TheKingInTheNorth wrote:
Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\)). Is the distance between point K and G prime?
(1) A^2 – 5A – 6 = 0
(2) A > 2
I suggest my students to use statements when you do not know how to proceed and this is one such scenario...
If you do not get hang of it or are short of time, -- There is one variable-A and
if we know A, we can answer for sure. Statement I gives
two values of A as it is a quadratic equation. It may be possible that both lead to prime or non-prime, but in most cases it will not.
Statement II gives
a range, so insuff
Combined,
I says -1 and 6, II says>2, so a unique value C
Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\))..
Two ways..
(I) Don't waste time finding an equation in variable here
We know that to know the distance here, we will require to know the value of A, so let us check the statements
(1) \(A^2 – 5A – 6 = 0...(A-6)(A+1)=0\) A =-1 or A=6..
A is -1.. \(K=(-1,0)\) and G\(=(2,\sqrt{7}\)...Distance = \(\sqrt{9+7}=\sqrt{16}=4\)..NO
A is 6.. \(K=(6,0)\) and G\(=(16,\sqrt{21}\)...Distance = \(\sqrt{100+21}=\sqrt{121}=11\).. Yes
Insuff
(2) A > 2
Clearly insuff. There may be cases where it is true and places where it may not be true.
Combined..ans is 11 and YES
Suff
C
(II) Simplify the question stem
Distance = \(\sqrt{(2A+4-A)^2+\sqrt{2A+9}^2}=\sqrt{A^2+10A+25}=\sqrt{(A+5)}^2=A+5\)
So, if A=2, 7 etc ans is YES and if A=3, 4 etc ans is NO
(1) A^2 – 5A – 6 = 0
A= -1 or 6
Insuff
(2) A > 2
Insuff
Combined A=6
Suff
C