Question

A and b are two points (3,4) and (5,-2). find the coordinates of point p such that pa=pb and area of triangle pab= 10

Answer 1

Given: A and b are two points (3,4) , (5,-2) and PA=PB and area of triangle PAB= 10 square units.

To find: The coordinates of P

Solution: Let the coordinate P be (x,y)

Since it is given that PA = PB

So, firstly we will calculate the distance PA.

PA = (x,y) (3,4)

Distance PA = $\sqrt{(3-x)^{2}+(4-y)^{2}}$

PB=(x,y) (5,-2)

Distance PB = $\sqrt{(5-x)^{2}+(-2-y)^{2}}$

So, $\sqrt{(3-x)^{2}+(4-y)^{2}}=\sqrt{(5-x)^{2}+(-2-y)^{2}}$

Squaring both the sides in the above equation,

${(3-x)^{2}+(4-y)^{2}}={(5-x)^{2}+(-2-y)^{2}}$

$9+x^{2}-6x+16+y^{2}-8y=25+x^{2}-10x+4+y^{2}+4y$

$-6x-8y=-10x+4+4y$

$4x-12y=4$

$x-3y=1$   (Equation 1)

Now,it is given that Area of triangle PAB = 10

Area of triangle of (3,4) (5,-2) and (x,y)

Area of triangle is given by the formula= $\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Area of triangle PAB = $\frac{1}{2}[3(-2-y)+5(y-4)+x(4+2)]=10$

$-6+2y-20+6x=20$

$46=2y+6x$

$3x+y=23$           (Equation 2)

Now, solving equations 1 and 2.

Since $x-3y=1$

therefore, x = 3y+1

Equation 2 implies,

$3(3y+1)+y=23$

$9y+3+y=23$

$10y=20$

$y= 2$

$x=3y+1$

$x=(3 \times 2)+1$

$x= 7$

Therefore, the coordinates are (7,2).

Answer 2

Answer:

Step-by-step explanation:

Brainly.in

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Secondary SchoolMath 13+7 pts

A and b are two points (3,4) and (5,-2). find the coordinates of point p such that pa=pb and area of triangle pab= 10

Report by Babupatel3723 20.08.2018

Answers

Apriyanshu885 · Ambitious

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Pinquancaro Ambitious

Given: A and b are two points (3,4) , (5,-2) and PA=PB and area of triangle PAB= 10 square units.

To find: The coordinates of P

Solution: Let the coordinate P be (x,y)

Since it is given that PA = PB

So, firstly we will calculate the distance PA.

PA = (x,y) (3,4)

Distance PA = \sqrt{(3-x)^{2}+(4-y)^{2}}

PB=(x,y) (5,-2)

Distance PB = \sqrt{(5-x)^{2}+(-2-y)^{2}}

So, \sqrt{(3-x)^{2}+(4-y)^{2}}=\sqrt{(5-x)^{2}+(-2-y)^{2}}

Squaring both the sides in the above equation,

{(3-x)^{2}+(4-y)^{2}}={(5-x)^{2}+(-2-y)^{2}}

9+x^{2}-6x+16+y^{2}-8y=25+x^{2}-10x+4+y^{2}+4y

-6x-8y=-10x+4+4y

4x-12y=4

x-3y=1 (Equation 1)

Now,it is given that Area of triangle PAB = 10

Area of triangle of (3,4) (5,-2) and (x,y)

Area of triangle is given by the formula= \frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]

Area of triangle PAB = \frac{1}{2}[3(-2-y)+5(y-4)+x(4+2)]=10

-6+2y-20+6x=20

46=2y+6x

3x+y=23 (Equation 2)

Now, solving equations 1 and 2.

Since x-3y=1

therefore, x = 3y+1

Equation 2 implies,

3(3y+1)+y=23

9y+3+y=23

10y=20

y= 2

x=3y+1

x=(3 \times 2)+1

x= 7

Therefore, the coordinates are (7,2).