Question

P(3, 1), Q(6, 5) and R(x, y) are three points such that the angle PRQ is a right angle and the area of DRQP = 7, then the number of such points R is-

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A)

0

Answer 2

$\frac { ( y - 1 ) } { ( x - 3 ) } \times \left( \frac { y - 5 } { x - 6 } \right) = - 1$

y² – 6y + 5 = – [x² – 9x + 18]

x² + y² – 9x – 6y + 23 = 0 … (1)

radius of circle = 5/2 and area of DRPQ = 7

\ $\frac { 1 } { 2 }$ $\left| \begin{array} { l l l } x & y & 1 \\ 3 & 1 & 1 \\ 6 & 5 & 1 \end{array} \right|$ = ± 7

Ž –4x + 3y + 9 = ±14

Ž –4x + 3y + 23 = 0 … (2)

Ž –4x + 3y – 5 = 0 … (3)

We will have to solve equation (1) and equation (2) and also equation (1) and equation (3)

So, we can check whether straight lines (2) and (3) are cutting the circle (1). Distance of centre of circle from St. Line (1)

= $\frac { - 4 \times \frac { 9 } { 2 } + 3 \times 3 + 23 } { \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } }$ = $\frac { 14 } { 5 }$

$\frac { 14 } { 5 }$ > radius of circle

Hence no. solution.

Distance of centre of circle from St. Line (2)

= $\frac { - 14 } { 5 }$ $\left| \frac { - 14 } { 5 } \right|$ > radius of circle

Hence no solution in both case . So answer is (1) .