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3 years ago

Which is not an equation of the line that passes through the points (1, 1) and (5, 5)?

1 answer:

8 0

$(1,\ 1)\to x_1=1,\ y_1=1\to x_1=y_1\\\\(5,\ 5)\to x_2=5,\ y_2=5\to x_2=y_2\\\\\text{Therefore we have the equation}\ y=x\\\\\text{subtract y from both sides}\\\\\boxed{x-y=0\to\boxed{B}}\to\boxed{y-x=0\to\boxed{A.}}\\\\Answer:\ C.\ -x-y=0$

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3 years ago

Find an equation of the circle that satisfies the given conditions. Endpoints of a diameter are P(−2, 1) and Q(8, 9)

IrinaK [193]

Answer:

Equation of the circle (x-3)²+(y-5)²=(6.4)²

x² -6x +9 +y² -10y +25 = 40.96

Step-by-step explanation:

Step(i):-

Given endpoints of diameter P(−2, 1) and Q(8, 9)

Centre of circle = midpoint of diameter

Centre = $(\frac{-2+8}{2} ,\frac{1+9}{2} )$

Centre (h, k) = (3 , 5)

Step(ii):-

The distance of two end points

PQ = $\sqrt{(x_{2}-x_{1} )^{2} +(y_{2} -y_{1} )^{2} }$

$PQ= \sqrt{(8+2 )^{2} +(8 )^{2} }$

PQ = √164 = 12.8

Diameter d = 2r

radius r = d/2

Radius r = 6.4

Final answer:-

Equation of the circle

(x-h)²+(y-k)² = r²

(x-3)²+(y-5)²=(6.4)²

x² -6x +9 +y² -10y +25 = 40.96

x² -6x +y² -10y = 40.96-34

x² -6x +y² -10y -7= 0

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