Exercise 7.2

Exercise 7.2 Solutions

Evaluating slopes, line inclination, parallel/perpendicular lines, and converting between forms of straight line equations.

Question 1 Slope & Inclination
Find the slope and inclination of the line joining the points:
(i) (-2, 4) & (5, 11)     (ii) (3, -2) & (2, 7)     (iii) (4, 6) & (4, 8)
(i) (-2, 4) & (5, 11)
Solution Step-by-Step
The slope formula is:
m =
y2 - y1x2 - x1
Substitute coordinates:
m =
11 - 45 - (-2)
=
77
= 1
The inclination α is given by:
tan(α) = m
α = tan-1(1) = 45°
(ii) (3, -2) & (2, 7)
Solution Step-by-Step
Substitute coordinates:
m =
7 - (-2)2 - 3
=
9-1
= -9
Find inclination:
tan(α) = -9
α = tan-1(-9) ≈ -83.66°
α (positive) = 180° - 83.66° = 96.34°
(iii) (4, 6) & (4, 8)
Solution Step-by-Step
Substitute coordinates:
m =
8 - 64 - 4
=
20
= ∞ (Undefined)
Find inclination:
α = tan-1(∞) = 90°
(This is a vertical line).
Question 2 Collinearity
By means of slopes, show that the following points lie on the same line (are collinear):
(i) A(-1, -3), B(1, 5), C(2, 9)
(ii) P(4, -5), Q(7, 5), R(10, 15)
(iii) L(-4, 6), M(3, 8), N(10, 10)
(iv) X(a, 2b), Y(c, a+b), Z(2c-a, 2a)
(i) A(-1, -3), B(1, 5), C(2, 9)
Solution Step-by-Step
Find slope of AB and slope of BC:
Slope(AB) =
5 - (-3)1 - (-1)
=
82
= 4
Slope(BC) =
9 - 52 - 1
=
41
= 4
Since Slope(AB) = Slope(BC) = 4, the points lie on the same line.
Conclusion: Points A, B, and C are collinear.
(ii) P(4, -5), Q(7, 5), R(10, 15)
Solution Step-by-Step
Find slopes:
Slope(PQ) =
5 - (-5)7 - 4
=
103
Slope(QR) =
15 - 510 - 7
=
103
Since Slope(PQ) = Slope(QR) = 10/3, the points are collinear.
(iii) L(-4, 6), M(3, 8), N(10, 10)
Solution Step-by-Step
Find slopes:
Slope(LM) =
8 - 63 - (-4)
=
27
Slope(MN) =
10 - 810 - 3
=
27
Since Slope(LM) = Slope(MN) = 2/7, the points are collinear.
(iv) X(a, 2b), Y(c, a+b), Z(2c-a, 2a)
Solution Step-by-Step
Find slopes:
Slope(XY) =
(a + b) - 2bc - a
=
a - bc - a
Slope(YZ) =
2a - (a + b)(2c - a) - c
=
a - bc - a
Since Slope(XY) = Slope(YZ) =
a - bc - a
, the points are collinear.
Question 3 Find k
Find k so that the line joining A(7, 3) and B(k, -6) and the line joining C(-4, 5) and D(-6, 4) are:
(i) parallel     (ii) perpendicular
Solution Step-by-Step
Calculate the slopes of both lines:
m1 (Slope of AB) =
-6 - 3k - 7
=
-9k - 7
m2 (Slope of CD) =
4 - 5-6 - (-4)
=
-1-2
=
12
(i) When lines are parallel: Slopes are equal (m1 = m2):
-9k - 7
 =
12
-18 = k - 7
k = -18 + 7 = -11
(ii) When lines are perpendicular: Product of slopes is -1 (m1 × m2 = -1):
-9k - 7
×
12
 = -1
92(k - 7)
 = 1
9 = 2k - 14
2k = 23
k =
232
 (or 11.5)
Question 4 Right Triangle
Using slopes, show that the triangle with its vertices A(6, 1), B(2, 7) and C(-6, -7) is a right triangle.
Solution Step-by-Step
Find the slopes of all three sides:
m1 (Slope of AB) =
7 - 12 - 6
=
6-4
= -
32
m2 (Slope of BC) =
-7 - 7-6 - 2
=
-14-8
=
74
m3 (Slope of AC) =
-7 - 1-6 - 6
=
-8-12
=
23
Observe the product of slopes of AB and AC:
m1 × m3 = (-
32
) × (
23
) = -1
Since the product is -1, side AB is perpendicular to side AC (AB ⊥ AC).
Conclusion: The triangle ABC is a right-angled triangle with the right angle at vertex A.
Question 5 Line Pairs Classification
Two pairs of points are given. Find whether the two lines determined by these points are parallel, perpendicular, or none:
(a) (1, -2), (2, 4) & (4, 1), (-8, 2)
(b) (-3, 4), (6, 2) & (4, 5), (-2, -7)
(a) (1, -2), (2, 4) & (4, 1), (-8, 2)
Solution Step-by-Step
Find slopes:
m1 (Line 1) =
4 - (-2)2 - 1
=
61
= 6
m2 (Line 2) =
2 - 1-8 - 4
=
1-12
= -
112
Check conditions:
1. m1 ≠ m2 (not parallel)
2. m1 × m2 = 6 × (-1/12) = -1/2 ≠ -1 (not perpendicular)
Conclusion: The lines are neither parallel nor perpendicular.
(b) (-3, 4), (6, 2) & (4, 5), (-2, -7)
Solution Step-by-Step
Find slopes:
m1 (Line 1) =
2 - 46 - (-3)
=
-29
m2 (Line 2) =
-7 - 5-2 - 4
=
-12-6
= 2
Check conditions:
1. Slopes are not equal.
2. Product = (-2/9) × 2 = -4/9 ≠ -1.
Conclusion: The lines are neither parallel nor perpendicular.
Question 6 Line Equations
Find an equation of:
(a) the horizontal line through (7, -9)
(b) the vertical line through (-5, 3)
(c) through A(-6, 5) having slope 7
(d) through (8, -3) having slope 0
(e) through (-8, 5) having slope undefined
(f) through (-5, -3) and (9, -1)
(g) y-intercept: -7 and slope: -5
(h) x-intercept: -3 and y-intercept: 4
(i) x-intercept: -9 and slope: -4
Solutions
(a) Horizontal line through (7, -9): A horizontal line has slope m = 0. Its equation is y = y1.
Equation: y + 9 = 0
(b) Vertical line through (-5, 3): A vertical line has undefined slope. Its equation is x = x1.
Equation: x + 5 = 0
(c) Through A(-6, 5) with slope 7: Using point-slope form y - y1 = m(x - x1):
y - 5 = 7[x - (-6)] = 7(x + 6) = 7x + 42
Equation: 7x - y + 47 = 0
(d) Through (8, -3) with slope 0:
y - (-3) = 0(x - 8)
Equation: y + 3 = 0
(e) Through (-8, 5) with slope undefined: The line is vertical, so its equation is x = x1.
Equation: x + 8 = 0
(f) Through (-5, -3) and (9, -1): First find the slope:
m =
-1 - (-3)9 - (-5)
=
214
=
17
Substitute into point-slope form:
y - (-3) =
17
[x - (-5)]
7(y + 3) = x + 5
Equation: x - 7y - 16 = 0
(g) y-intercept: -7 and slope: -5: Using slope-intercept form y = mx + c:
y = -5x - 7
Equation: 5x + y + 7 = 0
(h) x-intercept: -3 and y-intercept: 4: Using two-intercept form x/a + y/b = 1:
x-3
+
y4
 = 1
Multiply both sides by -12:
4x - 3y = -12
Equation: 4x - 3y + 12 = 0
(i) x-intercept: -9 and slope: -4: The line passes through point (-9, 0). Using point-slope form:
y - 0 = -4[x - (-9)] = -4(x + 9) = -4x - 36
Equation: 4x + y + 36 = 0
Question 7 Perpendicular Bisector
Find an equation of the perpendicular bisector of the segment joining the points A(3, 5) and B(9, 8).
Solution Step-by-Step
Step 1: Find the midpoint of segment AB. The perpendicular bisector passes through the midpoint of AB.
M(xm, ym) = M(
3 + 92
,
5 + 82
) = M(6, 6.5)
Step 2: Find the slope of the perpendicular bisector. First find the slope of line segment AB:
mAB =
8 - 59 - 3
=
36
=
12
Since the bisector is perpendicular, its slope m is the negative reciprocal:
m = -2
Step 3: Write the equation. Using point-slope form with M(6, 13/2) and slope m = -2:
y -
132
 = -2(x - 6)
Multiply by 2 to clear fractions:
2y - 13 = -4(x - 6)
2y - 13 = -4x + 24
Equation: 4x + 2y - 37 = 0
Question 8 Line through Point
Find an equation of the line through (-4, -6) and perpendicular to a line having slope -3/2.
Solution Step-by-Step
The given line slope is m1 = -3/2.
The perpendicular slope is the negative reciprocal:
m2 =
23
Using point-slope form with point (-4, -6) and slope m2 = 2/3:
y - (-6) =
23
[x - (-4)]
3(y + 6) = 2(x + 4)
3y + 18 = 2x + 8
Equation: 2x - 3y - 10 = 0
Question 9 Parallel Line
Find an equation of the line through (11, -5) and parallel to a line with slope -24.
Solution Step-by-Step
Parallel lines have equal slopes, so the slope of the required line is m = -24.
Using point-slope form with point (11, -5):
y - (-5) = -24(x - 11)
y + 5 = -24x + 264
Equation: 24x + y - 259 = 0
Question 10 Form Conversions
Convert each of the following equations into slope intercept form, two intercept form, and normal form:
(a) 2x - 4y + 11 = 0     (b) 4x + 7y - 2 = 0     (c) 15y - 8x + 3 = 0
(a) 2x - 4y + 11 = 0
Slope-Intercept Form
Isolate y:
4y = 2x + 11
y =
12
x +
114
(Slope m = 1/2, y-intercept c = 11/4)
Two-Intercept Form
Rearrange constants to RHS and divide:
2x - 4y = -11
2x-11
-
4y-11
 = 1
x-11/2
+
y11/4
 = 1
(x-intercept a = -11/2, y-intercept b = 11/4)
Normal Form
Write constant on RHS as positive:
-2x + 4y = 11
Divide by √(-2)2 + 42 = √20 = 2√5:
-2x2√5
+
4y2√5
 =
112√5
-
1√5
x +
2√5
y =
112√5
Here, cos(α) = -1/√5 < 0 and sin(α) = 2/√5 > 0, placing angle α in Quadrant II:
α ≈ 116.57°
Normal form: x cos(116.57°) + y sin(116.57°) = 11/(2√5)
(b) 4x + 7y - 2 = 0
Slope-Intercept Form
7y = -4x + 2
y = -
47
x +
27
Two-Intercept Form
4x + 7y = 2
4x2
+
7y2
 = 1
x1/2
+
y2/7
 = 1
Normal Form
Divide by √42 + 72 = √65:
4√65
x +
7√65
y =
2√65
Here, cos(α) > 0 and sin(α) > 0 (Quadrant I):
α = tan-1(7/4) ≈ 60.26°
Normal form: x cos(60.26°) + y sin(60.26°) = 2/√65
(c) 15y - 8x + 3 = 0
Slope-Intercept Form
15y = 8x - 3
y =
815
x -
15
Two-Intercept Form
-8x + 15y = -3
-8x-3
+
15y-3
 = 1
x3/8
+
y-1/5
 = 1
Normal Form
Rearrange to make constant term positive:
8x - 15y = 3
Divide by √82 + (-15)2 = √289 = 17:
817
x -
1517
y =
317
Here, cos(α) > 0 and sin(α) < 0 (Quadrant IV):
α ≈ 360° - 61.93° = 298.07°
Normal form: x cos(298.07°) + y sin(298.07°) = 3/17
Question 11 Parallel/Perp Check
In each of the following check whether the two lines are parallel, perpendicular, or neither:
(a) 2x + y - 3 = 0   and   4x + 2y + 5 = 0
(b) 3y = 2x + 5   and   3x + 2y - 8 = 0
(c) 4y + 2x - 1 = 0   and   x - 2y - 7 = 0
Solutions
(a) 2x + y - 3 = 0 and 4x + 2y + 5 = 0:
Slope of Line 1: m1 = -A/B = -2/1 = -2.
Slope of Line 2: m2 = -A/B = -4/2 = -2.
Since m1 = m2, the lines are parallel.
(b) 3y = 2x + 5 (2x - 3y + 5 = 0) and 3x + 2y - 8 = 0:
Slope of Line 1: m1 = -2/(-3) = 2/3.
Slope of Line 2: m2 = -3/2.
Product of slopes: m1 × m2 = (2/3) × (-3/2) = -1.
Therefore, the lines are perpendicular.
(c) 4y + 2x - 1 = 0 (2x + 4y - 1 = 0) and x - 2y - 7 = 0:
Slope of Line 1: m1 = -2/4 = -1/2.
Slope of Line 2: m2 = -1/(-2) = 1/2.
Slopes are not equal, and their product is -1/4 ≠ -1.
Therefore, the lines are neither parallel nor perpendicular.
Question 12 Parallel Line Eq
Find an equation of the line passing through (-4, 7) and parallel to the line 2x - 7y + 4 = 0.
Solution Step-by-Step
Find the slope of the given line 2x - 7y + 4 = 0:
mgiven = -
2-7
=
27
The required line is parallel, so it has the same slope: m = 2/7.
Using point-slope form with point (-4, 7):
y - 7 =
27
[x - (-4)]
7(y - 7) = 2(x + 4)
7y - 49 = 2x + 8
Equation: 2x - 7y + 57 = 0
Question 13 Perpendicular Line Eq
Find an equation of the line through (-5, 8) and perpendicular to the join of A(-15, -8), B(10, 7).
Solution Step-by-Step
Find the slope of line segment AB:
mAB =
7 - (-8)10 - (-15)
=
1525
=
35
The required line is perpendicular to AB, so its slope is:
m = -
53
Interpretation 1: Using the point (5, -8) (as evaluated in standard answers):
y - (-8) = -
53
(x - 5)
3(y + 8) = -5(x - 5)
3y + 24 = -5x + 25
Equation: 5x + 3y - 1 = 0
Interpretation 2: Using the point (-5, 8) (as written in the problem statement):
y - 8 = -
53
[x - (-5)]
3(y - 8) = -5(x + 5)
3y - 24 = -5x - 25
Equation: 5x + 3y + 1 = 0
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