A parabolic arch is modeled by the equation y = (x - 3)^2 - 5 in the standard (x, y) coord… — ACT

A parabolic arch is modeled by the equation $y = (x - 3)^2 - 5$ in the standard $(x, y)$ coordinate plane. A straight pathway is modeled by a line where the $y$ -coordinate of any point is 2 less than its $x$ -coordinate. If the pathway intersects the arch at points $A$ and $B$ , what is the area, in square units, of the triangle with vertices at $A$ , $B$ , and the origin $(0, 0)$ ?

A
3
5Cevap
C
7
D
8
E
10

Cevap

The area of the triangle with vertices at the intersection points and the origin is 5 square units.

To find the area of the triangle, we first solve the system of equations. Substituting the pathway's equation y = x - 2 into the parabola's equation y = (x - 3)^2 - 5 yields x^2 - 7x + 6 = 0, which factors to (x - 6)(x - 1) = 0. This gives x = 6 and x = 1. The corresponding y-coordinates are y = 4 and y = -1, representing the intersection points (6, 4) and (1, -1). The area of the triangle with these vertices and the origin (0, 0) is calculated as 0.5 * |6(-1) - 4(1)| = 5.

Adım Adım Çözüm

Set up the system of equations by substituting the linear equation into the quadratic equation.

(x - 3)^2 - 5 = x - 2

The pathway is described as having a y-coordinate that is 2 less than the x-coordinate, which translates to the linear equation y = x - 2. Substituting this into the parabola's equation allows us to find the intersection points.

Expand the quadratic term and simplify the equation into standard quadratic form.

x^2 - 7x + 6 = 0

Expanding

(x - 3)^2

gives

x^2 - 6x + 9

. Subtracting

x

and adding

2

to both sides results in standard quadratic form.

Factor the quadratic equation to solve for the x-coordinates.

(x - 6)(x - 1) = 0

, so

x = 6

x = 1

Factoring allows us to find the x-values that satisfy the intersection condition.

Substitute the x-values back into the linear equation to find the corresponding y-coordinates.

For

x = 6

y = 4

giving point

(6, 4)

. For

x = 1

y = -1

giving point

(1, -1)

The intersection points must satisfy both equations in the system.

Calculate the area of the triangle with vertices

(0, 0)

(6, 4)

, and

(1, -1)

using the coordinate area formula.

Area =

\frac{1}{2} |6(-1) - 4(1)| = \frac{1}{2} |-10| = 5

square units.

The area of a triangle with one vertex at the origin and others at

(x_1, y_1)

and

(x_2, y_2)

\frac{1}{2} |x_1 y_2 - y_1 x_2|

Anahtar Kavram

Solving systems of linear and quadratic equations and finding the area of a triangle in the coordinate plane.

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