✨ Complete Math Magic ✨

All 12 Straight Line Problems Solved Beautifully!
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Finding Slopes of Straight Lines
Find the slope of the following straight lines:
(i) 5y - 3 = 0
(ii) 7x - 3/17 = 0
1
Part (i): 5y - 3 = 0
First, let's rewrite the equation in slope-intercept form (y = mx + b)
2
5y - 3 = 0
Add 3 to both sides: 5y = 3
Divide both sides by 5: y = 3/5
3
This is a horizontal line (y = constant).
The slope (m) of any horizontal line is 0.
4
Part (ii): 7x - 3/17 = 0
Rewrite the equation to solve for x:
5
7x - 3/17 = 0
Add 3/17 to both sides: 7x = 3/17
Divide both sides by 7: x = 3/119
6
This is a vertical line (x = constant).
The slope (m) of any vertical line is undefined.

Visualization

[Graph showing a horizontal line at y=3/5 and vertical line at x=3/119]
Final Answers:
(i) Slope = 0
(ii) Slope = undefined
Finding Parallel and Perpendicular Slopes
Find the slope of the line which is:
(i) parallel to y = 0.7x - 11
(ii) perpendicular to the line x = -11
1
Part (i): Parallel to y = 0.7x - 11
Parallel lines have identical slopes.
2
The given line is in slope-intercept form y = mx + b, where m is the slope.
Here, m = 0.7.
3
Therefore, any line parallel to this will have the same slope:
m = 0.7
4
Part (ii): Perpendicular to x = -11
First, identify the slope of the given line.
5
x = -11 is a vertical line.
Vertical lines have undefined slopes.
6
Lines perpendicular to vertical lines are horizontal lines.
Horizontal lines have slope 0.

Visualization

[Graph showing a vertical line at x=-11 and a horizontal line perpendicular to it]
Final Answers:
(i) Slope = 0.7
(ii) Slope = 0
Checking Parallel and Perpendicular Lines
Check whether the given lines are parallel or perpendicular:
(i) x/3 + y/4 + 1/7 = 0 and 2x/3 + y/2 + 1/10 = 0
(ii) 5x + 23y + 14 = 0 and 23x - 5y + 9 = 0
Key Formulas:
• Parallel lines have equal slopes (m₁ = m₂)
• Perpendicular lines have slopes that are negative reciprocals (m₁ × m₂ = -1)
1
Part (i):
First, find the slope of both lines by rewriting in slope-intercept form (y = mx + c)
2
First line: x/3 + y/4 + 1/7 = 0
Multiply all terms by 84 (LCM of denominators): 28x + 21y + 12 = 0
Solve for y: 21y = -28x - 12
y = (-28/21)x - 12/21
Simplify: y = (-4/3)x - 4/7
Slope (m₁) = -4/3
3
Second line: 2x/3 + y/2 + 1/10 = 0
Multiply all terms by 30: 20x + 15y + 3 = 0
Solve for y: 15y = -20x - 3
y = (-20/15)x - 3/15
Simplify: y = (-4/3)x - 1/5
Slope (m₂) = -4/3
4
Comparison:
m₁ = -4/3 and m₂ = -4/3
Since m₁ = m₂, the lines are parallel.
5
Part (ii):
Find slopes of both lines:
6
First line: 5x + 23y + 14 = 0
Solve for y: 23y = -5x - 14
y = (-5/23)x - 14/23
Slope (m₁) = -5/23
7
Second line: 23x - 5y + 9 = 0
Solve for y: -5y = -23x - 9
y = (23/5)x + 9/5
Slope (m₂) = 23/5
8
Check perpendicularity:
m₁ × m₂ = (-5/23) × (23/5) = -1
Since the product is -1, the lines are perpendicular.
Final Answers:
(i) Parallel
(ii) Perpendicular
Finding Parameter for Perpendicular Lines
If the straight lines 12y = -(p + 3)x + 12 and 12x - 7y = 16 are perpendicular, then find 'p'.
Key Concept:
Two lines are perpendicular if the product of their slopes is -1 (m₁ × m₂ = -1)
1
First, find the slope of both lines by rewriting in slope-intercept form (y = mx + c)
2
First line: 12y = -(p + 3)x + 12
Divide both sides by 12: y = [-(p + 3)/12]x + 1
Slope (m₁) = -(p + 3)/12
3
Second line: 12x - 7y = 16
Solve for y: -7y = -12x + 16
y = (12/7)x - 16/7
Slope (m₂) = 12/7
4
For perpendicular lines: m₁ × m₂ = -1
[-(p + 3)/12] × [12/7] = -1
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Simplify the equation:
[-(p + 3) × 12] / [12 × 7] = -1
The 12's cancel out: -(p + 3)/7 = -1
6
Multiply both sides by 7: -(p + 3) = -7
Multiply both sides by -1: p + 3 = 7
Subtract 3 from both sides: p = 4
Final Answer:
p = 4
Equation of Line Parallel to Another Line
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3,-2) and R(-5,4).
Key Formulas:
• Slope between two points (x₁,y₁) and (x₂,y₂): m = (y₂ - y₁)/(x₂ - x₁)
• Equation of line through (x₀,y₀) with slope m: y - y₀ = m(x - x₀)
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First, find the slope of line QR joining points Q(3,-2) and R(-5,4):
2
m = (y₂ - y₁)/(x₂ - x₁) = (4 - (-2))/(-5 - 3) = 6/-8 = -3/4
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The required line is parallel to QR, so it has the same slope: m = -3/4
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Now, use point-slope form with point P(-5,2) and slope m = -3/4:
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y - y₁ = m(x - x₁)
y - 2 = (-3/4)(x - (-5))
y - 2 = (-3/4)(x + 5)
6
Convert to standard form:
Multiply both sides by 4: 4(y - 2) = -3(x + 5)
4y - 8 = -3x - 15
3x + 4y + 7 = 0

Visualization

[Graph showing points Q, R, line QR, point P, and the parallel line through P]
Final Answer:
3x + 4y + 7 = 0
Equation of Perpendicular Line
Find the equation of a line passing through (6,-2) and perpendicular to the line joining the points (6,7) and (2,-3).
Key Concepts:
• Slope of perpendicular line is negative reciprocal of original slope
• If original slope is m, perpendicular slope is -1/m
1
First, find slope of line joining (6,7) and (2,-3):
2
m = (y₂ - y₁)/(x₂ - x₁) = (-3 - 7)/(2 - 6) = (-10)/(-4) = 5/2
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Slope of perpendicular line is negative reciprocal: m⊥ = -2/5
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Now use point-slope form with point (6,-2) and slope m = -2/5:
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y - y₁ = m(x - x₁)
y - (-2) = (-2/5)(x - 6)
y + 2 = (-2/5)x + 12/5
6
Convert to standard form:
Multiply all terms by 5: 5y + 10 = -2x + 12
2x + 5y - 2 = 0
Final Answer:
2x + 5y - 2 = 0
Equations of Altitudes in a Triangle
A(-3,0), B(10,-2) and C(12,3) are the vertices of ΔABC. Find the equation of the altitude through A and B.
Key Concept:
An altitude is perpendicular to the opposite side. To find its equation:
  1. Find slope of the opposite side
  2. Take negative reciprocal for altitude's slope
  3. Use point-slope form with the vertex point
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Altitude through A: Perpendicular to BC
First find slope of BC:
2
m_BC = (3 - (-2))/(12 - 10) = 5/2
Slope of altitude through A is negative reciprocal: m⊥ = -2/5
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Using point A(-3,0) and slope -2/5:
y - 0 = (-2/5)(x - (-3))
y = (-2/5)x - 6/5
Standard form: 2x + 5y + 6 = 0
4
Altitude through B: Perpendicular to AC
First find slope of AC:
5
m_AC = (3 - 0)/(12 - (-3)) = 3/15 = 1/5
Slope of altitude through B is negative reciprocal: m⊥ = -5
6
Using point B(10,-2) and slope -5:
y - (-2) = -5(x - 10)
y + 2 = -5x + 50
Standard form: 5x + y - 48 = 0

Triangle Visualization

[Graph showing triangle ABC with altitudes through A and B]
Final Answers:
Altitude through A: 2x + 5y + 6 = 0
Altitude through B: 5x + y - 48 = 0
Perpendicular Bisector of a Line Segment
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6,-4).
Steps to Find Perpendicular Bisector:
  1. Find midpoint of AB
  2. Find slope of AB
  3. Take negative reciprocal for perpendicular slope
  4. Use point-slope form with midpoint
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Find midpoint of AB:
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Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2) = ((-4 + 6)/2, (2 + (-4))/2) = (1, -1)
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Find slope of AB:
4
m_AB = (-4 - 2)/(6 - (-4)) = -6/10 = -3/5
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Slope of perpendicular bisector is negative reciprocal: m⊥ = 5/3
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Using midpoint (1,-1) and slope 5/3:
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y - (-1) = (5/3)(x - 1)
y + 1 = (5/3)x - 5/3
Multiply by 3: 3y + 3 = 5x - 5
Standard form: 5x - 3y - 8 = 0

Visualization

[Graph showing points A, B, line AB, midpoint, and perpendicular bisector]
Final Answer:
5x - 3y - 8 = 0
Line Through Intersection and Parallel
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x - 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Key Steps:
  1. Find intersection point of the two lines
  2. Find slope of the parallel line
  3. Use point-slope form with intersection point and parallel slope
1
Find intersection of 7x + 3y = 10 and 5x - 4y = 1:
2
Solve the system of equations:
Multiply first equation by 4: 28x + 12y = 40
Multiply second equation by 3: 15x - 12y = 3
Add them: 43x = 43 ⇒ x = 1
3
Substitute x=1 into first equation: 7(1) + 3y = 10 ⇒ y = 1
Intersection point: (1,1)
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Find slope of parallel line 13x + 5y + 12 = 0:
5
Rewrite in slope-intercept form: 5y = -13x - 12 ⇒ y = (-13/5)x - 12/5
Slope (m) = -13/5
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Using point (1,1) and slope -13/5:
7
y - 1 = (-13/5)(x - 1)
Multiply by 5: 5y - 5 = -13x + 13
Standard form: 13x + 5y - 18 = 0
Final Answer:
13x + 5y - 18 = 0
Line Through Intersection and Perpendicular
Find the equation of a straight line through the intersection of lines 5x - 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x - 7y + 13 = 0.
Key Steps:
  1. Find intersection point of the two lines
  2. Find slope of given line and its perpendicular slope
  3. Use point-slope form with intersection point and perpendicular slope
1
Find intersection of 5x - 6y = 2 and 3x + 2y = 10:
2
Solve the system of equations:
Multiply second equation by 3: 9x + 6y = 30
Add to first equation: 14x = 32 ⇒ x = 16/7
3
Substitute x=16/7 into second equation: 3(16/7) + 2y = 10 ⇒ y = 11/7
Intersection point: (16/7, 11/7)
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Find slope of line 4x - 7y + 13 = 0:
5
Rewrite in slope-intercept form: -7y = -4x - 13 ⇒ y = (4/7)x + 13/7
Slope (m) = 4/7
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Perpendicular slope is negative reciprocal: m⊥ = -7/4
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Using point (16/7, 11/7) and slope -7/4:
8
y - 11/7 = (-7/4)(x - 16/7)
Multiply by 28 (LCM of denominators): 28y - 44 = -49x + 112
Standard form: 49x + 28y - 156 = 0
Simplified: 7x + 4y - 156/7 = 0
Final Answer:
49x + 28y - 156 = 0 or 7x + 4y - 156/7 = 0
Line Connecting Two Intersection Points
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x - 2y - 4 = 0 to the point of intersection of 7x - 3y = -12 and 2y = x + 3.
Key Steps:
  1. Find first intersection point (P)
  2. Find second intersection point (Q)
  3. Find equation of line PQ
1
First intersection point (P):
Solve 3x + y + 2 = 0 and x - 2y - 4 = 0
2
From first equation: y = -3x - 2
Substitute into second equation: x - 2(-3x - 2) - 4 = 0 ⇒ x + 6x + 4 - 4 = 0 ⇒ 7x = 0 ⇒ x = 0
Then y = -3(0) - 2 = -2
Point P: (0, -2)
3
Second intersection point (Q):
Solve 7x - 3y = -12 and 2y = x + 3
4
Rewrite second equation: x = 2y - 3
Substitute into first equation: 7(2y - 3) - 3y = -12 ⇒ 14y - 21 - 3y = -12 ⇒ 11y = 9 ⇒ y = 9/11
Then x = 2(9/11) - 3 = -15/11
Point Q: (-15/11, 9/11)
5
Equation of line PQ:
Points P(0,-2) and Q(-15/11,9/11)
Slope (m) = (9/11 - (-2))/(-15/11 - 0) = (31/11)/(-15/11) = -31/15
6
Using point P(0,-2) and slope -31/15:
7
y - (-2) = (-31/15)(x - 0)
y + 2 = (-31/15)x
Multiply by 15: 15y + 30 = -31x
Standard form: 31x + 15y + 30 = 0

Visualization

[Graph showing both pairs of intersecting lines and the line connecting their intersection points]
Final Answer:
31x + 15y + 30 = 0
Line Through Intersection and Midpoint
Find the equation of a straight line through the point of intersection of the lines 8x + 3y = 18, 4x + 5y = 9 and bisecting the line segment joining the points (5,-4) and (-7,6).
Key Steps:
  1. Find intersection point of the two lines (P)
  2. Find midpoint of given segment (Q)
  3. Find equation of line PQ
1
Intersection point (P):
Solve 8x + 3y = 18 and 4x + 5y = 9
2
Multiply second equation by 2: 8x + 10y = 18
Subtract first equation: 7y = 0 ⇒ y = 0
Substitute y=0 into first equation: 8x = 18 ⇒ x = 9/4
Point P: (9/4, 0)
3
Midpoint (Q) of (5,-4) and (-7,6):
4
Q = ((5 + (-7))/2, (-4 + 6)/2) = (-1, 1)
5
Equation of line PQ:
Points P(9/4,0) and Q(-1,1)
Slope (m) = (1 - 0)/(-1 - 9/4) = 1/(-13/4) = -4/13
6
Using point P(9/4,0) and slope -4/13:
7
y - 0 = (-4/13)(x - 9/4)
y = (-4/13)x + 9/13
Multiply by 13: 13y = -4x + 9
Standard form: 4x + 13y - 9 = 0

Visualization

[Graph showing intersecting lines, points (5,-4) and (-7,6), their midpoint, and the resulting line]
Final Answer:
4x + 13y - 9 = 0