Question

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10. Find a relation between x and y such that the point (x, y) is equidistant from the sia
(3,6) and (-3,4).​

Answer 1

Answer:

3x + y = 5

Step-by-step explanation:

Using distance formula, which says if there are two points (x₁, y₁) and (x₂, y₂) then the distance between them is √(x₁ - x₂) + (y₁ - y₂).

 Here,

⇒ distance b/w (x,y) and (3.6) is equal to the distance b/w (x, y) and (-3,4).

⇒ √(3 - x)² + (6 - y)² = √(-3 - x)² + (4 - y)²

⇒ (3 - x)² + (6 - y)² = (-3-x)² + (4 - y)²

⇒ (3 - x)² - (- 3- x)² = (4 - y)² - (6 - y)²

⇒ (3 - x)² - (3 + x)² = (4 - y)² - (6 - y)²

⇒ (3 - x + 3 + x)(3 - x - 3 - x) = (4 - y + 6 - y)(4 - y - 6 + y)

⇒ 6(- 2x) = (10 - 2y)(- 2)

⇒ - 12x = - 20 + 4y

⇒ - 3x = - 5 + y

⇒ 3x + y = 5

Answer 2

$\bf{\blue{\underline{\underline{\pink{GIVEN:-}}}}}$

• The point (x , y) is equidistant from the points (3 , 6) and (-3 , 4) .

$\bf{\blue{\underline{\underline{\pink{TO\: FIND:-}}}}}$

• The relation between x and y .

$\bf{\blue{\underline{\underline{\pink{SOLUTION:-}}}}}$

Let,

• A be the point whose coordinates are (x , y) .

• B be the point whose coordinates are (3 , 6) .

• C be the point whose coordinates are (-3 , 4) .

☃️ According to the question, A is equidistant from B and C .

• => AB = AC ------(1)

$\orange\star\:\bf{\gray{\underline{\pink{\boxed{\purple{AB\:=\:\sqrt{(x_2\:-\:x_1)^2\:+\:(y_2\:-\:y_1)^2}\:}}}}}}$

Where,

• $\bf\red{x_1}$ = x

• $\bf\red{x_2}$ = 3

• $\bf\red{y_1}$ = y

• $\bf\red{y_2}$ = 6

$\bf\green{\implies\:AB\:=\:\sqrt{(3\:-\:x)^2\:+\:(6\:-\:y)^2}\:}$

Similarly,

$\green\star\:\bf{\gray{\underline{\pink{\boxed{\purple{AC\:=\:\sqrt{(x_2\:-\:x_1)^2\:+\:(y_2\:-\:y_1)^2}\:}}}}}}$

Where,

• $\bf\red{x_1}$ = x

• $\bf\red{x_2}$ = -3

• $\bf\red{y_1}$ = y

• $\bf\red{y_2}$ = 4

$\bf\green{\implies\:AC\:=\:\sqrt{(-3\:-\:x)^2\:+\:(4\:-\:y)^2}\:}$

⚡ From Equation 1 :-

$\rm{\implies\:\sqrt{(3\:-\:x)^2\:+\:(6\:-\:y)^2}\:=\:\sqrt{(-3\:-\:x)^2\:+\:(4\:-\:y)^2}\:}$

$\rm{\implies\:(9\:+\:x^2\:-\:6x)\:+\:(36\:+\:y^2\:-\:12y)\:=\:(9\:+\:x^2\:+\:6x)\:+\:(16\:+\:y^2\:-\:8y)\:}$

$\rm{\implies\:x^2\:+\:y^2\:-\:6x\:-\:12y\:+\:45\:=\:x^2\:+\:y^2\:+\:6x\:-\:8y\:+\:25\:}$

$\rm{\implies\:-12x\:-\:4y\:+\:20\:=\:0\:}$

$\rm{\implies\:3x\:+\:y\:=\:5\:}$

$\rm{\implies\:3x\:=\:5\:-\:y\:}$

$\bf\pink{\implies\:x\:=\:\dfrac{5\:-\:y}{3}\:}$