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Zorluk: ZorSystems of Linear and Non-Linear Equations

The equation of a circle is x2+y2=25x^2 + y^2 = 25, and the equation of a line is 2xy=52x - y = 5. If the line intersects the circle at the points PP and QQ, what is the sum of the yy-coordinates of PP and QQ?

  1. A
    -18
  2. B
    -6
  3. -2Cevap
  4. D
    2
  5. E
    4

Cevap

The sum of the yy-coordinates is 2-2.
Substituting the linear equation into the circle equation yields a quadratic equation with roots x=0x = 0 and x=4x = 4. Evaluating the linear equation at these values gives the yy-coordinates 5-5 and 33. Adding these yy-coordinates results in 2-2.

Adım Adım Çözüm

1
Rearrange the linear equation to express yy in terms of xx.
y=2x5y = 2x - 5
Expressing one variable in terms of the other allows for substitution into the quadratic circle equation.
2
Substitute y=2x5y = 2x - 5 into the circle equation x2+y2=25x^2 + y^2 = 25 and simplify.
x2+(2x5)2=25    x2+4x220x+25=25    5x220x=0x^2 + (2x - 5)^2 = 25 \implies x^2 + 4x^2 - 20x + 25 = 25 \implies 5x^2 - 20x = 0
This substitution reduces the system of equations to a single quadratic equation in terms of xx.
3
Solve the quadratic equation 5x220x=05x^2 - 20x = 0 for xx.
5x(x4)=0    x1=05x(x - 4) = 0 \implies x_1 = 0 and x2=4x_2 = 4
Finding the roots of this quadratic equation gives the xx-coordinates of the intersection points.
4
Substitute the xx-values back into the linear equation y=2x5y = 2x - 5 to find the corresponding yy-coordinates.
For x1=0x_1 = 0, y1=2(0)5=5y_1 = 2(0) - 5 = -5. For x2=4x_2 = 4, y2=2(4)5=3y_2 = 2(4) - 5 = 3. The intersection points are P(0,5)P(0, -5) and Q(4,3)Q(4, 3).
This step determines the coordinates of the two points of intersection.
5
Calculate the sum of the yy-coordinates of the points PP and QQ.
y1+y2=5+3=2y_1 + y_2 = -5 + 3 = -2
This yields the final value requested by the question.

Anahtar Kavram

Solving a system of linear and circular equations by substitution
Tahmini Süre:2m 0s
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