You visited us 1 times! Enjoying our articles? Unlock Full Access!

Condition of Concurrency of 3 Straight Lines

Two points ...

Question

Two points $A (- 1, 0)$ and $B (1, 0)$ are given, locus of $P$ when

Open in App

Solution

A) $P A = 3 P B$
$\Rightarrow \sqrt{{(x + 1)}^{2} + y^{2}} = 3 \sqrt{{(x - 1)}^{2} + y^{2}}$
$\Rightarrow 4 x^{2} + 4 y^{2} - 10 x + 4 = 0$
B) $3 P A = P B$
$\Rightarrow 3 \sqrt{{(x + 1)}^{2} + y^{2}} = \sqrt{{(x - 1)}^{2} + y^{2}}$
$\Rightarrow 4 x^{2} + 4 y^{2} + 10 x + 4 = 0$
C) $(P A)^{2} + (P B)^{2} = 24$
$\Rightarrow {(x + 1)}^{2} + y^{2} + {(x - 1)}^{2} + y^{2} = 24$
$\Rightarrow x^{2} + y^{2} = 11$
D) $(P A)^{2} - (P B)^{2} = 24$
$\Rightarrow {(x + 1)}^{2} + y^{2} - {(x - 1)}^{2} - y^{2} = 24$
$\Rightarrow x = 6$

Suggest Corrections

Similar questions

A (- 9, 0), B = (- 1, 0)

P

(x, y)

3 P B = P A

P

P = (1, 0); Q = (- 1, 0)

R = (2, 0)

S Q^{2} + S R^{2} = 2 S P^{2}

P

(1, 0)

Q

y^{2} = 8 x

P Q

P

(1, 0)

Q

y^{2} = 8 x

P Q

A = (1, 0)

B = (- 1, 0), C = (2, 0)

P

P B^{2} + P C^{2} = 2 P A^{2}

Join BYJU'S Learning Program