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

Answer 1

Given :-

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

To Find :-

Find the relation between x and y

Solution :-

We know that

$\sf D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let the distance be PR and PQ

For PR

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

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

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

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

$\sf PR=\sqrt{x^2+6x+9+y^2-8y+16}$

$\sf PR=\sqrt{x^2+6x+25+y^2-8y}$

For PQ

$\sf PQ=\sqrt{(x-3)^2+(y-6)^2}$

$\sf PQ=\sqrt{x^2-2(3)(x)+(3)^2+y^2-2(y)(6) + (6)^2}$

$\sf PQ=\sqrt{x^2-6x+9+y^2-12y+36}$

$\sf PQ=\sqrt{x^2-6x+45+y^2-12y}$

On squaring

We get

$\sf x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 - 8y$

$\sf 9 - 6x+y^2+36-12y=9+6x+y^2+16-8y$

$\sf 45-6x+y^2-12y=25+6x+y^2-8y$

$\sf 45-25=12x+4y$

$\sf 20=12x+4y$

Divide by 4

$\sf\dfrac{20}{4}=\dfrac{12x+4y}{4}$

$\sf 5=3x+y$

Answer 2

$\rm{ {3x+y−5 = 0}}$

Step-by-step explanation:

★ Given -

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

★To find -

• Relation between x and y.

★Solution -

We are give that the point (x, y) is equidistant from the point (3,6) and (-3, 4).

$\therefore \rm{PA = PB}$

$\boxed{\rm{Distance \: formula = \sqrt{{(x {\tiny{2}} - x {\tiny{1}}) ^{2} }{(y {\tiny{2}} - y {\tiny{1}}) ^{2} }}}}$

putting values in the formula we get,

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

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

now squaring both sides,

$\boxed{ \rm \longmapsto{x}^{2}−6x+9+ {y}^{2} −12y+36= {x}^{2} +6x+9+ {y}^{2} −8y+16}$

$\boxed{ \rm \longmapsto{x}^{2}−6x+9+ {y}^{2} −12y+36 - {x}^{2} -6x-9- {y}^{2} +8y-16 = 0 }$

$\rm{ \longmapsto −12x−4y+20=0}$

$\rm{ \longmapsto−4(3x+y−5) = 0}$

$\rm{ \longmapsto \bf\green {3x+y−5 = 0}}$

★More to know -

• An equidistant point is a point that is an equal distance from two other points, but is not necessarily in the middle of two points.