Consider the points A(0,1) and B(2,0), and P be a point on the line 4 x+3 y+9=0. The coordinates of P such that |P A-P B| is maximum are

Question

Consider the points A(0,1) and B(2,0), and P be a point on the line 4 x+3 y+9=0. The coordinates of P such that |P A-P B| is maximum are (A) (-12 / 5,17 / 5) (B) (-84 / 5,13 / 5) (C) (-6 / 5,17 / 5) (D) (0,-3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/dd1e33068716200805f6a2b4d7c97406-.png" /> The equation of A B is (x/2)+(y/1)=1 or x+2 y-2=0 |P A-P B| ≤ A B Thus, |P A-P B| is maximum if the points A, B, and P are collinear. Hence, solving x+2 y-2=0 and 4 x+3 y+9=0, we get point P ≡(-84 / 5,13 / 5)

Q. Consider the points $A (0, 1)$ and $B (2, 0)$ , and $P$ be a point on the line $4 x + 3 y + 9 = 0$ . The coordinates of $P$ such that $∣ P A - PB ∣$ is maximum are

$(- 12/5, 17/5)$

$(- 84/5, 13/5)$

$(- 6/5, 17/5)$

$(0, - 3)$

The equation of $A B$ is
$\frac{x}{2} + \frac{y}{1} = 1$
or $x + 2 y - 2 = 0$
$∣ P A - PB ∣ \leq A B$
Thus, $∣ P A - PB ∣$ is maximum if the points $A, B$ , and $P$ are collinear.
Hence, solving $x + 2 y - 2 = 0$ and
$4 x + 3 y + 9 = 0$ , we get point $P \equiv (- 84/5, 13/5)$

Q. Consider the points A(0,1) and B(2,0), and P be a point on the line 4x+3y+9=0. The coordinates of P such that ∣PA−PB∣ is maximum are

(−12/5,17/5)

(−84/5,13/5)

(−6/5,17/5)

(0,−3)