Exercises Coordinate Geometry 2

Solution to 7

Exercises Coordinate Geometry 1

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Solution to 5

(a) To show that all four sides are equal in length, we need to calculate their distances (length), using the Distance Formula.

d = √(changing in x)² + (changing in y)² d = √(x2 − x1)² + (y2 − y1)²

1. Distance between A(− 3, 2) and B(4, 3)
d_AB = √(4 − (− 3))² + (3 − 2)² = √7² + 1² = √50 = √2 · 5² = 5√2


2. Distance between B(4, 3) and C(9, − 2)
d_BC = √(9 − 4)² + (− 2 − 3)² = √5² + (−5)² = √50 = √2 · 5² = 5√2


3. Distance between C(9, − 2) and D(2, − 3)
d_CD = √(2 − 9)² + (− 3 − (− 2))² = √(−7)² + (−1)² = √50 = √2 · 5² = 5√2


4. Distance between D(2, − 3) and A(− 3, 2)
d_DA = √(2 − (−3))² + (− 3 − 2)² = √5² + (−5)² = √50 = √2 · 5² = 5√2

(b) As you can notice:
      - All for sides are equal in length
      - Opposite sides are parallel
but
      - The four angles are not right
      - The diagonals are not congruent

Therefore, the figure is still a quadrilateral, not a square, but a parallelogram.

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Solution to 6

(a) First solve the line eqaution 3x + 4y = 16 for y:
      that way you'll come to l₁: y = ⁻³/₄·x + 4, with the slope m₁ = ⁻³/₄.

      If l₂ passes through P and perpendicular to l₁, it means that the slope of m₂ ist negative       reciprocal to the slope of m₁
      ⇒ m₁ = ⁻¹/_m2 ⇔ m₂ = ⁻¹/_m1 = ⁻¹/_⁻³/4 = ⁴/₃, m₂ = ⁴/₃
      l₂ passes through P
      ⇒ 5 = ⁴/₃ − 7 + b, with b to be found!
      5 = ²⁸/₃ + b and b = ⁻¹³/₃

      Therefore, the equation of l₂: y = ⁴/₃·x − ¹³/₃

(b) Point of intersection of l₁ and l₂:
      this means l₁ = l₂
      ⇒ ⁻³/₄·x + 4 = ⁴/₃·x − ¹³/₃, solve for x:
      ⇒ x = 4
      then plot x into either l₁ or l₂:
      ⇒ y = ⁻³/₄·4 + 4 =
      In conclusion, the points l₁ and l₂ intersect at the point (4, 1)

(c) The perpendicular distance of P from the line l₁ represents the Hypotenuse of the slope       with the values x = 3 and y = 4
      therefore, we will use the Pythagoras formula for right triangle to find out the distance      concerned:

      so, d² = 3² + 4²    |   √ (square root it)
      ⇒ d = √3² + 4² = √5²
      ⇒ d = 5

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Exercises Coordinate Geometry

Solutions

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2. y = 4x − 8 meets the x-axis means that y = 0
by solving 4x − 8 = 0, it appears that x = 2
⇒ the line meets the x-axis at the point A(2, 0)

We'll now find the equation of the line with gradient 3, passing through the point A
The gradient m = ^{change in y}/_{change in x} = ³/₁, which means the line passes through another point B(1, 3) with A(2, 0) as origin. (see Plot 1)



Plot 1




3. y = − 2x + 8 meets the y-axis means that x = 0
⇒ y = − 2(0) + 8 and y = 8
The point met is B(0, 8)
The other line passing, with gradient 2 has B as origin. (see Plot 2)



Plot 2


4. y = ¹/₂·x + 6 meets the x-axis means that y = 0
then, y = ¹/₂·x + 6 = 0 ⇒ x = − 12
if another line passes through C, its equation should be written y = ²/₃·x + 6 (see Plot 3)
The second equation colud the be written: ²/₃·x − y + 6 = 0, with a = ²/₃, b = −, c = 6.



Plot 3



5. Just do it like the previous exercises !!!




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6. Just do it like the previous exercises !!!

More exercises

Coordinate Geometry

Perpendicular lines
Two lines or linear equations are perpendicular if their slopes are negative reciprocals of each other (opposite signs and upside down):

m1 = − 1/m2

Example: 5x − y = 8 and 5x = − x + 3

Let's check if these two functions are Perpendicular:
- first turn the fucntions into a slope intercept form (regular form of linear equations, in oder words solve for y): y = mx + b with m = slope

5x − y = 8 − y = 8 − 5x y = 5x − 8 The slope of this line is m1 = 5 or m1 = 5/1

5y = − x + 3 y = − 1/5·x − 3/5 The slope of this line is m2 = − 1/5

- then compare their slopes or gradients:
As you can notice m₁ = ⁵/₁ is exactly the negative reciprocal of m² = − ¹/₅ and vice versa.

⇒ Therefore, the lines 5x − y = 8 and 5x = − x + 3 are perpendicular!

Parallel lines
Two lines or linear equations are parallel if their slopes are equal (they have the same slopes). Since they'll never intersect, they continue forever without touching (assuming that these lines are on the same plane).

m1 = m2

Example:
One line passes through the points p₁(–1, –2) and p₂(1, 2); another line passes through the points q₁(–2, 0) and q₂(0, 4).

Our answer to this question is based on the calculation of slopes of both lines (functions).

⇒ With identical slopes, these lines are parallel!

Neither Parallel nor Perpendicular lines
Two lines or linear equations are neither parallel nor perpendicular if their slopes are neither the same and nor negative reciprocals of each other.

m1 ≠ m2 and m1 ≠ − 1/m2

Example:
A line passes through the points p₁(–4, 2) and p₂(0, 3); another line passes through the points q₁(–3, –2) and q₂(3, 2). Are these lines parallel, perpendicular, or neither?

These slope values are not the same, so the lines are not parallel. The slope values are not negative reciprocals either, so the lines are not perpendicular. ⇒ Then the answer is "neither".!

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