421/HK17. problem / Math - Solved Math Problems

MR-421 / 16. problem

There are two points on the coordinate plane: A (2; 6) and B (4; -2).
a) Write the equation of the perpendicular to the section AB.
b) Write down the equation of the circle centered on A through B.
Given y equals is a 3 x straight line and {formula_3698} a circle.
c) Give the coordinates of the line and the circle with their coordinates.

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Source: Magyarország: Unified final exam for high schools - Mathematics - intermediate level [09.05.2017.] - 16..)

Algebra → Exponential equations and inequalities → Exponential equations

a) Írja fel az AB szakasz felezőmerőlegesének egyenletét!

1.svg

x_{F} = \frac{x_{A} + x_{B}}{2} = \frac{2 + 4}{2} = \frac{6}{2} = 3

y_{F} = \frac{y_{A} + y_{B}}{2} = \frac{6 + (- 2)}{2} = \frac{4}{2} = 2

F_{A B} (3; 2)

\vec{n} = \vec{A B} = (x_{B} - x_{A}; y_{B} - y_{A})

\vec{n} = (4 - 2; - 2 - 6)

\vec{n} = (2, - 8)

e : A x + B y + C = 0, \vec{n} (A; B)

e : 2 x - 8 y + C = 0

F_{A B} \in e \Rightarrow 2 \cdot 3 - 8 \cdot 2 + C = 0

6 - 16 + C = 00

C = 10

2 x - 8 y = - 10

b) Írja fel az A ponton átmenő, B középpontú kör egyenletét!

2.svg

r = \bar{A B} = \sqrt{{(x_{B} - x_{A})}^{2} + {(y_{B} - y_{A})}^{2}}

r = \sqrt{{(4 - 2)}^{2} + {(- 2 - 6)}^{2}} = \sqrt{2^{2} + {(- 8)}^{2}}

r = \sqrt{4 + 64} = \sqrt{68}

k : {(x - x_{B})}^{2} + {(y - y_{B})}^{2} = r^{2}

k : {(x - 4)}^{2} + {(y - (- 2))}^{2} = {(\sqrt{68})}^{2}

{(x - 4)}^{2} + {(y + 2)}^{2} = 68

c) Adja meg koordinátáikkal az egyenes és a kör közös pontjait!

(1) y = 3 x

(2) x^{2} + 8 x + y^{2} - 4 y = 48

(1) \to (2) x^{2} + 8 x + {(3 x)}^{2} - 4 \cdot (3 x) = 48

x^{2} + 8 x + 9 x^{2} - 12 x = 48

10 x^{2} - 4 x - 48 = 0

5 x^{2} - 2 x - 24 = 0

a = 5; b = - 2; c = - 24

x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 \cdot a \cdot c}}{2 \cdot a} = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 \cdot 5 \cdot (- 24)}}{2 \cdot 5}

x_{1, 2} = \frac{2 \pm \sqrt{4 + 480}}{10} = \frac{2 \pm 22}{10} = \frac{1 \pm 11}{5}

x_{1} = \frac{1 + 11}{5} = \frac{12}{5} = 2, 4

x_{2} = \frac{1 - 11}{5} = \frac{- 10}{5} = - 2

y_{1} = 3 \cdot x_{1} = 3 \cdot 2, 4 = 7, 2

y_{2} = 3 \cdot x_{2} = 3 \cdot (- 2) = - 6 x

3.svg

P (- 2; - 6), Q (2, 4; 7, 2)

2 x - 8 y = - 10 {(x - 4)}^{2} + {(y + 2)}^{2} = 68 P (- 2; - 6), Q (2, 4; 7, 2)

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