A line intersects a parabola at two distinct points in the standard (x, y) coordinate plan… — ACT

A line intersects a parabola at two distinct points in the standard $(x, y)$ coordinate plane. The system of equations representing these curves is given by:

\begin{aligned} x^2 - 7x + 3y &= 6 \\ 3y - 4x &= 6 \end{aligned}

What is the distance between the two intersection points?

Answer: 5

Answer

Solving the system of equations yields the intersection points

(0, 2)

and

(3, 6)

. The distance between these points is computed using the distance formula:

\sqrt{(3 - 0)^2 + (6 - 2)^2} = \sqrt{25} = 5

Step-by-Step Solution

Isolate the

3y

term in the linear equation.

3y = 4x + 6

This allows for a direct substitution into the quadratic equation without introducing fractional coefficients.

Substitute

3y

into the first equation and simplify.

x^2 - 3x = 0

Substituting

4x + 6

for

3y

yields

x^2 - 7x + (4x + 6) = 6

. Subtracting 6 from both sides and combining like terms simplifies this to a basic quadratic equation.

Solve the quadratic equation for

x

and determine the corresponding

y

-coordinates.

The intersection points are

(0, 2)

and

(3, 6)

Factoring gives

x(x - 3) = 0

, so the

x

-coordinates are

0

and

3

. Substituting

x = 0

into the linear relation gives

3y = 6 \implies y = 2

. Substituting

x = 3

gives

3y = 18 \implies y = 6

Calculate the distance between the two coordinates.

Using the distance formula:

d = \sqrt{(3 - 0)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5

Key Concept

Systems of Linear and Non-Linear Equations

Question

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