Question

the coordinates of two points p and q are (2,3) and (5,2) respectively.If R is any point such that PR=QR and area of ∆PQR =7 sq. units, then find the coordinates of R. ​

Answer 1

co-ordinates of R = (49/10, 67/10) or (21/10, -17/10)

given, two co-ordinates of triangle ∆PQR are P(2, 3) and Q(5, 2) respectively.

R is any point such that, PR = QR

let R (x, y)

now PR = √{(x - 2)² + (y - 3)²}

QR = √{(x - 5)² + (y - 2)²}

now PR = QR

⇒ √{(x - 2)² + (y - 3)²} = √{(x - 5)² + (y - 2)²}

⇒(x - 2)² + (y - 3)² = (x - 5)² + (y - 2)²

⇒(x - 2)² - (x - 5)² = (y - 2)² - (y - 3)²

⇒(x - 2 - x + 5)(x - 2 + x - 5) = (y - 2 - y + 3)(y - 2 + y - 3)

⇒3(2x - 7) = 1(2y - 5)

⇒6x - 21 = 2y - 5

⇒6x - 2y = 16

⇒3x - y = 8..........(1)

also area of ∆PQR = 7 sq unit

⇒1/2 |2(2 - y) + 5(y - 3) + x(3 - 2)| = 7

⇒|4 - 2y + 5y - 15 + x| = 14

⇒|x + 3y - 11| = 14

⇒x + 3y - 11 = ±14

⇒x + 3y = 25.....(2) or x + 3y = -3 ......(2)

case 1 : from equation (1) and (2) we get,

x = 49/10 , y = 67/10

case 2 : from equation (1) and (2) we get,

x = 21/10 , y = -17/10

Answer 2

hope it is helpful to you