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 |
- Distance between A(− 3, 2) and B(4, 3) dAB = √(4 − (− 3))² + (3 − 2)² = √7² + 1² = √50 = √2 · 5² = 5√2
- Distance between B(4, 3) and C(9, − 2) dBC = √(9 − 4)² + (− 2 − 3)² = √5² + (−5)² = √50 = √2 · 5² = 5√2
- Distance between C(9, − 2) and D(2, − 3) dCD = √(2 − 9)² + (− 3 − (− 2))² = √(−7)² + (−1)² = √50 = √2 · 5² = 5√2
- Distance between D(2, − 3) and A(− 3, 2) dDA = √(2 − (−3))² + (− 3 − 2)² = √5² + (−5)² = √50 = √2 · 5² = 5√2
- All for sides are equal in length
- Opposite sides are parallel
but
- The four angles are not right
- The diagonals are not congruent
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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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continue... |
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