Math Problem Statement

Aprēķināt laukumu trijstūrim, kura virsotnes ir koordinātu sākumpunkts un taisnes y = -2x + 6 krustpunkti ar Ox un Oy asīm. Uzrakstīt šī trijstūra tās viduslīnijas vienādojumu, kas paralēla garākajai trijstūra malai.

Solution

The image appears to contain a problem in Latvian about a triangle whose vertices are defined by the coordinate origin and the intersection points of the line y=2x+6y = -2x + 6 with the OxOx and OyOy axes. Here's a step-by-step solution to the task:


1. Find the vertices of the triangle

The line y=2x+6y = -2x + 6 intersects the axes as follows:

Intersection with OyOy-axis:

  • At the OyOy-axis, x=0x = 0.
  • Substituting into the equation y=2(0)+6y = -2(0) + 6: y=6y = 6 So the intersection point is (0,6)(0, 6).

Intersection with OxOx-axis:

  • At the OxOx-axis, y=0y = 0.
  • Solving 0=2x+60 = -2x + 6: x=3x = 3 So the intersection point is (3,0)(3, 0).

Thus, the vertices of the triangle are:

  • A=(0,0)A = (0, 0) (origin),
  • B=(3,0)B = (3, 0),
  • C=(0,6)C = (0, 6).

2. Calculate the area of the triangle

The formula for the area of a triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), (x3,y3)(x_3, y_3) is: Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Substituting A=(0,0)A = (0, 0), B=(3,0)B = (3, 0), and C=(0,6)C = (0, 6): Area=120(06)+3(60)+0(00)\text{Area} = \frac{1}{2} \left| 0(0 - 6) + 3(6 - 0) + 0(0 - 0) \right| Area=120+18+0=1218=9\text{Area} = \frac{1}{2} \left| 0 + 18 + 0 \right| = \frac{1}{2} \cdot 18 = 9

So, the area of the triangle is 9square units9 \, \text{square units}.


3. Write the equation of the midline parallel to the longest side

The longest side of the triangle is BCBC, which connects B=(3,0)B = (3, 0) and C=(0,6)C = (0, 6). The slope of BCBC is: slope of BC=6003=2\text{slope of } BC = \frac{6 - 0}{0 - 3} = -2

The midline parallel to BCBC passes through the midpoint of the opposite side ABAB, whose midpoint is: Midpoint of AB=(0+32,0+02)=(32,0)\text{Midpoint of } AB = \left( \frac{0 + 3}{2}, \frac{0 + 0}{2} \right) = \left( \frac{3}{2}, 0 \right)

Using the point-slope form of a line, yy1=m(xx1)y - y_1 = m(x - x_1), with slope 2-2 and point (32,0)\left( \frac{3}{2}, 0 \right): y0=2(x32)y - 0 = -2\left(x - \frac{3}{2}\right) y=2x+3y = -2x + 3

Thus, the equation of the midline is: y=2x+3y = -2x + 3


Final Results:

  1. Area of the triangle: 9square units9 \, \text{square units}.
  2. Equation of the midline parallel to the longest side: y=2x+3y = -2x + 3.

Would you like further clarification or details on any of the steps? 😊


Related Questions:

  1. How do you calculate the distance between two points in a coordinate plane?
  2. Can you prove why the midline is always parallel to one of the triangle’s sides?
  3. What are the general properties of triangles formed by intersecting a line with the coordinate axes?
  4. Can you derive the equation of a line given two points on it?
  5. How would the problem change if the line equation were y=mx+cy = mx + c?

Tip:

Always sketch the triangle to visualize relationships between its sides, vertices, and midlines!

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Triangles
Lines and Slopes

Formulas

Area of a triangle: 1/2 * |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
Slope of a line: (y2-y1)/(x2-x1)
Equation of a line: y - y1 = m(x - x1)

Theorems

Basic properties of triangles
Midline theorem for triangles

Suitable Grade Level

Grades 8-10