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What is the slope of a line whose inclination with positive direction of x-axis is:
(i) 90° (ii) 0°
Concept: Slope (m) = tan(θ), where θ is the angle of inclination.
(i) For θ = 90°:
m = tan(90°) = undefined
This is a vertical line (parallel to y-axis).
(ii) For θ = 0°:
m = tan(0°) = 0
This is a horizontal line (parallel to x-axis).
What is the inclination of a line whose slope is:
(i) 0 (ii) 1
Concept: θ = tan⁻¹(m), where θ is the angle of inclination and m is the slope.
(i) For m = 0:
θ = tan⁻¹(0) = 0°
The line is horizontal.
(ii) For m = 1:
θ = tan⁻¹(1) = 45°
The line makes a 45° angle with the positive x-axis.
Find the slope of a line joining the points:
(i) (5, √5) with the origin
(ii) (sinθ, -cosθ) and (-sinθ, cosθ)
Formula: Slope (m) = (y₂ - y₁)/(x₂ - x₁)
(i) Points: (5, √5) and (0, 0):
m = (√5 - 0)/(5 - 0) = √5/5
(ii) Points: (sinθ, -cosθ) and (-sinθ, cosθ):
m = (cosθ - (-cosθ))/(-sinθ - sinθ) = (2cosθ)/(-2sinθ) = -cotθ
What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4)?
Step 1: Find midpoint P:
P = ((4 + (-6))/2, (2 + 4)/2) = (-1, 3)
Step 2: Find slope of AP:
m₁ = (3 - 1)/(-1 - 5) = 2/-6 = -1/3
Step 3: Find perpendicular slope:
For perpendicular lines, m₁ × m₂ = -1
m₂ = -1/m₁ = -1/(-1/3) = 3
Show that the given points are collinear: (-3, -4), (7,2) and (12,5)
Method 1: Slope Method
If slopes between all pairs of points are equal, points are collinear.
Slope between (-3,-4) and (7,2):
m₁ = (2 - (-4))/(7 - (-3)) = 6/10 = 3/5
Slope between (7,2) and (12,5):
m₂ = (5 - 2)/(12 - 7) = 3/5
Conclusion:
Since m₁ = m₂, all three points lie on the same straight line.
If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Method: Area of triangle formed by three collinear points is zero.
Area = ½[x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)] = 0
Apply formula:
½[3(3-(-3)) + a(-3-(-1)) + 1(-1-3)] = 0
½[18 + a(-2) + (-4)] = 0
½[14 - 2a] = 0
Solve for a:
14 - 2a = 0 ⇒ 2a = 14 ⇒ a = 7
The line through the points (-2, a) and (9, 3) has slope -½. Find the value of a.
Slope formula: m = (y₂ - y₁)/(x₂ - x₁)
Apply values:
-½ = (3 - a)/(9 - (-2)) ⇒ -½ = (3 - a)/11
Solve for a:
-½ × 11 = 3 - a ⇒ -5.5 = 3 - a ⇒ a = 3 + 5.5 = 8.5
The line through (-2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24). Find x.
Find slope of first line (m₁):
m₁ = (8 - 6)/(4 - (-2)) = 2/6 = 1/3
Find slope of second line (m₂):
m₂ = (24 - 12)/(x - 8) = 12/(x - 8)
Condition for perpendicularity: m₁ × m₂ = -1
(1/3) × (12/(x - 8)) = -1 ⇒ 12/(3(x - 8)) = -1 ⇒ 4/(x - 8) = -1
Solve for x:
4 = - (x - 8) ⇒ 4 = -x + 8 ⇒ x = 8 - 4 = 4
Show that the given points form a right-angled triangle and check Pythagoras theorem:
(i) A(1, -4), B(2, -3), C(4, -7)
(ii) L(0, 5), M(9, 12), N(3, 14)
(i) Points A(1,-4), B(2,-3), C(4,-7)
Find slopes:
AB: (-3 - (-4))/(2 - 1) = 1/1 = 1
BC: (-7 - (-3))/(4 - 2) = -4/2 = -2
AC: (-7 - (-4))/(4 - 1) = -3/3 = -1
Check perpendicularity:
AB × AC = 1 × -1 = -1 ⇒ AB ⊥ AC
Thus, right-angled at A.
Verify Pythagoras:
AB² = (2-1)² + (-3-(-4))² = 1 + 1 = 2
AC² = (4-1)² + (-7-(-4))² = 9 + 9 = 18
BC² = (4-2)² + (-7-(-3))² = 4 + 16 = 20
AB² + AC² = 2 + 18 = 20 = BC² ✔️
(ii) Points L(0,5), M(9,12), N(3,14)
Find slopes:
LM: (12-5)/(9-0) = 7/9
MN: (14-12)/(3-9) = 2/-6 = -1/3
LN: (14-5)/(3-0) = 9/3 = 3
Check perpendicularity:
LN × MN = 3 × (-1/3) = -1 ⇒ LN ⊥ MN
Thus, right-angled at N.
Verify Pythagoras:
LN² = (3-0)² + (14-5)² = 9 + 81 = 90
MN² = (3-9)² + (14-12)² = 36 + 4 = 40
LM² = (9-0)² + (12-5)² = 81 + 49 = 130
LN² + MN² = 90 + 40 = 130 = LM² ✔️
Show that the given points form a parallelogram:
A(2.5,3.5), B(10,-4), C(2.5,-2.5) and D(-5,5)
Property: In a parallelogram, midpoints of diagonals coincide.
Find midpoint of AC:
((2.5 + 2.5)/2, (3.5 + (-2.5))/2) = (2.5, 0.5)
Find midpoint of BD:
((10 + (-5))/2, (-4 + 5)/2) = (2.5, 0.5)
Conclusion:
Since midpoints of both diagonals are the same, ABCD is a parallelogram.
If the points A(2,2), B(-2,-3), C(1,-3) and D(x,y) form a parallelogram, find x and y.
Method: In parallelogram, vector AB = vector DC
Find vector AB:
B - A = (-2-2, -3-2) = (-4, -5)
Set vector DC = vector AB:
C - D = (-4, -5) ⇒ (1 - x, -3 - y) = (-4, -5)
Solve equations:
1 - x = -4 ⇒ x = 5
-3 - y = -5 ⇒ y = 2
D(5, 2)
Show that ABCD is a trapezium where:
A(3,-4), B(9,-4), C(5,-7) and D(7,-7)
Property: A trapezium has exactly one pair of parallel sides.
Find slopes:
Slope of AB = (-4 - (-4))/(9 - 3) = 0 (horizontal line)
Slope of CD = (-7 - (-7))/(7 - 5) = 0 (horizontal line)
Slope of AD = (-7 - (-4))/(7 - 3) = -3/4
Slope of BC = (-7 - (-4))/(5 - 9) = -3/-4 = 3/4
Conclusion:
AB ∥ CD (same slope), AD ∦ BC (different slopes)
Thus, ABCD is a trapezium with AB parallel to CD.
Show that the midpoints of quadrilateral ABCD form a parallelogram where:
A(-4,-2), B(5,-1), C(6,5) and D(-7,6)
Find all midpoints:
Midpoint of AB: ((-4+5)/2, (-2-1)/2) = (0.5, -1.5)
Midpoint of BC: ((5+6)/2, (-1+5)/2) = (5.5, 2)
Midpoint of CD: ((6-7)/2, (5+6)/2) = (-0.5, 5.5)
Midpoint of DA: ((-7-4)/2, (6-2)/2) = (-5.5, 2)
Find midpoints of midpoint quadrilateral:
Midpoint of (AB mid and CD mid): ((0.5-0.5)/2, (-1.5+5.5)/2) = (0, 2)
Midpoint of (BC mid and DA mid): ((5.5-5.5)/2, (2+2)/2) = (0, 2)
Conclusion:
Since midpoints of both diagonals coincide, the midpoint quadrilateral is a parallelogram.