The points P is equidistant from A(1,3), B (-3,5) and C(5,-1). Then PA = [EAMCET 2003]

The points P is equidistant from A(1,3), B (-3,5) and C(5,-1). Then PA

1.	The points P is equidistant from A(1,3), B (-3,5) and C(5,-1). Then PA = [EAMCET 2003]
A.	5
B.	\[5\sqrt{5}\]
C.	25
D.	\[5\sqrt{10}\]
Answer» E.

Discussion

No Comment Found

Related MCQs

For a series the value of mean deviation is 15. The most likely value of its quartile deviation is
If the plane \[2ax-3ay+4az+6=0\] passes through the midpoint of the line joining the centres of the spheres \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+6x-8y-2z=13\] and \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-10x+4y-2z=8\], then \[a\] equals [AIEEE 2005]
Three persons A, B and C are to speak at a function along with five others. If they all speak in random order, the probability that A speaks before B and B speaks before C, is
The average marks obtained by the students in a class are 43. If the average marks obtained by 25 boys are 40 and the average marks obtained by the girl students are 48, then what is the number of girl students in the class?
In a class 30% students like tea, 20% like coffee and 10% like both tea and coffee. A student is selected at random then what is the probability that he does not like tea if it is known that he likes coffee?
The system of equations \[\lambda x+y+z=0,\] \[-x+\lambda y+z=0,\] \[-x-y+\lambda z=0\], will have a non zero solution if real values of \[\lambda \]are given by [IIT 1984]
A coin is tossed and a dice is rolled. The probability that the coin shows the head and the dice shows 6 is
A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is
For the three events A, B and C, P (exactly one of the events A or B occurs)=P (exactly one of the two events B or C occurs)=P(exactly one of the events C or A occurs)=P and P (all the three events occur simultaneously)\[={{P}^{2}}\], where 0
The number of values of k for which the system of equations \[(k+1)x+8y=4k,\] \[kx+(k+3)y=3k-1\] has infinitely many solutions, is [IIT Screening 2002]

The points P is equidistant from A(1,3), B (-3,5) and C(5,-1). Then PA = [EAMCET 2003]

Discussion

No Comment Found

Related MCQs

Reply to Comment