To find the points of intersection between the line $4x - 3y - 10 = 0$ and the circle $x^2 + y^2 - 2x + 4y - 20 = 0$, we follow these steps:

Step 1: Express $y$ from the line equation

Given the line equation: $4x - 3y - 10 = 0$ We can solve for $y$: $3y = 4x - 10$ $y = \frac{4x - 10}{3}$

Step 2: Substitute $y$ in the circle equation

Substitute $y = \frac{4x - 10}{3}$ into the circle equation: $x^2 + y^2 - 2x + 4y - 20 = 0$ Substitute $y$: $x^2 + \left(\frac{4x - 10}{3}\right)^2 - 2x + 4\left(\frac{4x - 10}{3}\right) - 20 = 0$ Now, let's simplify and solve for $x$.

Step 3: Simplify the equation

First, expand and simplify the terms: $x^2 + \frac{(4x - 10)^2}{9} - 2x + \frac{16x - 40}{3} - 20 = 0$ Expand $(4x - 10)^2$: $x^2 + \frac{16x^2 - 80x + 100}{9} - 2x + \frac{16x - 40}{3} - 20 = 0$ Multiply the entire equation by 9 to eliminate the fractions: $9x^2 + 16x^2 - 80x + 100 - 18x + 48x - 120 - 180 = 0$ Simplify: $25x^2 - 50x - 200 = 0$

Step 4: Solve the quadratic equation

Divide the entire equation by 25: $x^2 - 2x - 8 = 0$ Factor the quadratic equation: $(x - 4)(x + 2) = 0$ So, $x = 4$ or $x = -2$.

Step 5: Find corresponding $y$ values

For $x = 4$: $y = \frac{4(4) - 10}{3} = \frac{16 - 10}{3} = \frac{6}{3} = 2$ For $x = -2$: $y = \frac{4(-2) - 10}{3} = \frac{-8 - 10}{3} = \frac{-18}{3} = -6$

Final Answer

The points of intersection are $(4, 2)$ and $(-2, -6)$.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. How do you find the distance between two points of intersection?
2. What is the length of the chord formed by the intersection of the line and the circle?
3. How do you determine if a line is tangent to a circle?
4. What is the slope of the line perpendicular to the given line at the point of intersection?
5. How do you find the equation of a circle given its center and a radius?

Tip: When solving the intersection of a line and a circle, substituting one equation into the other reduces the problem to solving a quadratic equation!