Find an equation of the line that satisfies the given condition. (Let x be the independent variable and y be the dependent variable.)
The line passing through the origin and parallel to the line passing through the points
(2, 5) and (4, 6)

Question

Find an equation of the line that satisfies the given condition. (Let x be the independent variable and y be the dependent variable.)
The line passing through the origin and parallel to the line passing through the points
(2, 5) and (4, 6)

anonymous · Answer

for da slope i got -1/3?

campbell_st · Answer

lets check the slope calculation 

( 6 - 5)/(4 - 2) = 1/2


does that make sense..

anonymous · Answer

i dd it 5-6 i got -1 den i did 2- (-1) i got 3

campbell_st · Answer

(x, y)
(4, 6)
(2, 5)

well the y values are   6 and 5
             x values are  4 and 2

is does that make sense...

anonymous · Answer

lol oh darn i just read wat i ahd put up. sorry its da wrong question!

campbell_st · Answer

so if you did

5 - 6 = -1    for the y values ( rise)

you need to do  2 - 4 = -2 for the x values ( run)

slope is rise/run = -1/-2  = 1/2

campbell_st · Answer

ok...

anonymous · Answer

yh i got -1/2

whpalmer4 · Answer

Look, plot the two points on your paper.  Draw a line through them.  Does the line go up, or down?  If up, the slope is positive.  If down, the slope is negative.

anonymous · Answer

when i put the answer in, it tells me its wrong @ campbell

whpalmer4 · Answer

Then, write x1 = 4, y1 = 6, x2 = 2, y2 = 5 and plug into the slope formula:

$$m=\frac{y_2-y_1}{x_2-x_1}$$

$$m=\frac{5-6}{2-4} = \frac{-1}{-2} = \frac{1}{2}$$

You have the slope, and you need to force the line to go through a known point, the origin (0,0).

$$y-y_0 = m(x-x_0)$$$$y-0 = \frac{1}{2}*(x-0)$$$$y = \frac{1}{2}x$$

anonymous · Answer

thank u, i forgot to put in da "y" and the "x"