Question

If the co-ordinates of two points A and B are (3,4) and (5,-2) respectively, find the co-ordinates of any point P if PA=PB and Area of triangle PAB =10

• A. (2,7)
• B. (0,1)
• C. (7,2)
• D. (1,0)

Answer 1

Answer: (C), (D)

The correct options are
C

(7,2)

D

(1,0)

Let the co-ordinates of P be (x,y). Then

PA=PB

$P{A}^2$ = $P{B}^2$

$(x-3{)}^2+(y-4{)}^2=(x-5{)}^2+(y+2{)}^2$

$x^2+9-6x+y^2+16-8y=x^2+25-10x+y^2+4+4y$

x - 3y - 1 = 0---------------(1)

now,

$areaof△PAB=\frac{1}{2}\left|\begin{array}{ccc}5& 2& 1\\ 4& 7& 1\\ 7& 4& 1\end{array}\right|$ = 10

$\frac{1}{2}|3(-2-y)+5(y-4)+x(4+2)|$

$\frac{1}{2}|-6-3y+5y-20+6x|=10$

$\frac{1}{2}|6x+2y-26|=\pm 20$

6x + 2y - 26 = 20 or 6x + 2y - 6 = 0

3x + y - 23 = 0 or 3x + y - 3 = 0

3x + y - 23 = 0 ------------------(2)

3x + y - 3 = 0 ------------------(2)

Solving equation 1 and 2

x - 3y - 1 = 0

3x + y - 23 = 0

Multiplying 3 in equation (2) and add it to equation (1)

10x = 70 $\to x=7$

Substituting x = 7 in equation 2, we get, y=2

Solving equation 1 and 3

x - 3y - 1 = 0

3x + y - 3 = 0

Multiply 3 in equation 3 and add it to equation 2,

10x = 10 $\to x=1$

Substituting x = 1, in equation 2

We get, y = 0

Thus, the co-ordinates of P is (7,2) and (1,0)

So, option C and D are correct.