Question

11. Find the locus of the point whose distances from the point (4,5) and (-4,-3) equal. are​

Answer 1

ㅤㅤ✠Given :-

The distance from point (4,5) and (-4, -3) are equal

ㅤㅤ✠ To find :-

Equation of the locus

ㅤㅤㅤ✠Solution :-

Let us consider From the point P=(x,y) is equal from the point A=(4,5) and pointB (-4,-3) So,

Thus According to the Question,

➲PA = PB

Here,

• PA = Distance between P and A
• PB = Distance between P and B

We find the distance between the points by using Distance formula

ㅤㅤ✠Distance formula:- √(x₁-x₂)² +(y₁-y₂)²

So, here

P= (x,y) = (x₁ , y₁)

A = (4,5) = (x₂, y₂)

PA = √(x-4)² + (y-5)²

P= (x,y) = (x₁ , y₁)

B = (-4, -3) = (x₂, y₂)

PB = √[x-(-4)]² + [y-(-3)]²

PB = √(x+4)² + (y+3)²

ㅤㅤㅤ➲According to the Question,

PA = PB

√(x-4)² + (y-5)² = √(x+4)² + (y+3)²

➲Squaring on both sides

(x-4)² + (y-5)² = (x+4)² + (y+3)²

➲Transposing all terms to L.H.S

(x-4)² + (y-5)² - (x+4)² -(y +3)² = 0

x² -8x + 16 + y² -10y + 25 - (x² +8x+ 16) -(y² + 6y+9) = 0

x² -8x + 16 + y² -10y + 25 -x² -8x -16 -y² -6y -9 = 0

➲Keeping like terms together

x² -x² -8x -8x + 16 -16 + y² -y² -10y -6y +25-9 =0

-16x -16y +16= 0

➲Take common -

-16[x + y -1] = 0

x + y -1 = 0

x + y = 1

➲So, the equation of locus is x+y = 1