i just deleted
can you help me on the other question i posted
Indicate in standard form the equation of the line through the given points. P(0, -4), Q(5, 1)
Im sorry but i must go now
dont go lol
its about 11 pm so I got go :( sorry....but if you ever need help in the future and see me on just tag me in your question ^^
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