The line above goes through points (-5,0) and (5,-3). What is the equation of the line? A. y=(-3/10)x+1.5 B. y=(-10/3)x+1.5 C. y=(-3/10)x-1.5 D. y=(-10/3)x-1.5
first you must calculate slope this is change in Y over change in X
Solve the following for y:\[\frac{y-0}{x+5}=\frac{y+3}{x-5}\]\[y=-\frac{3 x}{10}-\frac{3}{2} \]
Im confused.. @robtobey I get yours some what @garrett_payne I don't get what you're saying
the slope is rise over run That means how much does it go up (or down) over the distance in x
so for two points it's (Y2-Y1)/ (x2-x1) what do you get? It doesn't really matter which point is X1,Y1 and which is X2,Y2
I still don't get it. (Note: Im really bad at slope)
it's ok. say point one is (-5,0) and point two is (5,-3) what would the slope equation look like? remember it's always written (x,y)
I still have no clue. I don't understand slope at all
\[slope=(Y_2 -Y_1)/(X_2 - X_1) \]
it will look like \[\frac{ y2-y1 }{x2-x1 }\] where y2 = -3, y1= 0, x2=5 and x1 = -5
ok so it would be -3 - 0 ------- 5 - -5
@kaek98 I think that @garrett_payne is saying that the following is a starter for the line equation.\[y =\frac{\text{y2}-\text{y1}}{\text{x2}-\text{x1}}x+b \]b is still unknown, but can be solved for by replacing x an y with the coordinate values of either given point and then solving for b.
Exactly, thanks @robtobey
Ok so can you walk me through on how to get to the answer of it?
you have -3/10 as slope also called m the equation of a line is y =mx+b
you now have m, next solve for b to do this plug in either point (x1,Y1) or (x2, y2) and find b
1.5
good job so with a slope of -3/10 and an intercept of 1.5 you now know which answer to select
c
C. is correct.
sorry, 1.5+b= 0 so b= -1.5 then yes C
I think that equating m1 to m2 might be quicker in that when one solves for y, the result is the line equation answer.
might be too much at once for her
Yes, you may be correct. I rely on Mathematica for calculations. This program can derive the solution in one statement as using the statement below:\[\text{Solve}\left[\frac{y-0}{x+5}==\frac{y+3}{x-5},y\right]\text{//}\text{Expand }\text{//}\text{Flatten }\text{//}\text{ TraditionalForm} \]
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