IDZ Ryabushko 3.2 Option 6

No. 1 To solve the problem, we need a formula for finding the distance between two points on a plane:

d = √((x₂-x₁)² + (y₂-y₁)²)

a) To find the equation of side AB, find the coordinates of points A and B:

A(-2, -3); B(1, 6)

Let's find the distance between points A and B:

d = √((1-(-2))² + (6-(-3))²) = √(3² + 9²) = √90

The equation for side AB is:

(x₁ - x₂)y + (y₂ - y₁)x + (x₁y₂ - x₂y₁) = 0

Substitute the coordinates of points A and B:

(-3 - 6)y + (1 - (-2))x + (-3×6 - 1×(-2)) = 0

-9y + 3x + 20 = 0

b) To find the equation for the height of CH, we find the coordinates of point H, which is the intersection of the height of CH and side AB. To do this, find the equation of line AB and the coordinates of point H:

(-3 - 6)y + (1 - (-2))x + (-3×6 - 1×(-2)) = 0

-9y + 3x + 20 = 0

The equation of line AB is:

y = (3/9)x + (20/9)

Since the height CH passes through point C and is perpendicular to side AB, the slope of the height is -3/9 = -1/3. Using the coordinates of point C and the angular coefficient of the height, we find the equation for the height of CH:

y - 1 = (-1/3)(x - 6)

y = (-1/3)x + 7

c) To find the equation of the median AM, let’s find the coordinates of point M, which is the midpoint of side AB. To do this, we find the arithmetic mean of the coordinates of points A and B:

x = (-2 + 1)/2 = -0.5; y = (-3 + 6)/2 = 1.5

Point M has coordinates (-0.5, 1.5). The equation of the median AM passes through points A and M, so we use the formula to find the equation of a straight line passing through two given points:

y - (-3) = ((1.5 - (-3))/(-0.5 - (-2)))(x - (-2))

y + 3 = (4.5/1.5)(x + 2)

y = 3x - 9

d) To find the intersection point of the median AM and the height CH, we solve the system of equations:

y = (-1/3)x + 7

y = 3x - 9

Solving the system, we obtain the coordinates of point N:

x = 1; y = 4

Point N has coordinates (1, 4).

e) To find the equation of a straight line passing through vertex C and parallel to side AB, we find the slope of side AB:

k = (6 - (-3))/(1 - (-2)) = 3

Since the desired line passes through point C, the equation of the line has the form:

y - 1 = 3(x - 6)

y = 3x - 17

f) To find the distance from point C to line AB, we use the formula:

d = |ax₀ + by₀ + c|/√(a² + b²), where x₀ and y₀ are the coordinates of point C; a, b and c are the coefficients of the equation of the line AB.

Substitute the values ​​into the formula:

d = |(-9)×(-1) + 3×(-3) + 20|/√(9 + 81) = 6/√90 = 2√10/3.

Thus, the distance from point C to line AB is 2√10/3.

No. 2 To prove that the quadrilateral ABCD is a trapezoid, it is necessary to check that the angles between parallel sides are equal (this property is called the property of the bases of a trapezoid), that is, ∠A = ∠C and ∠B = ∠D.

Let's find the equations of sides AD and BC using the formula for finding the distance between two points on a plane:

AD: d = √((x₂-x₁)² + (y₂-y₁)²) = √((-5-3)² + (5-6)²) = √74

BC: d = √((x₂-x₁)² + (y₂-y₁)²) = √((5-(-1))² + (2-(-3))²) = √65

Since the sides AD and BC are not equal, the quadrilateral ABCD is not an isosceles trapezoid.

It remains to check that the angles between parallel sides are equal:

∠А = arctg((5-6)/(-5-3)) ≈ 128.66°

∠С = arctg((2-(-3))/(5-(-1))) ≈ 128.66°

∠В = arctg((5-6)/(3-(-1))) ≈ -41.19°

∠D = arctg((2-(-3))/(5-3)) ≈ -41.19°

Thus, ∠A = ∠C and ∠B = ∠D, which confirms that ABCD is a trapezoid.

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