PS - Rectangle Coord.

karthikpandian19 wrote:Rectangle PQRS lies in the first quadrant of the xy-plane, with point P diagonally opposite point R and point Q diagonally opposite point S. The coordinates of points P and Q are (a + 2, 2a) and (a – 3, a), respectively. If Point R lies at (11, 2), what is the value of a?

Draw a line joining the diagonal PR. Since PQRS is a rectangle, we can use pythagoras theorem in triangle PQR (with 90 at Q).
We have
PQ^2 + QR^2 = PR^2
Now P=(a+2, 2a), Q=(a-3,a), and R=(11,2)
Then PQ^2 + QR^2 = PR^2 becomes:

=> (a+2-a+3)^2+(2a-a)^2 + (a-3-11)^2 + (a-2)^2 = (a+2-11)^2 + (2a-2)^2
=> 5^2 + a^2 + (a-14)^2 + (a-2)^2 = (a-9)^2 + 4(a-1)^2
=> 25+ a^2 + a^2+196-28a + a^2 +4 -4a = a^2 + 81 -18a +4a^2 +4 -8a
=> a^2*(5-3) + a*(32-26) + (81 - 196 - 25) = 0
=> 2a^2 + 6a - 140 = 0
=> a^2 + 3a = 70

Check from the options, for middle value 5, 5^2 + 5*3 = 40, smaller than 70, so try next one (7):
7^2+3*7 = 49+21 = 70. Awesome.
Hence D is the correct answer.

Let me know if this helps

I solved it like this:

Since PQRS is a rectangle and PR,QS are the diagonals
Therefore PQ is to be perpendicular to QR

Find Slope of PQ:
P(a + 2, 2a) and Q(a - 3, a)
= a-2a/(a-3-(a+2))
= a/5 is the slope

Find Slope of QR:
Q(a - 3, a) and R(11,2)
=2-a/(11-(a-3))
=2-a/(14-a) is the slope

Since both are perpendicular:
2-a/(14-a) = -5/a
2a-a^2=-70+5a
0=a^2+3a-70
(a+10)(a-7)=0
So "a" as per the available answer should be 7
Answer is D

eagleeye wrote:

karthikpandian19 wrote:Rectangle PQRS lies in the first quadrant of the xy-plane, with point P diagonally opposite point R and point Q diagonally opposite point S. The coordinates of points P and Q are (a + 2, 2a) and (a – 3, a), respectively. If Point R lies at (11, 2), what is the value of a?

Draw a line joining the diagonal PR. Since PQRS is a rectangle, we can use pythagoras theorem in triangle PQR (with 90 at Q).
We have
PQ^2 + QR^2 = PR^2
Now P=(a+2, 2a), Q=(a-3,a), and R=(11,2)
Then PQ^2 + QR^2 = PR^2 becomes:

=> (a+2-a+3)^2+(2a-a)^2 + (a-3-11)^2 + (a-2)^2 = (a+2-11)^2 + (2a-2)^2
=> 5^2 + a^2 + (a-14)^2 + (a-2)^2 = (a-9)^2 + 4(a-1)^2
=> 25+ a^2 + a^2+196-28a + a^2 +4 -4a = a^2 + 81 -18a +4a^2 +4 -8a
=> a^2*(5-3) + a*(32-26) + (81 - 196 - 25) = 0
=> 2a^2 + 6a - 140 = 0
=> a^2 + 3a = 70

Check from the options, for middle value 5, 5^2 + 5*3 = 40, smaller than 70, so try next one (7):
7^2+3*7 = 49+21 = 70. Awesome.
Hence D is the correct answer.

Let me know if this helps

karthikpandian19 wrote:I solved it like this:

Since PQRS is a rectangle and PR,QS are the diagonals
Therefore PQ is to be perpendicular to QR

Find Slope of PQ:
P(a + 2, 2a) and Q(a - 3, a)
= a-2a/(a-3-(a+2))
= a/5 is the slope

Find Slope of QR:
Q(a - 3, a) and R(11,2)
=2-a/(11-(a-3))
=2-a/(14-a) is the slope

Since both are perpendicular:
2-a/(14-a) = -5/a
2a-a^2=-70+5a
0=a^2+3a-70
(a+10)(a-7)=0
So "a" as per the available answer should be 7
Answer is D

eagleeye wrote:

karthikpandian19 wrote:Rectangle PQRS lies in the first quadrant of the xy-plane, with point P diagonally opposite point R and point Q diagonally opposite point S. The coordinates of points P and Q are (a + 2, 2a) and (a – 3, a), respectively. If Point R lies at (11, 2), what is the value of a?

Draw a line joining the diagonal PR. Since PQRS is a rectangle, we can use pythagoras theorem in triangle PQR (with 90 at Q).
We have
PQ^2 + QR^2 = PR^2
Now P=(a+2, 2a), Q=(a-3,a), and R=(11,2)
Then PQ^2 + QR^2 = PR^2 becomes:

=> (a+2-a+3)^2+(2a-a)^2 + (a-3-11)^2 + (a-2)^2 = (a+2-11)^2 + (2a-2)^2
=> 5^2 + a^2 + (a-14)^2 + (a-2)^2 = (a-9)^2 + 4(a-1)^2
=> 25+ a^2 + a^2+196-28a + a^2 +4 -4a = a^2 + 81 -18a +4a^2 +4 -8a
=> a^2*(5-3) + a*(32-26) + (81 - 196 - 25) = 0
=> 2a^2 + 6a - 140 = 0
=> a^2 + 3a = 70

Check from the options, for middle value 5, 5^2 + 5*3 = 40, smaller than 70, so try next one (7):
7^2+3*7 = 49+21 = 70. Awesome.
Hence D is the correct answer.

Let me know if this helps

Your way is definitely less error-prone and faster. Awesome job !