Let's analyze the given quadrilateral [tex]\(ABCD\)[/tex] with vertices [tex]\(A(4,0)\)[/tex], [tex]\(B(14,11)\)[/tex], [tex]\(C(-10,6)\)[/tex], and [tex]\(D(-10,-5)\)[/tex].
### Part (i): Show that [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] bisect each other at right angles.
To show that [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] bisect each other, we need to find the midpoints of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] and verify that they coincide. Then, we need to verify that the slopes of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are negative reciprocals, indicating they are perpendicular.
#### Midpoints Calculation
The midpoint of a line segment with endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
For segment [tex]\(AC\)[/tex]:
- Coordinates of [tex]\(A\)[/tex]: [tex]\( (4, 0) \)[/tex]
- Coordinates of [tex]\(C\)[/tex]: [tex]\((-10, 6)\)[/tex]
[tex]\[
\text{Midpoint of } AC = \left( \frac{4 + (-10)}{2}, \frac{0 + 6}{2} \right) = \left( \frac{-6}{2}, \frac{6}{2} \right) = (-3, 3)
\][/tex]
For segment [tex]\(BD\)[/tex]:
- Coordinates of [tex]\(B\)[/tex]: [tex]\( (14, 11) \)[/tex]
- Coordinates of [tex]\(D\)[/tex]: [tex]\((-10, -5)\)[/tex]
[tex]\[
\text{Midpoint of } BD = \left( \frac{14 + (-10)}{2}, \frac{11 + (-5)}{2} \right) = \left( \frac{4}{2}, \frac{6}{2} \right) = (2, 3)
\][/tex]
Since the midpoints of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are [tex]\((-3, 3)\)[/tex] and [tex]\( (2, 3)\)[/tex], respectively, they do not coincide, so [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] do not bisect each other at the same midpoint. Hence, [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] do not bisect each other.
#### Slopes Calculation
The slope [tex]\(m\)[/tex] of a line segment with endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For segment [tex]\(AC\)[/tex]:
[tex]\[
\text{Slope of } AC = \frac{6 - 0}{-10 - 4} = \frac{6}{-14} = -\frac{3}{7} = -0.42857142857142855
\][/tex]
For segment [tex]\(BD\)[/tex]:
[tex]\[
\text{Slope of } BD = \frac{-5 - 11}{-10 - 14} = \frac{-16}{-24} = \frac{2}{3} = 0.6666666666666666
\][/tex]
The product of the slopes:
[tex]\[
(-0.42857142857142855) \times (0.6666666666666666) = -0.2857142857142857
\][/tex]
Since the product of the slopes is not -1, [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are not perpendicular. Thus, they do not bisect each other at right angles.
### Part (ii): Calculate the ratio [tex]\(\lvert BD \rvert : \lvert AC \rvert \)[/tex].
#### Lengths Calculation
The length of a line segment with endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
For segment [tex]\(AC\)[/tex]:
[tex]\[
\lvert AC \rvert = \sqrt{(-10 - 4)^2 + (6 - 0)^2} = \sqrt{(-14)^2 + (6)^2} = \sqrt{196 + 36} = \sqrt{232} = 15.231546211727817
\][/tex]
For segment [tex]\(BD\)[/tex]:
[tex]\[
\lvert BD \rvert = \sqrt{(-10 - 14)^2 + (-5 - 11)^2} = \sqrt{(-24)^2 + (-16)^2} = \sqrt{576 + 256} = \sqrt{832} = 28.844410203711913
\][/tex]
#### Ratio Calculation
The ratio [tex]\(\frac{\lvert BD \rvert}{\lvert AC \rvert}\)[/tex] is therefore:
[tex]\[
\text{Ratio} = \frac{28.844410203711913}{15.231546211727817} = 1.8937283058959973
\][/tex]
So the ratio of [tex]\(\lvert BD \rvert\)[/tex] to [tex]\(\lvert AC \rvert\)[/tex] is approximately [tex]\(1.8937\)[/tex].
