Question

4.10 Unit Test: Analytic Geometry - Part 1

Answer 1

The correct responses for the analytical geometry questions are;

1. 50 units²
2. 12·√10 unit²
3. 15.2 units
4. Parallel: $line \ q \ with \ slope \ -\frac{3}{4}$; Perpendicular: $line \ n \ with \ slope \ \frac{4}{3}$Neither: $Line \ m \ with \ slope \ \frac{3}{4}$ and $line \ p \ with \ slope \ -\frac{4}{3}$
6. $\frac{5}{4}$
7. 3·y - x = 12
8. y + 2·x = -3
9. C(2, 2), ΔABC is a Right Triangle; C(0, 4), ΔABC is a Not a Right Triangle; C(-2, 1), ΔABC is a Right Triangle
10. The slope of $\overline{AB}$ is -2, the slope of $\overline{BC}$ is $\underline{\frac{1}{2}}$, the slope of $\overline{CD}$ is $\underline{-\frac{3}{4}}$, the slope of $\overline{AD}$ is a trapezoid because only one pair of opposite sides is parallel.
11. $\underline{(3\cdot a, \, a)}$
12. The coordinates of the rectangle ABCD are A(0, 0), B(a, 0), $\underline{C(a, \, b)}$, and D(0, b); The length of $\overline{AC}$ is equal to $\underline{\sqrt{ a^2 + b^2}}$; The length of $\overline{BD}$ is equal to  

Reasons:

1. The length of the sides of the rectangle are;

Width = √(4² + 2²) = 2·√5

Length = √(10² + 5²) = 5·√5

Area = Length × Width

∴ Area = 5·√5 × 2·√5 = 50

Area = 50 units²

2. Width = √((-2 - (-2))² + (-5 - 1)²) = 6

Length = √((4 - (-2))² + (3 - 1)²) = 2·√10

∴ Area = 2·√10 × 6 = 12·√10

The area of the rectangle = 12·√10 unit²

3. The lengths of the sides of the polygon are;

Side 1 = √(1² + 2²) = √5

Side 2 = √(2² + 2²) = 2·√2

Side 3 = √(3² + 1²) = √10

Side 4 = 3

Side 5 = 4

The perimeter = √5 + 2·√2 + √10 + 3 + 4 ≈ 15.2 units

5. The slope of ║ lines are equal,  the slope of the line ║ to the line whose slope is $-\dfrac{3}{4}$, is also

The ║ line is $line \ q \ with \ slope \ -\dfrac{3}{4}$.

The slope of a line ⊥ to another line that has a slope of m is $-\dfrac{1}{m}$

Slope of the ⊥ line is $-\dfrac{1}{-\dfrac{3}{4}} =\dfrac{4}{3}$, which is  $line \ n \ with \ slope \ \dfrac{4}{3}$

Neither are;

$Line \ m \ with \ slope \ \dfrac{3}{4}$ and $line \ p \ with \ slope \ -\dfrac{4}{3}$

6. The equation of line m is 4·x + 5·y = -2

Slope of the line m, is $-\dfrac{4}{5}$

Slope of the ⊥ line is $\dfrac{5}{4}$

7. Equation of line is -x + 3·y = 6

Point through which the line passes is (3, 5)

Slope of the line is $\dfrac{1}{3}$

Slope of the ║ line is $\dfrac{1}{3}$

The equation of the ║ line is $y - 5 = \dfrac{1}{3} \cdot (x - 3)$

Which gives;

3·y - 15 = x - 3

3·y - x = -3 + 15 = 12

The equation of the ║ line is; 3·y - x = 12

8. Equation of line is -x + 2·y = 4

Point ⊥ line passes = (-2, 1)

Slope is $\dfrac{1}{2}$

Slope of the ⊥ line is -2

Which gives; (y - 1) = -2·(x - (-2))

(y - 1) = -2·x - 4 + 1

The equation of the ⊥ line is; y + 2·x = -3

9.  $Slope, \ of \ \overline{AB} \, =\dfrac{1-4}{-1-(-2)} = -3$

Slope of BC is $\dfrac{1}{3}$

At C(2, 2), we have;

$Slope, \ of \ \overline{BC} \, =\dfrac{1-2}{-1-2} = \dfrac{1}{3}$

• C(2, 2), ΔABC is a Right Triangle

At C(0, 4), we have;

$Slope, \ of \ \overline{BC} \, =\dfrac{4-1}{0-(-1)} =3$

Therefore;

• C(0, 4), ΔABC is a Not a Right Triangle

At C(-2, 1), we have;

$Slope, \ of \ \overline{BC} \, =\dfrac{1-1}{-2-(-1)} =0$

However, we have;

$Slope, \ of \ \overline{AC} \, =\dfrac{1-(-2)}{-2-(-2)} = \infty$

Therefore $\overline{BC}$ ⊥ $\overline{AC}$.

• C(-2, 1), ΔABC is a Right Triangle

10. Given quadrilateral ABCD;

$Slope, \ of \ \overline{AB} \, =\dfrac{3 -(-1)}{-3-(-1)} = -2$

$Slope, \ of \ \overline{BC} \, =\dfrac{3 -5}{-3-1} = \dfrac{1}{2}$

$Slope, \ of \ \overline{CD} \, =\dfrac{2 -5}{5-1} = -\dfrac{3}{4}$

$Slope, \ of \ \overline{AD} \, =\dfrac{2 -(-1)}{5-(-1)} = \dfrac{1}{2}$

Sides  $\overline{BC}$ ║ $\overline{AD}$, sides  $\overline{AB}$ ∦ $\overline{CD}$, sides  

∴ ABCD is a right trapezoid.

Which gives;

The slope of $\overline{AB}$ is -2, the slope of $\overline{BC}$ is $\underline{\frac{1}{2}}$, the slope of $\overline{CD}$ is $\underline{-\frac{3}{4}}$, the

slope of $\overline{AD}$ is $\underline{\frac{1}{2}}$. Quadrilateral ABCD is a trapezoid because only one pair

of opposite sides is ║.

11. Shorter side = a

∴ Longer side = 3·a (║to the x-axis)

Top right corner is $\underline {(3\cdot a, \, a)}$.

12. The coordinates of the point C is (a, b)

According to Pythagoras's theorem, we have;

$\overline{AC}$ = √(a² + b²)

$\overline{BD}$ = √(b² + a²)

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