Question 1

The distance between the parallel lines 3x + 4y = 10 and 3x + 4y = 25 is:

Accepted Answer

Setup: The distance between two parallel lines ax + by = c₁ and ax + by = c₂ is given by the formula: d = |c₁ - c₂| / √(a² + b²)

Solution:
Step 1: Identify coefficients.
Both lines have the form 3x + 4y = c, so a = 3, b = 4.
c₁ = 10, c₂ = 25

Step 2: Apply the distance formula.
d = |10 - 25| / √(3² + 4²)
d = |-15| / √(9 + 16)
d = 15 / √25
d = 15 / 5
d = 3

Verification: The lines are indeed parallel (same coefficients for x and y). The perpendicular distance is the shortest distance between them, which is 3 units.

Question 2

The angle between the lines y = 2x + 3 and y = -x/2 + 5 is:

Accepted Answer

Setup: The angle θ between two lines with slopes m₁ and m₂ is given by: tan(θ) = |(m₁ - m₂)/(1 + m₁m₂)|

Solution:
Step 1: Identify the slopes.
Line 1: y = 2x + 3 → m₁ = 2
Line 2: y = -x/2 + 5 → m₂ = -1/2

Step 2: Check if lines are perpendicular.
m₁ · m₂ = 2 × (-1/2) = -1
Since the product of slopes equals -1, the lines are perpendicular.

Step 3: Determine the angle.
When two lines are perpendicular, the angle between them is 90° or π/2 radians.

Verification: The denominator in the angle formula becomes 1 + m₁m₂ = 1 + (-1) = 0, which indicates the angle is 90°.

Question 3

The foot of the perpendicular from the point (2, 3) to the line x + 2y = 5 is:

Accepted Answer

Setup: The foot of the perpendicular from a point to a line is the point on the line closest to the given point. It lies on both the original line and the perpendicular line through the given point.

Solution:
Step 1: Find the slope of the given line x + 2y = 5.
Rewrite as 2y = -x + 5 → y = (-1/2)x + 5/2
Slope of given line = -1/2

Step 2: Find the slope of the perpendicular line.
Perpendicular slope = -1/(-1/2) = 2

Step 3: Write the equation of the perpendicular line through (2, 3).
y - 3 = 2(x - 2)
y - 3 = 2x - 4
y = 2x - 1

Step 4: Find the intersection of y = 2x - 1 and x + 2y = 5.
Substitute y = 2x - 1 into x + 2y = 5:
x + 2(2x - 1) = 5
x + 4x - 2 = 5
5x = 7
x = 7/5

Step 5: Find y-coordinate.
y = 2(7/5) - 1 = 14/5 - 5/5 = 9/5

Foot of perpendicular = (7/5, 9/5)

Verification: 
Point (7/5, 9/5) on line x + 2y = 5: 7/5 + 2(9/5) = 7/5 + 18/5 = 25/5 = 5 ✓
Slope from (2, 3) to (7/5, 9/5): (9/5 - 3)/(7/5 - 2) = (-6/5)/(-3/5) = 2 ✓ (perpendicular slope)

Question 4

The point of intersection of the lines x + 2y = 5 and 2x - y = 5 is:

Accepted Answer

Setup: To find the intersection of two lines, solve the system of linear equations simultaneously.

Solution:
Step 1: Write the system.
x + 2y = 5  ... (1)
2x - y = 5  ... (2)

Step 2: Use elimination method. Multiply equation (1) by 2.
2x + 4y = 10  ... (1')
2x - y = 5    ... (2)

Step 3: Subtract equation (2) from equation (1').
(2x + 4y) - (2x - y) = 10 - 5
5y = 5
y = 1

Step 4: Substitute y = 1 into equation (1).
x + 2(1) = 5
x + 2 = 5
x = 3

Step 5: Verify in both equations.
Equation (1): 3 + 2(1) = 5 ✓
Equation (2): 2(3) - 1 = 6 - 1 = 5 ✓

The point of intersection is (3, 1).

Question 5

The equation of the line passing through the points (1, 2) and (3, 6) is:

Accepted Answer

Setup: Two-point form of a line: for points (x₁, y₁) and (x₂, y₂), the equation is (y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁), or equivalently y - y₁ = m(x - x₁) where m = (y₂ - y₁)/(x₂ - x₁).

Solution:
Step 1: Calculate the slope.
m = (6 - 2)/(3 - 1) = 4/2 = 2

Step 2: Use point-slope form with point (1, 2).
y - 2 = 2(x - 1)
y - 2 = 2x - 2
y = 2x

Alternatively: 2x - y = 0

Step 3: Verify with both points.
Point (1, 2): 2(1) - 2 = 0 ✓
Point (3, 6): 2(3) - 6 = 0 ✓

Both points satisfy the equation.

Question 6

The angle between the lines 2x + y = 5 and x - 2y = 3 is:

Accepted Answer

Setup: For two lines with slopes m₁ and m₂, the acute angle θ between them is given by tan(θ) = |m₁ - m₂|/(1 + m₁m₂), provided the lines are not perpendicular.

Solution:
Step 1: Find the slope of the first line 2x + y = 5.
Rewrite: y = -2x + 5
Slope m₁ = -2

Step 2: Find the slope of the second line x - 2y = 3.
Rewrite: 2y = x - 3 → y = (1/2)x - 3/2
Slope m₂ = 1/2

Step 3: Check if lines are perpendicular.
m₁ · m₂ = (-2) · (1/2) = -1
Yes, the lines are perpendicular. ✓

Step 4: When lines are perpendicular, the angle between them is 90° or π/2 radians.

Verification: The condition for perpendicularity is m₁ · m₂ = -1, which is satisfied.

Question 7

A line passes through the point (4, 5) and makes an angle of 45° with the positive x-axis. Its equation is:

Accepted Answer

Setup: When a line makes an angle θ with the positive x-axis, its slope is m = tan(θ). Using point-slope form y - y₁ = m(x - x₁) gives the equation.

Solution:
Step 1: Find the slope from the angle.
Angle θ = 45°
Slope m = tan(45°) = 1

Step 2: Apply point-slope form with point (4, 5) and slope m = 1.
y - 5 = 1(x - 4)
y - 5 = x - 4
y = x + 1

Alternatively: x - y + 1 = 0

Step 3: Verify with the given point.
Substitute (4, 5): 5 = 4 + 1 = 5 ✓

Step 4: Verify the slope.
Rewrite as y = x + 1. Slope = 1 = tan(45°) ✓

Question 8

The equation of the straight line passing through the point (2, 3) and perpendicular to the line 3x + 4y = 7 is:

Accepted Answer

Setup: A line perpendicular to a given line has slope that is the negative reciprocal of the original slope.

Solution:
Step 1: Find the slope of the given line 3x + 4y = 7.
Rewrite in slope-intercept form: 4y = -3x + 7 → y = (-3/4)x + 7/4
Slope of given line = -3/4

Step 2: Find the slope of the perpendicular line.
Slope of perpendicular line = -1/(-3/4) = 4/3

Step 3: Use point-slope form with point (2, 3) and slope 4/3.
y - 3 = (4/3)(x - 2)
3(y - 3) = 4(x - 2)
3y - 9 = 4x - 8
4x - 3y + 1 = 0

Verification: Coefficient product test: (3)(4) + (4)(-3) = 12 - 12 = 0 ✓ (perpendicular condition satisfied)

Question 9

The distance between the parallel lines 2x + 3y = 5 and 2x + 3y = 11 is:

Accepted Answer

Setup: Distance between two parallel lines ax + by = c₁ and ax + by = c₂ is given by the formula: d = |c₁ - c₂|/√(a² + b²)

Solution:
Step 1: Verify the lines are parallel.
Both lines have the form 2x + 3y = constant, so they have the same normal vector (2, 3). They are parallel. ✓

Step 2: Identify coefficients.
a = 2, b = 3, c₁ = 5, c₂ = 11

Step 3: Apply the distance formula.
d = |5 - 11|/√(2² + 3²)
d = |-6|/√(4 + 9)
d = 6/√13

Step 4: Rationalize (if needed for answer form).
d = 6√13/13

Verification: Both lines can be checked by substituting a point. Line 1 passes through (1, 1): 2(1) + 3(1) = 5 ✓. The perpendicular distance from (1,1) to line 2 is |2(1) + 3(1) - 11|/√13 = 6/√13 ✓

Question 10

The slope of the line passing through the points A(−1, 4) and B(3, −2) is:

Accepted Answer

Setup: The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by m = (y₂ - y₁)/(x₂ - x₁).

Solution:
Step 1: Identify the coordinates.
Point A: (x₁, y₁) = (−1, 4)
Point B: (x₂, y₂) = (3, −2)

Step 2: Apply the slope formula.
m = (y₂ - y₁)/(x₂ - x₁)
m = (−2 - 4)/(3 - (−1))
m = (−6)/(3 + 1)
m = −6/4
m = −3/2

Step 3: Verify by checking the direction.
As x increases from −1 to 3 (moving right), y decreases from 4 to −2 (moving down).
This confirms a negative slope. ✓

The slope is −3/2.

Question 11

The equation of the line with slope 2 and y-intercept −3 is:

Accepted Answer

Setup: The slope-intercept form of a line is y = mx + c, where m is the slope and c is the y-intercept.

Solution:
Step 1: Identify given values.
Slope m = 2
y-intercept c = −3

Step 2: Substitute into slope-intercept form.
y = mx + c
y = 2x + (−3)
y = 2x − 3

Step 3: Convert to standard form (if required).
2x − y − 3 = 0
Or equivalently: 2x − y = 3

Step 4: Verify using the y-intercept.
When x = 0: y = 2(0) − 3 = −3 ✓
The line crosses the y-axis at (0, −3).

Step 5: Verify using the slope.
Choose another point: when x = 1, y = 2(1) − 3 = −1.
Slope between (0, −3) and (1, −1) = (−1 − (−3))/(1 − 0) = 2/1 = 2 ✓

The equation is y = 2x − 3 or 2x − y − 3 = 0.

Question 12

The equation of the straight line passing through the point (2, 3) and perpendicular to the line 2x + 3y = 5 is:

Accepted Answer

Setup: A line perpendicular to a given line has slope that is the negative reciprocal of the original slope.

Solution:
Step 1: Find the slope of the given line 2x + 3y = 5.
Rewrite in slope-intercept form: 3y = -2x + 5 → y = (-2/3)x + 5/3
Slope of given line = -2/3

Step 2: Find the slope of the perpendicular line.
If m₁ = -2/3, then m₂ = -1/m₁ = -1/(-2/3) = 3/2

Step 3: Use point-slope form with point (2, 3) and slope 3/2.
y - 3 = (3/2)(x - 2)
y - 3 = (3/2)x - 3
y = (3/2)x

Step 4: Convert to standard form.
2y = 3x
3x - 2y = 0

Verification: Slope of 3x - 2y = 0 is 3/2. Product of slopes: (-2/3) × (3/2) = -1 ✓
Point (2, 3): 3(2) - 2(3) = 6 - 6 = 0 ✓

Question 13

The distance between the parallel lines 3x + 4y - 10 = 0 and 3x + 4y + 5 = 0 is:

Accepted Answer

Step 1: Verify that the lines are parallel.
Both lines have the form 3x + 4y + c = 0 with the same coefficients for x and y, confirming they are parallel.

Step 2: Apply the distance formula for parallel lines.
For two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0, the distance is:
d = |c₁ - c₂| / √(a² + b²)

Step 3: Substitute values.
Here, a = 3, b = 4, c₁ = -10, c₂ = 5
d = |-10 - 5| / √(3² + 4²)
d = |-15| / √(9 + 16)
d = 15 / √25
d = 15 / 5
d = 3

Question 14

The angle between the lines 2x + y - 4 = 0 and x - 2y + 3 = 0 is:

Accepted Answer

Step 1: Find the slopes of both lines.
Line 1: 2x + y - 4 = 0 → y = -2x + 4, so m₁ = -2
Line 2: x - 2y + 3 = 0 → 2y = x + 3 → y = (1/2)x + 3/2, so m₂ = 1/2

Step 2: Check if the lines are perpendicular.
m₁ · m₂ = (-2) · (1/2) = -1
Since the product of slopes is -1, the lines are perpendicular.

Step 3: Determine the angle.
When two lines are perpendicular, the angle between them is 90° or π/2 radians.

Question 15

The foot of the perpendicular from the point (3, 4) to the line 2x - y + 1 = 0 is:

Accepted Answer

Step 1: Find the equation of the perpendicular line through (3, 4).
The given line 2x - y + 1 = 0 can be written as y = 2x + 1, so its slope is 2.
The perpendicular line has slope m = -1/2.
Equation of perpendicular: y - 4 = (-1/2)(x - 3)
2(y - 4) = -(x - 3)
2y - 8 = -x + 3
x + 2y - 11 = 0

Step 2: Find the intersection of the perpendicular with the given line.
Solve 2x - y + 1 = 0 and x + 2y - 11 = 0 simultaneously.
From the first equation: y = 2x + 1
Substitute into the second: x + 2(2x + 1) - 11 = 0
x + 4x + 2 - 11 = 0
5x - 9 = 0
x = 9/5

Step 3: Find y-coordinate.
y = 2(9/5) + 1 = 18/5 + 5/5 = 23/5

Therefore, the foot of the perpendicular is (9/5, 23/5).

Question 16

The equation of the line passing through the point of intersection of the lines 2x + 3y - 5 = 0 and x - 2y + 1 = 0, and perpendicular to the line 3x - 4y + 7 = 0 is:

Accepted Answer

Step 1: Find the point of intersection of 2x + 3y - 5 = 0 and x - 2y + 1 = 0.
From the second equation: x = 2y - 1.
Substitute into the first: 2(2y - 1) + 3y - 5 = 0 → 4y - 2 + 3y - 5 = 0 → 7y = 7 → y = 1.
Then x = 2(1) - 1 = 1.
Point of intersection: (1, 1).

Step 2: Find the slope of the required line.
The given line 3x - 4y + 7 = 0 can be written as y = (3/4)x + 7/4, so its slope is 3/4.
For a line perpendicular to this, the slope m = -4/3 (negative reciprocal).

Step 3: Write the equation using point-slope form.
Using y - y₁ = m(x - x₁) with (x₁, y₁) = (1, 1) and m = -4/3:
y - 1 = (-4/3)(x - 1)
3(y - 1) = -4(x - 1)
3y - 3 = -4x + 4
4x + 3y - 7 = 0.

Question 17

The distance from the point (2, 3) to the line 3x - 4y + 5 = 0 is:

Accepted Answer

Step 1: Recall the distance formula from a point (x₀, y₀) to a line ax + by + c = 0:
Distance = |ax₀ + by₀ + c| / √(a² + b²).

Step 2: Identify the values.
Point: (x₀, y₀) = (2, 3).
Line: 3x - 4y + 5 = 0, so a = 3, b = -4, c = 5.

Step 3: Substitute into the formula.
Distance = |3(2) + (-4)(3) + 5| / √(3² + (-4)²)
         = |6 - 12 + 5| / √(9 + 16)
         = |-1| / √25
         = 1 / 5.

Question 18

If the lines ax + by + c = 0 and bx + ay + c = 0 are perpendicular to each other, then the relationship between a, b, and c is:

Accepted Answer

Step 1: Find the slopes of both lines.
Line 1: ax + by + c = 0 → by = -ax - c → y = (-a/b)x - c/b, so slope m₁ = -a/b (assuming b ≠ 0).
Line 2: bx + ay + c = 0 → ay = -bx - c → y = (-b/a)x - c/a, so slope m₂ = -b/a (assuming a ≠ 0).

Step 2: Apply the perpendicularity condition.
For perpendicular lines: m₁ · m₂ = -1.
(-a/b) · (-b/a) = -1
(ab)/(ab) = -1
1 = -1.

Step 3: Analyze the result.
This is a contradiction. The equation 1 = -1 is never true, which means the lines cannot be perpendicular for any real values of a, b, and c (with a, b ≠ 0).

However, if the question intends to ask when they ARE perpendicular under special conditions, we reconsider: if a = 0 or b = 0, the analysis changes. But the standard interpretation is that there is NO relationship that makes them perpendicular.

Question 19

The equation of the line that passes through the point (3, 4) and makes equal intercepts on the coordinate axes is:

Accepted Answer

Step 1: Set up the intercept form of a line.
A line making equal intercepts a on both axes has the equation: x/a + y/a = 1, which simplifies to x + y = a.

Step 2: Use the condition that the line passes through (3, 4).
Substitute (3, 4) into x + y = a:
3 + 4 = a
a = 7.

Step 3: Write the equation.
The equation is x + y = 7, or x + y - 7 = 0.

Step 4: Verify.
Check intercepts: When x = 0, y = 7 (y-intercept = 7). When y = 0, x = 7 (x-intercept = 7). ✓
Check point: 3 + 4 = 7. ✓

Question 20

The angle between the lines 2x - y + 3 = 0 and x + 2y - 5 = 0 is:

Accepted Answer

Step 1: Find the slopes of both lines.
Line 1: 2x - y + 3 = 0 → y = 2x + 3, so m₁ = 2.
Line 2: x + 2y - 5 = 0 → 2y = -x + 5 → y = (-1/2)x + 5/2, so m₂ = -1/2.

Step 2: Check if the lines are perpendicular.
m₁ · m₂ = 2 · (-1/2) = -1.
Since the product of slopes equals -1, the lines are perpendicular.

Step 3: Determine the angle.
When two lines are perpendicular, the angle between them is 90° or π/2 radians.

Alternative verification using the angle formula:
tan(θ) = |(m₁ - m₂) / (1 + m₁m₂)| = |(2 - (-1/2)) / (1 + 2(-1/2))| = |(5/2) / (1 - 1)| = |(5/2) / 0|.
The denominator is 0, which indicates θ = 90°.

AFCAT Straight Lines

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Key Points to Remember

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Straight Lines — Practice Questions

60-Second Revision — Straight Lines