Question

The diagram shows a rectangle ABCD. The point A is (2,14). B is (-2,8) and C lies on the x-axis (?,0)
Find the equation of BC.
the coordinates of C and D.

Answer 1

One second, I have done some work, just trying to find the coordinates for C and D. Missing X values.

Answer 2

y-8= -2/3 (x+2) for BC equation. C (?,0) D (?,6) Not sure what to do. I did slope equation.
14-8/2+2 = 6/4 3/2 then opp slope recip. -2/3

Answer 3

okay @.@ may i ask what are the question that you want to ask @.@

Answer 4

How to find the x values for those coordinates. :p

Answer 5

okay can you show me how you got those answers up there? cuz i confuse by your own answers and question @.@

Answer 6

Apologies, I have a packet with an image. My answers are accurate, which I am completely confident of.

Answer 7

@iYuko

Answer 8

The equation of line BC, in point slope form, is \(\Large y - 8 = -\frac{2}{3}(x+2)\). In slope-intercept form, the equation is \(\Large y= -\frac{2}{3}x+\frac{20}{3}\)


It doesn't matter which equation you use. Plug in y = 0 and solve for x to get the x coordinate of point C.

Answer 9

Can you explain to me how you got 20/3? @jim_thompson5910

Answer 10

Oh, I see, for some reason I attempted to use outside numbers to distribute, I completely understand what I did wrong, thank you so much!

Answer 11

Sure. Here are the steps I did to convert from point slope form to slope intercept form.
\[\Large y - 8 = -\frac{2}{3}(x+2)\]
\[\Large y - 8 = -\frac{2}{3}(x)-\frac{2}{3}(2)\]
\[\Large y - 8 = -\frac{2}{3}x-\frac{4}{3}\]
\[\Large y - 8+8 = -\frac{2}{3}x-\frac{4}{3}+8\]
\[\Large y = -\frac{2}{3}x-\frac{4}{3}+8\]
\[\Large y = -\frac{2}{3}x-\frac{4}{3}+8*\frac{3}{3}\]
\[\Large y = -\frac{2}{3}x-\frac{4}{3}+\frac{24}{3}\]
\[\Large y = -\frac{2}{3}x+\frac{-4+24}{3}\]
\[\Large y = -\frac{2}{3}x+\frac{20}{3}\]