Consider two points A - -3-0 and B - 0-4- A point P on line 2x-3y-12-0 is such that -PA-PB- is maximum- Then P is

|PA PB| will be maximum if P,A,B are collinear. Hence, equation of AB is x 3+ y4=1 ⋯(1) and given line nbsp;2x 3y 12=0 ⋯(2) On solving equation (1) ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to Be Parallel

Consider two ...

Question

Consider two points $A \equiv (- 3, 0)$ and $B \equiv (0, 4) .$ A point $P$ on line $2 x - 3 y - 12 = 0$ is such that $| P A - P B |$ is maximum. Then $P$ is

(- 12, - 12)

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(6, 0)

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(0, - 4)

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None of these

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Solution

The correct option is A $(- 12, - 12)$
$| P A - P B |$ will be maximum if $P, A, B$ are collinear.

Hence, equation of $A B$ is
$\frac{x}{- 3} + \frac{y}{4} = 1 \dots (1)$
and given line $2 x - 3 y - 12 = 0 \dots (2)$
On solving equation $(1)$ and $(2),$ we get
$P \equiv (- 12, - 12)$

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Similar questions

Q. Consider two points

A \equiv (- 3, 0)

and

B \equiv (0, 4) .

A point

P

on line

2 x - 3 y - 12 = 0

is such that

| P A - P B |

is maximum. Then

P

Q. Consider the point

A \equiv (0, 1)

and

B \equiv (2, 0) .^{'} P^{'}

be a point on the line

4 x + 3 y + 9 = 0.

Coordinate of the point

P

such that

| P A - P B |

is maximum, is

Q. Let

A \equiv (- 3, 0)

and

B \equiv (3, 0)

be two fixed points and

P

moves on a plane such that

P A = n \times P B (n > 0)

. If

n \neq 1

, then locus of a point

P

Q. Let

A \equiv (- 3, 0)

and

B \equiv (3, 0)

be two fixed points and

P

moves on a plane such that

P A = n \times P B (n > 0)

. If

n > 1

, then-

Q. Let

A \equiv (- 3, 0)

and

B \equiv (3, 0)

be two fixed points and

P

moves on a plane such that

P A = n \times P B (n > 0)

. If

0 < n < 1

, then the locus of point

P

is a circle and

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Consider two points A≡(−3,0) and B≡(0,4). A point P on line 2x−3y−12=0 is such that |PA−PB| is maximum. Then P is

The correct option is A (−12,−12)|PA−PB| will be maximum if P,A,B are collinear. Hence, equation of AB is x−3+y4=1 ⋯(1) and given line 2x−3y−12=0 ⋯(2) On solving equation (1) and (2), we get P≡(−12,−12)

Consider two points $A \equiv (- 3, 0)$ and $B \equiv (0, 4) .$ A point $P$ on line $2 x - 3 y - 12 = 0$ is such that $| P A - P B |$ is maximum. Then $P$ is

The correct option is A $(- 12, - 12)$
$| P A - P B |$ will be maximum if $P, A, B$ are collinear.

Hence, equation of $A B$ is
$\frac{x}{- 3} + \frac{y}{4} = 1 \dots (1)$
and given line $2 x - 3 y - 12 = 0 \dots (2)$
On solving equation $(1)$ and $(2),$ we get
$P \equiv (- 12, - 12)$