Question

If the points (2,1) and( 1,- 2) are equidistant from point (x,y) ,show that x+3y = 0​

Answer 1

$\huge\sf\blue{Given}$

✭ Points (2,1) and(1,-2) are equidistant from (x,y)

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$\huge\sf\gray{To \:Find}$

◈ x+3y = 0

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$\huge\sf\purple{Steps}$

According to the distance formula,

$\underline{\boxed{\sf\sqrt{(x_{2} -x_{1})^2+{(y_{2}-y_{1})^2}}}}$

Distance between (2,1) & (x,y)

≫ $\sf \sqrt{(x-2)^2 + (y-1)^2} \:\:\: -eq(1)$

Distance between (1,-2) & (x,y)

≫ $\sf \sqrt{(x-1)^2+(y+2)^2} \:\:\: -eq(2)$

Now that we are given the distance between these points are equal then on equating eq(1) & eq(2)

➝ $\sf \sqrt{(x-2)^2 + (y-1)^2} = \sqrt{(x-1)^2+(y+2)^2}$

➝ $\sf (x-2)^2+ (y-1)^2 = (x-1)^2+(y+2)^2$

$\sf \bigg\lgroup (x+y)^2 = x^2+y^2+2xy\bigg\rgroup$

Ans also,

$\sf \bigg\lgroup (x-y)^2 = x^2-2xy+y^2\bigg\rgroup$

➝ $\sf \bigg\{ x^2+(-2)^2+2(x)(-2)\bigg\} + \bigg\{ y^2+(-1)^2+2(y)(-1)\bigg\} = \bigg\{ x^2+(-1)^2+2(x)(-1)\bigg\} + \bigg\{y^2 + 2^2 + 2(y)(2)\bigg\}$

➝ $\sf x^2+4-4x+y^2+1-2y = x^2+1-2x+y^2+4+4y$

➝ $\sf \cancel{x^2}+\cancel{4}-4x+\cancel{y^2}+\cancel{1}-2y = \cancel{x^2} + \cancel{1} -2x + \cancel{y^2}+\cancel{4} +4y$

➝ $\sf -4x-2y = -2x+4y$

➝ $\sf -4x-2y+2x-4y = 0$

➝ $\sf -4x+2x-2y-4y = 0$

➝ $\sf -2x-6y=0$

«« Multiple by $\sf \dfrac{-1}{2}$ »»

➝ $\sf -\dfrac{1}{2} (-2x-6y) = 0$

➝ $\sf \orange{x+3 = 0}$

$\sf Hence \ Proved!!$

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