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Difficulty: HardSystems of Linear and Non-Linear Equations

A system of equations consists of the circle defined by x2+(y−4)2=10x^2 + (y - 4)^2 = 10 and the line defined by y=2x−1y = 2x - 1. The two points of intersection of this system and the origin, (0,0)(0, 0), form the vertices of a triangle in the standard (x,y)(x, y) coordinate plane. What is the area of this triangle?

  1. A
    2
  2. B
    7\sqrt{7}
  3. 1Answer
  4. D
    4
  5. E
    5

Answer

The correct area of the triangle is 1.
The correct answer of 1 is found by substituting the linear equation into the circle's equation, solving the resulting quadratic equation to find the coordinates of the two intersection points, (1,1)(1, 1) and (3,5)(3, 5), and then applying the coordinate area formula for a triangle with a vertex at the origin.

Step-by-Step Solution

1
Substitute the linear equation into the circle's equation to form a quadratic equation in terms of xx.
x2+(2x−5)2=10x^2 + (2x - 5)^2 = 10
This allows us to solve for the x-coordinates of the intersection points by replacing yy with 2x−12x - 1 in (y−4)2(y - 4)^2 to get (2x−5)2(2x - 5)^2.
2
Expand the squared binomial and simplify the quadratic equation.
5x2−20x+15=05x^2 - 20x + 15 = 0
Expanding (2x−5)2(2x - 5)^2 yields 4x2−20x+254x^2 - 20x + 25. Adding x2x^2 and subtracting 1010 from both sides simplifies the equation to standard quadratic form.
3
Divide the quadratic equation by 5 and solve for xx by factoring.
x=1x = 1 and x=3x = 3
Dividing the equation by 5 yields x2−4x+3=0x^2 - 4x + 3 = 0, which factors as (x−1)(x−3)=0(x - 1)(x - 3) = 0.
4
Find the corresponding y-coordinates by substituting the xx-values back into the linear equation y=2x−1y = 2x - 1.
The intersection points are (1,1)(1, 1) and (3,5)(3, 5).
For x=1x = 1, y=2(1)−1=1y = 2(1) - 1 = 1. For x=3x = 3, y=2(3)−1=5y = 2(3) - 1 = 5.
5
Calculate the area of the triangle with vertices at the origin (0,0)(0, 0) and the two intersection points (1,1)(1, 1) and (3,5)(3, 5).
Area = 1
Using the coordinate area formula for a triangle with one vertex at the origin, Area=12∣x1y2−x2y1∣=12∣1(5)−3(1)∣=12∣2∣=1\text{Area} = \frac{1}{2} |x_1 y_2 - x_2 y_1| = \frac{1}{2} |1(5) - 3(1)| = \frac{1}{2} |2| = 1.

Key Concept

Solving a system of linear and non-linear equations and using coordinate geometry to find the area of a triangle.

Alternative Method

Instead of using the coordinate area formula, we can find the distance between the two intersection points to serve as the base of the triangle (b=(3−1)2+(5−1)2=25b = \sqrt{(3-1)^2 + (5-1)^2} = 2\sqrt{5}), and find the perpendicular distance from the origin to the line 2x−y−1=02x - y - 1 = 0 to serve as the height (h=∣−1∣22+(−1)2=15h = \frac{|-1|}{\sqrt{2^2 + (-1)^2}} = \frac{1}{\sqrt{5}}). The area is then 12×25×15=1\frac{1}{2} \times 2\sqrt{5} \times \frac{1}{\sqrt{5}} = 1.
Estimated Time:2m 30s
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Topics

Systems of Linear and Non-Linear Equations

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