A B are two points 3, 4 and (5 2 find a point P such that PA = PB and area of apab 10)

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If the coordinates of two points A and B are 3,4 and 5, 2 respectively. Then, the coordinates of any point P, if P A=P B and area of Δ P A B=10, areA. 7,5,1,0B. 2,7,1,0C. 7,2,1,0D. 7,2, 1,0

Let the coordinates of P be (x, y). Then,

PA = PB

`⇒ PA^2 = PB^2`

`⇒ (x – 3)^2 + (y– 4)^2 = (x – 5)^2 + (y + 2)^2`

⇒ x – 3y – 1 = 0 ….(1)

Now, Area of ∆PAB = 10

`⇒ \frac { 1 }{ 2 } |(4x + 3 × (–2) + 5y) – (3y + 20 – 2x)| = 10`

⇒ |(4x + 5y – 6) – (–2x + 3y + 20)| = 20

⇒ |6x + 2y – 26| = ± 20 ⇒ 6x + 2y – 26 = ± 20

⇒ 6x + 2y – 46 = 0 or, 6x + 2y – 6 = 0

⇒ 3x + y – 23 = 0 or, 3x + y – 3 = 0

Solving x – 3y – 1 = 0 and 3x + y – 23 = 0

we get x = 7, y = 2.

Solving x – 3y – 1 = 0 and 3x + y – 3 = 0,

we get x = 1, y = 0.

Thus, the coordinates of P are (7, 2) or (1, 0)

• determine graphically the vertices of the traingle 3x+3y=10
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• Find the coordinates of point C, which divides the line segment joining the points D (-2, 5) and E (4, 6) in the ratio 2 : 3.
• If the slope of line joining the points (2, 3) and (4, -5) is equal to the slope of line joining the points (x, 5) and (5, 6). Find the value of x.
• Find the slope of line which passes through the point (5, 6) and the mid-point of line joining the points (4, 6) and (-3, 7).
• Find the point which divides the line segment joining the points A (5, 7) and B (8, 4) in the ratio 3 : 5.
• If three points (3, 6), (-5, 7) and (x, 1) are collinear, find the value of x.
• Find a point on the Y-axis, which is equidistant from the points (-3, 4) and (5, 6).
• If the slope of the line joining the points A (5, 7) and B (4, 8) is
  
   of the slope of line joining the points P (2, 8) and Q (5, y), then find the value of y.

Text Solution

(1, 0) only (7, 2) only(1, 0) or (7, 2)Neither (1, 0) nor (7, 2)

Answer : C

Solution : Given A( 3, 4) and B( 5, -2) <br> Let, P (x, y) <br> Given that, `PA=PB` <br> `implies PA^(2)=PB^(2)` <br> `implies (x-3)^(2)+(y-4)^(2)=(x-5)^(2)+(y+2)^(2)` <br> `implies x^(2)-6x+9+y^(2)-8y+16` <br> `=x^(2)-10x+25+y^(2)+4y+4` <br> `implies 4x-12y=4` <br> `implies x-3y=1` ...(i) <br> `:'` Area of `Delta PAB=10` <br> `:. 1/2|(x,y,1),(3,4,1),(5,-2,1)|= pm 10` <br> `implies x(4+2)-y(3-5)+1(-6-20)= pm 20` <br> `implies 6x+2y-26= pm 20` <br> `implies 6x+2y-26=20` <br> or, `6x+2y-26=-20` <br> `implies 6x+2y=46` ...(ii) <br> or `6x+2y=6` ...(iii) <br> From eqs. (i) and (ii), we get <br> `x=7, y=2` <br> Similarly, from eqs. (i) and (iii), we get <br> `x=1, y=0` <br> Hence, coordinates of P are (7, 2) or (1, 0)