Find the coordinates of a point in 3-space

[CaliforniaRoll88]

You have 2 points P(1,2) and Q(3k, 4k). You require the distance between them to be 10. So what equation can you now write down and solve?

$d^2=(x_Q-x_P)^2+(y_Q-y_P)^2$
$100=(3k-1)^2+(4k-2)^2$
$100=(3k-1)^2+(4k-2)^2$
$(a-b)^2=a^2-2ab+b^2$
$(3k-1)^2=(3k)^2-2(3k)(-1)+(-1)^2=9k^2+6k+1$
$(4k-2)^2=(4k)^2-2(4k)(-2)+(-2)^2=16k^2+16k+4$
$(3k-1)^2+(4k-2)^2=25k^2+22k+5$
@Steve4Physics, I can see that I am using the quadratic formula method. How do I employ the Law of Cosines here?

 

[Steve4Physics]

$d^2=(x_Q-x_P)^2+(y_Q-y_P)^2$
$100=(3k-1)^2+(4k-2)^2$
$100=(3k-1)^2+(4k-2)^2$

You have repeated a line, but no matter.

$(a-b)^2=a^2-2ab+b^2$
$(3k-1)^2=(3k)^2-2(3k)(-1)+(-1)^2=9k^2+6k+1$

No. You have used b=-1. But b=1 because the minus sign is already taken into account in the term $a-b$.

Here's a simple example, to illustrate what you have done. We know $(3-1)^2 = 2^2 = 4$. What you have done is:
$(3 - 1)^2 = 3^2 - (2)(3)(-1) + (-1)^2 = 9 + 6 + 1 = 16$
This is wrong! It should be:
$(3 - 1)^2 = 3^2 - (2)(3)(1) + 1^2 = 9 - 6 + 1 = 4$.

This is very important basic algebra, so you need to make sure you master it.

$(4k-2)^2=(4k)^2-2(4k)(-2)+(-2)^2=16k^2+16k+4$

No. Same mistake as described above.

$(3k-1)^2+(4k-2)^2=25k^2+22k+5$

This expression (when corrected) must be equal to 100, This gives you a quadratic equation you can solve to find $k$. Then you know Q$(3k,4k)$ and the problem is done.

@Steve4Physics, I can see that I am using the quadratic formula method. How do I employ the Law of Cosines here?

You do not need the law of cosines if you use the above method.