Question

A(1, 2) and B(5,-2) are two given points on the xy-plane,on which C is such a moving point that the numerical value of the area of triangleCAB is 12 square unit. Find the equation to the locus of C
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Answer 1

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

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

★ To find: The coordinates of P

Let the coordinate P be (x,y)

• Since it is given that PA = PB

$\huge{\text{\underline{Explainationtion:-}}}$

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 × 2) + 1

x = 7

Therefore, the coordinates are (7,2).

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Answer 2

Step-by-step explanation:

I hope you will understand.