Question

if p(x,y) is equidistant from the points A(7,-2) and B(3,1) express y in terms of x

Answer 1

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Answer 2

Answer:

$y \:in \:terms \:x,\\ y=\frac{8x-43}{6}$

Step-by-step explanation:

$Distance \: between \:(x_{1},y_{1}),\:(x_{2},y_{2})\\=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$Distance \: between \:(x,y)\:and\:A(7,-2)\\=Distance \: between \:(x,y)\:and\:B(3,1)$

$\sqrt{(7-x)^{2}+(-2-y)^{2}}\\=\sqrt{(3-x)^{2}+(1-y)^{2}}$

$\implies (7-x)^{2}+(2+y)^{2}\\=(3-x)^{2}+(1-y)^{2}$

$\implies 7^{2}+x^{2}-14x+2^{2}+y^{2}+4y\\=3^{2}+x^{2}-6x+1^{2}+y^{2}-2y$

$\implies 49-14x+4+4y=9-6x+1-2y$

$\implies 4y+2y = -6x+14x+10-53$

$\implies 6y = 8x-43$

$\implies y =\frac{8x-43}{6}$

Therefore,

$y \:in \:terms \:x,\\ y=\frac{8x-43}{6}$

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