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Doss [256]
3 years ago
10

The line 2y - x + 5 = 0 and the curve (10 – x)^2+ (y – 5)^ 2= 10 intersect at points A and B.

Mathematics
1 answer:
DedPeter [7]3 years ago
7 0

Answer:

A(9, 2 ), B(13, 4 )

Step-by-step explanation:

Given the 2 equations

2y - x + 5 = 0 → (1)

(10 - x)² + (y - 5)² = 10 → (2)

Rearrange (1) expressing x in terms of y by adding x to both sides

x = 2y + 5 → (3)

Substitute x = 2y + 5 into (2)

(10 - (2y + 5))² + (y - 5)² = 10, that is

(10 - 2y - 5)² + (y - 5)² = 10

(5 - 2y)² + (y - 5)² = 10 ← expand the factors using FOIL

25 - 20y + 4y² + y² - 10y + 25 = 10

5y² - 30y + 50 = 10 ( subtract 10 from both sides )

5y² - 30y + 40 = 0 ( divide through by 5 )

y² - 6y + 8 = 0 ← in standard form

(y - 2)(y - 4) = 0 ← in factored form

Equate each factor to zero and solve for y

y - 2 = 0 ⇒ y = 2

y - 4 = 0 ⇒ y = 4

Substitute these values into (3) for corresponding values of x

y = 2 : x = 2(2) + 5 = 4 + 5 = 9 ⇒ (9, 2 )

y = 4 : x = 2(4) + 5 = 8 + 5 = 13 ⇒ (13, 4 )

Since we are not given the position of A or B in relation to each other then either of the points can represent A or B, that is

A(9, 2), B(13, 4)

OR

A(13, 4), B(9, 2)

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