Co-ordinate Geomatry, PLease Help

akpareek wrote:In triangle ABC, the co-ordinates of A, B and C are (4,3), (6,2) and (T,-3) respectively and triangle is right angled at A. Find T.

A. 11
B. 8/5
C. -5/8
D. -6
E. -11

AB^2 = 2^2 +1^2 = 5
AC ^2 = (4-T)^2 + 6^2 = (4-T)^2 + 36

BC^2 = (6-T)^2 + 5^2 = (6-T)^2 + 25

AB^2+AC^2 = BC^2 =>

(6-T)^2 + 25 = (4-T)^2 + 36 +5

(6-T)^2 - (4-T)^2 = 16

(6-T - (4-T)) (6-T+4-T) = 16

2*(10-2T) = 16

10-2T =8

T =1 [spoiler]{at least that is what I got}[/spoiler]

Thanks ajith !

Atleast i came to know that the way i was trying to solve this problem was right.
The problem is with options.

I have one more problem.. i hope u ll help me again.

At what point on line x-y=3 does a perpendicular drawn from the line x-y=1 at point (3,2) intersect ?

A. (4,1)
B. (1,4)
C. (1,3)
D. (1,2)
E. (3,1)

akpareek wrote:In triangle ABC, the co-ordinates of A, B and C are (4,3), (6,2) and (T,-3) respectively and triangle is right angled at A. Find T.

A. 11
B. 8/5
C. -5/8
D. -6
E. -11

slope of line AB = m = (3-2)/(4-6) =-1/2
slope of line AC = m' = (-3-3)/(T-4) = 6/(4-T)
since AB is perpendicular to AC so mm' = -1 --> (-1/2)(6/(4-T)) = -1 --> 3=4-T --> T=1

ajith wrote:

akpareek wrote:In triangle ABC, the co-ordinates of A, B and C are (4,3), (6,2) and (T,-3) respectively and triangle is right angled at A. Find T.

A. 11
B. 8/5
C. -5/8
D. -6
E. -11

AB^2 = 2^2 +1^2 = 5
AC ^2 = (4-T)^2 + 6^2 = (4-T)^2 + 36

BC^2 = (6-T)^2 + 5^2 = (6-T)^2 + 25

AB^2+AC^2 = BC^2 =>

(6-T)^2 + 25 = (4-T)^2 + 36 +5

(6-T)^2 - (4-T)^2 = 16

(6-T - (4-T)) (6-T+4-T) = 16

2*(10-2T) = 16

10-2T =8

T =1 [spoiler]{at least that is what I got}[/spoiler]

don't think this much step is at all required to solve this small problem..
Its a second's thing by right approach...anyways well tried..

akpareek wrote:Thanks ajith !

Atleast i came to know that the way i was trying to solve this problem was right.
The problem is with options.

I have one more problem.. i hope u ll help me again.

At what point on line x-y=3 does a perpendicular drawn from the line x-y=1 at point (3,2) intersect ?

A. (4,1)
B. (1,4)
C. (1,3)
D. (1,2)
E. (3,1)

any perpendicular to x-y = 1 can be expressed as x+y = t (product of the slopes is -1)
x+y =t goes through (3,2 )
3+2 = t
t= 5
so the equation of the line becomes x+y =5

To find the intersection we should solve
x-y = 3
x+y = 5
simultaneously

y = 1
x= 4
so (4,1) is the point of intersection

akpareek wrote:In triangle ABC, the co-ordinates of A, B and C are (4,3), (6,2) and (T,-3) respectively and triangle is right angled at A. Find T.

A. 11
B. 8/5
C. -5/8
D. -6
E. -11

slope of AB = (3-2)/(4-6) = -1/2

hence, slope of CA should be 2 ( perpendicular)

=> (-3-3)/(T-4) = 2
=> T-4 = -3
=> T = 1

At what point on line x-y=3 does a perpendicular drawn from the line x-y=1 at point (3,2) intersect ?

A. (4,1)
B. (1,4)
C. (1,3)
D. (1,2)
E. (3,1)

if the point lies on X-Y=3, it should satisfy the condition. In the answer choices, only A satisfies the condition.

akpareek wrote:Thanks ajith !

Atleast i came to know that the way i was trying to solve this problem was right.
The problem is with options.

I have one more problem.. i hope u ll help me again.

At what point on line x-y=3 does a perpendicular drawn from the line x-y=1 at point (3,2) intersect ?

A. (4,1)
B. (1,4)
C. (1,3)
D. (1,2)
E. (3,1)

see these two lines are paralle to each other i.e. x-y=3 (i)and x-y=1 (ii)as they have same slope = m = 1
so slope of line perpendicular to it = m' = -1/m = -1/1 = -1

so let eq of perpendicular line is y=m'x+c --> y=-x+c
since (3,2) passes through it so 3=-2+c --> c=5 so eq of line is y=-x+5 -----(iii)
solving (i) and (iii), we get x-3 = -x+5 --> x=4 --> y= x-3 =1
[spoiler]so point is (4,1) Ans A[/spoiler]

harshavardhanc wrote: if the point lies on X-Y=3, it should satisfy the condition. In the answer choices, only A satisfies the condition.

harshavardhanc your approach is very good for this question and will really help to solve problems faster in actual GMAT exam.
Well I have given solution with general approach..if in case two or more cases satisfies the line X-Y=3(though its not the case here..

shashank.ism wrote:

harshavardhanc wrote: if the point lies on X-Y=3, it should satisfy the condition. In the answer choices, only A satisfies the condition.

harshavardhanc your approach is very good for this question and will really help to solve problems faster in actual GMAT exam.
Well I have given solution with general approach..if in case two or more cases satisfies the line X-Y=3(though its not the case here..

yes, you are absolutely correct. A method will be required to actually "solve" these questions which are better framed than this one. The general approach will come in handy there.