First Order Linear Equation

Category: Differential Equations, Integral Calculus, Algebra

"Published in Newark, California, USA"

Find the general solution for the given equation

Solution:

Consider the given equation

Check if the given equation is exact or not exact as follows

Let 

Let

Since 

The given equation is not an exact equation. In this case, we need to find the integrating factor and then multiply it to the both sides of the equation. Let's consider again the given equation

Arrange the above equation into its standard form as follows

Divide both sides of the equation by dx

Divide both sides of the equation by (x + 1)²

As you notice that the above equation is now already in standard form

where

and

The integrating factor is

Therefore, the general solution for the above equation is

Area Derivation - Triangle, Three Vertices

Category: Analytic Geometry, Plane Geometry, Algebra

"Published in Newark, California, USA"

Given a triangle with vertices as shown below

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Prove that the area of a triangle is

Solution:

The first step is to draw vertical lines from the vertices to the x-axis as shown below

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Area of Triangle ABC = Area of Trapezoid ACDF - Area of Trapezoid ABEF - Area of Trapezoid BCDE

Label further the figure, we have

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Consider Trapezoid ACDF

Consider Trapezoid ABEF

Consider Trapezoid BCDE

Therefore

Using a distance of two points formula

The above equation becomes

You notice that the above equation looks like a matrix or determinant because of the sequence of x and y. The three positive terms are the principal diagonals (product from top left to bottom right) while the three negative terms are the secondary diagonals (product from bottom left to top right). The above equation can be written as

Therefore

Proving - Parallelogram, Triangles

Category: Plane Geometry

"Published in Newark, California, USA"

Given: Parallelogram CDEF; S and T are midpoints of EF and ED. 

Prove: SR ≅ FD

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Solution:

Consider the given figure

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Proof:

1. Statement: Parallelogram CDEF

    Reason: Given item.

2. Statement: S and T are the midpoints of EF and ED.

    Reason: Given item.

3. Statement: FS ≅ SE 

    Reason: Point S bisects the line segment EF into two equal parts. 

4. Statement: ET ≅ TD

    Reason: Point T bisects the line segment ED into two equal parts.

5. Statement: FE ║ CD

    Reason: The opposite sides of a parallelogram are parallel.

6. Statement: ∠ SET ≅ ∠ TDR

    Reason: If a transversal line (ED) passed the two parallel lines (EF and CD), then the alternating interior angles are congruent. 

7. Statement: ∠ STE ≅ ∠ DTR

    Reason: Vertical angles are congruent.

8. Statement: ∆SET ≅ ∆DTR

    Reason: Angle Side Angle (ASA) Postulate.

9. Statement: DR ≅ SE

    Reason: Since ∆SET ≅ ∆DTR, then all sides of a triangle are congruent to all sides of other triangle.

10. Statement: DR ≅ SE ≅ FS

      Reason: Transitive property of congruence. 

11. Statement: FS ║ DR

      Reason: Since Point R is colinear with CD and CD is parallel to EF, then it follows that FS is parallel to DR since Point S is colinear with EF.

12. Statement: FDRS is a parallelogram.

      Reason: Since FS and DR are parallel and congruent, then it follows that the figure formed by the points FDRS is a parallelogram.

13. Statement: FD ≅ SR

      Reason: The two opposite sides of a parallelogram are congruent and parallel.