Problem 1 :
Think of each segment in the diagram as part of a line.
Which of the lines appear to fit the descriptions given below?
(i) Parallel to AB and contains D
(ii) Perpendicular to AB and contains D
(iii) Skew to AB and contains D
(iv) Name the plane(s) that contain D and appear to be parallel to plane ABE.
Problem 2 :
In the diagram given below, lines m, n and k represent three of the oars. If m||n and n||k, then prove m||k.
Problem 3 :
In the diagram given below, find the slope of each line. Determine whether the lines j_{1 }and j_{2 }are parallel.
Problem 4 :
In the diagram given below, line n_{1} has the equation
y = -ˣ⁄₃ -1
Line n_{2} is parallel to the line n_{1} and passes through the point (3, 2).
Write the equation of the line n_{2}.
Problem 5 :
Decide whether the lines are perpendicular.
Line 1 : y = ³ˣ⁄₄ + 2
Line 2 : y = -⁴ˣ⁄₃ - 3
Problem 6 :
In the diagram given below, the equation y = ³ˣ⁄₂ + 3 represents a mirror. A ray of light hits the mirror at (-2, 0). What is the equation of the line p that is perpendicular to the mirror at this point?
Problem 7 :
Is it possible for two lines with negative slopes to be perpendicular?
1. Answer :
Which of the lines appear to fit the descriptions given below?
(i) Parallel to AB and contains D
(ii) Perpendicular to AB and contains D
(iii) Skew to AB and contains D
(iv) Name the plane(s) that contain D and appear to be parallel to plane ABE.
Solution (i) :
CD, GH and EF are all parallel to AB. But, only CD passes through D and is parallel to AB.
Solution (ii) :
BC, AD, AE and BF are all perpendicular to AB. But, only AD passes through D and is perpendicular to AB.
Solution (iii) :
DG, DH and DE all pass through D and are skew to AB.
Solution (iv) :
Only plane DCH contains D and is parallel to plane ABE.
2. Answer :
Statements m||n ∠1 ≅ ∠2 n||k ∠2 ≅ ∠3 ∠1 ≅ ∠3 m||k |
Reasons Given Corresponding angles postulate Given Corresponding angles postulate Transitive property of congruence Corresponding angle converse |
3. Answer :
Line j_{1} has a slope of
m_{2} = ⁴⁄₂ = 2
Line j_{2} has a slope of
m_{2} = ²⁄₁ = 2
Since the slope of the lines j_{1 }and j_{2 }are equal, the lines j_{1 }and j_{2 }are parallel.
4. Answer :
The slope of the line n_{1} is -⅓. Because the lines n_{1} and n_{2} are parallel, they have the same slope. So, the slope of the line n_{2} is also -1/3.
Slope-intercept form equation of a line :
y = mx + b ----(1)
Because the line n_{2} is passing through (3, 2), substitute (x, y) = (3, 2) and m = -1/3.
2 = (-⅓)(3) + b
2 = -1 + b
Add 1 to both sides.
3 = b
The equation of the required line is
(1) ----> y = (-⅓)x + 3
y = -ˣ⁄₃ + 3
5. Answer :
When we compare the given equations to slope intercept equation of a line y = mx + b, we get
slope of line 1 = ³⁄₄
slope of line 2 = -⁴⁄₃
Multiply the slopes :
The product is
= (³⁄₄)x (-⁴⁄₃)
= - 1
Since the product of slopes of the lines is -1, the given lines are perpendicular.
6. Answer :
The slope of the mirror is ³⁄₂. So, the slope of the line p is -⅔.
Let y = mx + b be the equation of the line p.
Substitute (x, y) = (-2, 0) and m = -⅔ to find the value of b.
0 = (-⅔)(-2) + b
0 = ⁴⁄₃ + b
Subtract 4/3 from both sides.
-⁴⁄₃ = b
So, the equation of the line p is
y = -²ˣ⁄₃ - ⁴⁄₃
7. Answer :
We already know that if two lines are perpendicular, then the product of the slopes is equal to -1.
If the slopes of two lines are negative, then their product can never be equal to -1.
So, it is NOT possible for two lines with negative slopes to be perpendicular.
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