Question

The line y = 3x/4 meet the lines x – y + 1 = 0 and 2x – y – 5 = 0 at points A and B respectively. If P on the line y = 3x/4 which satisfies the condition PA · PB = 25 then number of possible coordinate of P is-

• A. 3
• B. 2
• C. 1
• D. None of these

Answer 1

Answer: (A)

3

Answer 2

Point P which lies on the line y = can be chosen as P $\left( h , \frac { 3 h } { 4 } \right)$ . If q be the angle that the line y = makes with the +ve direction of the X-axis, then

tan q = $\frac { 3 } { 4 }$ Ž cos q = $\frac { 4 } { 5 }$ and sin q = $\frac { 3 } { 5 }$

Now, coordinates of points A and B which lie on the line

y = $\frac { 3 x } { 4 }$ can be chosen as

A ŗ and B ŗ

Since A lies on the line x – y + 1 = 0, therefore

– + 1 = 0 gives

r₁ = $\frac { - 5 } { 4 }$(h + 4)

and B lies on the line 2x – y – 5 = 0, therefore

2 $\left( \frac { 3 h } { 4 } + \frac { 3 r _ { 2 } } { 5 } \right)$ – 5 = 0 gives

r₂ = $\frac { - 5 } { 4 }$(h – 4)

According to the given condition, we have

PA · PB = 25

i.e. |r₁| · |r₂| = 25

i.e. $\frac { 25 } { 16 }$(h² – 16) = ± 25

i.e. h² = 16 ± 16 = 32, 0 gives h = ±4$\sqrt { 2 }$, 0

Hence, the required points, are (0, 0), (4$\sqrt { 2 }$) and

(–4$\sqrt { 2 }$).