- #1
riddle
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Ok, I'm new here, and I'm not sure if this is the right place to ask. So anyway here's the question:
OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AP and MQ meet at P. If OA=a and OC=c, express OP in terms of a and c.
The question's easy enough... until the point I get stuck.
Here's what I've got so far:
OA = a, MO = -0.5a, OC = c, CQ = .75a, MQ= .25a+c, AC = c-a
OP = OA + x(AC) = 0.5(OA) + y(MQ)
=> OP = a + x(c-a) = 0.5a + y(c+0.25a)
=> a + x(c-a) = 0.5a + y(c+0.25a)
Aaand this is where I get stuck. I've got the textbook. It mentions "forming simultaneous euqations by equating the coefficients", but I've got no idea how that works. It's the first time I've ever heard of solving equations like that. I've done some looking up, but I'm still confused. Can someone please help? Thanks
P.S.: Please excuse the lack of arrows on top of the lines. There should be "[tex]\rightarrow[/tex]" on top of them all, I just wasn't sure how to use Latex to do it.
OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AP and MQ meet at P. If OA=a and OC=c, express OP in terms of a and c.
The question's easy enough... until the point I get stuck.
Here's what I've got so far:
OA = a, MO = -0.5a, OC = c, CQ = .75a, MQ= .25a+c, AC = c-a
OP = OA + x(AC) = 0.5(OA) + y(MQ)
=> OP = a + x(c-a) = 0.5a + y(c+0.25a)
=> a + x(c-a) = 0.5a + y(c+0.25a)
Aaand this is where I get stuck. I've got the textbook. It mentions "forming simultaneous euqations by equating the coefficients", but I've got no idea how that works. It's the first time I've ever heard of solving equations like that. I've done some looking up, but I'm still confused. Can someone please help? Thanks
P.S.: Please excuse the lack of arrows on top of the lines. There should be "[tex]\rightarrow[/tex]" on top of them all, I just wasn't sure how to use Latex to do it.