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9. Which of the following is an equation of a line that is perpendicular to y = -2x + 5 and passes through the point (-2, 3)?

A) y = −1/2x + 2
B) y = -2x - 1
C) y = 2x + 7
D) y = 1/2x + 4

11. Find the missing side lengths, perimeter, and area of triangle ABC on the coordinate plane. All answers should be in simplest radical form, if necessary: Note: the formula for area of a triangle is A = bh/2, where b is the length of the base (BC), and h is the height from side BC to point A.
Point A - (-1, 4)
Point B - (-3, 2)
Point C - (0, 2)
Side Length AB: _______
Side Length BC: _______
Side Length AC: _______
Perimeter of Triangle ABC: _______
Area of Triangle ABC: _______
A) 7 + Sq8
B) 2
C) Sq8
D) Sq5
E) 5
F) 3
G) 2Sq2
H) 3 + 2Sq + Sq5
I) Sq3

1 answer

To solve the problems, we need to find the necessary equations and distances.

Problem 9: Equation of a Line Perpendicular to \( y = -2x + 5 \)

  1. Find the slope of the line \( y = -2x + 5 \):

    • The slope (m) is -2.

  2. Find the slope of the perpendicular line:

    • The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope.
    • Therefore, the slope of the perpendicular line is \( \frac{1}{2} \) (negative reciprocal of -2).

  3. Use the point-slope form of the line's equation with the point (-2, 3):

    • Point-slope form: \( y - y_1 = m(x - x_1) \)
    • Here, \( (x_1, y_1) = (-2, 3) \) and \( m = \frac{1}{2} \).
    • The equation becomes: \[ y - 3 = \frac{1}{2}(x + 2) \]
    • Simplifying this: \[ y - 3 = \frac{1}{2}x + 1 \] \[ y = \frac{1}{2}x + 4 \]

  4. Choose the correct option:

    • The equation is \( y = \frac{1}{2}x + 4 \), which corresponds with option D.

Answer for Question 9: D) \( y = \frac{1}{2}x + 4 \)

Problem 11: Triangle ABC with Points A (-1, 4), B (-3, 2), C (0, 2)

  1. Find side lengths:

    • Length AB: \[ d = \sqrt{(-3 - (-1))^2 + (2 - 4)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
    • Length BC: \[ d = \sqrt{(0 - (-3))^2 + (2 - 2)^2} = \sqrt{(3)^2 + (0)^2} = \sqrt{9} = 3 \]
    • Length AC: \[ d = \sqrt{(0 - (-1))^2 + (2 - 4)^2} = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

  2. Calculate the perimeter:

    • Perimeter = AB + BC + AC \[ = 2\sqrt{2} + 3 + \sqrt{5} \]

  3. Find the area:

    • Base \( BC = 3 \).
    • Height: From point A (-1, 4) to line BC (where y = 2), the height is the vertical distance, which is \( 4 - 2 = 2 \).
    • Area \( A = \frac{bh}{2} \): \[ A = \frac{3 \times 2}{2} = 3 \]

Summary of Answers for Question 11:

  • Side Length AB: \( 2\sqrt{2} \)
  • Side Length BC: \( 3 \)
  • Side Length AC: \( \sqrt{5} \)
  • Perimeter of Triangle ABC: \( 2\sqrt{2} + 3 + \sqrt{5} \)
  • Area of Triangle ABC: \( 3 \)

Answers:

  • AB: \( 2\sqrt{2} \)
  • BC: \( 3 \)
  • AC: \( \sqrt{5} \)
  • Perimeter: \( 2\sqrt{2} + 3 + \sqrt{5} \)
  • Area: 3

If there are specific answer choices to pick from for the side lengths and others, please provide them for a more tailored response.