My GMA Question Thread

[alchemy]

Haha, no that's not what I was implying. Yeah sure, sorry about that.

Would the answer be (x^2+8)(x+2sqrt2)(x-2sqrt2)?

Yep, reworked it and I think that's correct. Need someone to verify though.

EDIT: You mean z, not x.

[lzxnl]

I don't know how to type plus minus signs.

[Lasercookie]

I don't know how to type plus minus signs.

\pm

[alchemy]

Need to solve the attached question without using calculator :c

[lzxnl]

For those who ceebs opening the file, the question is to express in exact rectangular form.

Firstly let's deal with the power of 10. The thing inside the the bracket can be written as
If you multiply this by itself ten times, aka raising it to the power of ten, you're multiplying the argument of the complex number by ten, so

[alchemy]

The distance of a point from the line x=4 is equal to it's distance from the line y=1. Find the locus of this point P(x,y).

[alchemy]

The distance of a point from the line x=4 is equal to it's distance from the line y=1. Find the locus of this point P(x,y).

Can someone tell me if my first line of working is correct? Also, how would we depict this instance on the axis?
=

[lzxnl]

The distance of a point from the line x=4 is equal to it's distance from the line y=1. Find the locus of this point P(x,y).

The distance of a point from a line is defined to be the perpendicular distance; in this case, for a general point (x,y), the distance from that to x=4 is simply |x-4|. The y coordinate doesn't matter. Draw it out and you'll see what I mean.
Similarly, the distance from y=1 is |y-1|
|y-1|=|x-4|
(y-1)^2=(x-4)^2
You could leave it in this form, or you could write:
y-1 = x-4, y = x-3
AND y-1 = 4-x, y=5-x

Can someone tell me if my first line of working is correct? Also, how would we depict this instance on the axis?
=

This is correct; exactly what I did. You would draw a point on the axes and then label what the distance to each of the lines means.

[alchemy]

Question: The distance of a point from the y-axis is three times it's distance from the x-axis. Find the locus of this point.

Working:
If we let P(x,y) and calculate  the distance to (0,12), we can then let this distance be 3 times the distance from P to (4,0).
=
or
Why isn't this correct?

Answer:

[brightsky]

The SHORTEST distance from a point P (x,y) to the y-axis is simply the absolute value of the x-coordinate. The SHORTEST distance from P to the x-axis is simply the absolute value of the y-coordinate. So we have:

abs(x) = 3 abs(y)
abs(x) - 3 abs(y) = 0

Consider all cases. You will find that the equation above can be simplified to:

x +- 3y = 0

[alchemy]

AND y-1 = 4-x, y=5-x

Can you please explain why we need to do this and how such simplification works. It's just that I think I get it, but not quite. Also, I've seen these steps done before but never pondered over them.

[lzxnl]

Because of the absolute value signs, there are two cases. I've done y-1 = +(x-4) and y-1 = -(x-4)

[alchemy]

Because of the absolute value signs, there are two cases. I've done y-1 = +(x-4) and y-1 = -(x-4)

Oh right, now the following question makes sense as well : )

[alchemy]

I'm having a general problem in solving the following types of questions. I think I be making a silly error while completing the square or something. Can someone help me identify the error in my working?

Question: Find the equation of the locus of point P(x,y) which satisfy the property that the distance of P to the point F(2,5) is twice the distance PM, the perpendicular distance to the line with equation x=1.

My answer:










Answer:

[b^3]

It's going from the first line to the second line where you've made a mistake, when you take the 3 out of the denominator you need to square it and this cancels the 9 on the top. You should have