Coordinate Geometry Class 10 OP Malhotra Exe-11B ICSE Maths Solutions Ch-11 questions as latest prescribe guideline for upcoming exam. In this article you would learn how to find slope / gradient of two point and what is the slope of a line which is parallel / perpendicular to a given line. . Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Coordinate Geometry Class 10 OP Malhotra Exe-11B ICSE Maths Solutions Ch-11
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 10th |
| Chapter-11 | Coordinate Geometry |
| Writer | OP Malhotra |
| Exe-11B | slope / gradient |
| Edition | 2024-2025 |
Slope / Gradient of Two Point
Slope, m = Change in y-coordinates/Change in x-coordinates m = (y2 – y1) / (x2 – x1)
m = (y2 – y1)/(x2 – x1)
Slope of a Line Equation
y − y1 = m(x − x1)
Slope of a line which is parallel / perpendicular to a given line
if two lines are parallel then, Slope of both line is m1 = m2 and if two line are perpendicular to each other then, slope is m1 x m2 = -1
Exercise- 11B
( Coordinate Geometry Class 10 OP Malhotra Exe-11B ICSE Maths Solutions Ch-11 )
Que-1: Find the slope of a line whose inclination to the positive direction of x-axis in anticlockwise direction is given as : (a) 30° (b) 45° (c) 60° (d) 15° (e) 75°
Sol: (a) slope = tan 30° = 1/√3
(b) slope = tan 45° = 1
(c) slope = tan 60° = √3
(d) slope = tan 15° = 0.2679
(e) slope = tan 75° = 3.7231
Que-2: Find the slope and inclination of the line through each pair of the following points : (a) (1,2) and (5,6) (b) (0,0) and (√3,3)
Sol: (a) Slope (m) = {(y2−y1)/(x2−x1)}
= {(6−2)/(5−1)}
= 4/4 = 1
Also slope (m) = tan θ, where θ is the inclination of the line to the positive direction of the x-axis in the anticlockwise direction.
tan θ = 1 ⇒ θ = (tan^–1) 1 = 45º.
(b) Slope (m) = {(y2−y1)/(x2−x1)}
= {(3−0)/(√3−0)}
= 3/√3 = √3
Also slope (m) = tan θ, where θ is the inclination of the line to the positive direction of the x-axis in the anticlockwise direction.
tan θ = √3 ⇒ θ = (tan^–1) √3 = 60º.
Que-3: The side BC of an equilateral ΔABC is parallel to the x-axis. What are the slopes of its sides ?
Sol: We know that the slope of any line parallel to x-axis is 0.
Therefore, slope of BC = 0
Since, ABC is an equilateral triangle, ∠A = 60°
Slope of BA = tan 60° = √3
and Slope of CA = –tan 60° = -√3
Que-4: In a regular hexagon ABCDEF, AB || ED || the x-axis. What are the slopes of its sides?
Sol: ABCDEF is a regular hexagon in which sides AB and CD are parallel to x-axis
Now slope of AB = tan 0° = 0
Slope of BC = tan 60° = √3
,, = tan 120° = -√3
Slope of ED = tan 0° = 0
Slope of EF = tan 60° = √3
and Slope of FA = tan 120° = -√3
Que-5: Find y if the slope of the line joining (-8,11), (2,y) is -4/3.
Sol: Slope (m) = {(y2−y1)/(x2−x1)}
-4/3 = {(y−11)/(2−(-8))}
-4/3 = (y-11)/10
-40/3 = y-11
y = (-40/3) + 11
y = (-40+33)/3, y = -7/3
Que-6: Find the value of a, if the line passing through (-5,-8) and (3,0) is parallel to the line passing through (6,3) and (4,a).
Sol: Two lines are parallel if their slopes are equal
∴ {(0−(−8))/(3−(−5))} = {(a−3)/(4−6)}
⇒ 8/8 = (a−3)/−2
⇒ a – 3 = –2
Therefore a = 1.
Que-7: Find the slope of a line perpendicular to the line whose slope is (a) 1/3 (b) -5/6 (c) 5 (d) -5*(1/7) (e) 0 (f) Infinite
Sol: (a) Let m1 and m2 be the slopes of two lines then:
m1 × m2 = -1 ….(1)
Let m1 = 1/3.
From equation (1)
⇒ 1/3 × m2 = -1
m2 = -3
(b) Let m1 and m2 be the slopes of two lines then:
m1 × m2 = -1 ….(1)
Let m1 = -5/6.
From equation (1)
⇒ -5/6 × m2 = -1
m2 = 6/5
(c) Let m1 and m2 be the slopes of two lines then:
m1 × m2 = -1 ….(1)
Let m1 = 5.
From equation (1)
⇒ 5 × m2 = -1
m2 = -1/5.
(d) Let m1 and m2 be the slopes of two lines then:
m1 × m2 = -1 ….(1)
Let m1 = -5*(1/7) = -36/7.
From equation (1)
⇒ -36/7 × m2 = -1
m2 = 7/36
(e) Let m1 and m2 be the slopes of two lines then:
m1 × m2 = -1 ….(1)
Let m1 = 0.
From equation (1)
⇒ × m2 = -1
m2 = 1/0
which is infinite.
(f) Let m1 and m2 be the slopes of two lines then:
m1 × m2 = -1 ….(1)
Let m1 = ∞
From equation (1)
⇒ ∞ × m2 = -1
m2 = 1/∞
Which is 0.
Que-8: Find the slope of a line perpendicular to the line which passes through the pair of the following points : (a) (0,8) and (-5,2) (b) (1,-11) and (5,2) (c) (-k,h) and (b,-f) (d) (x1,y1) and (x2,y2)
Sol: (a) Slope (m) = {(y2−y1)/(x2−x1)}
m = {(2-8)/(-5-0)}
m = -6/-5 = 6/5
So, slope of line perpendicular to it is (-1/m)
= -5/6
(b) Slope (m) = {(y2−y1)/(x2−x1)}
m = {(2+11)/(5-1)}
m = 13/4
So, slope of line perpendicular to it is (-1/m)
= -4/13
(c) Slope (m) = {(y2−y1)/(x2−x1)}
m = {(-f-h)/(b+k)}
m = -(f+h)/(b+k)
So, slope of line perpendicular to it is (-1/m)
= (b+k)/(f+h).
Que-9: In rectangle ABCD, the slope of AB = 5/6. State the slope of (a) BC (b) CD (c) DA
Sol: In rectangle ABCD, the slope of AB = 5/6
(a) slope of BC = -1/m
= -6/5
(b) Slope of CD = m
= 5/6
(c) Slope of DA = -1/m
= -6/5.
Que-10: In parallelogram ABCD, slope of AB = -2, slope of BC = 3/5. State the slope of (a) AD (b) CD (c) the altitude of AD (d) the altitude of CD
Sol: In parallelogram ABCD, slope of AB = -2, slope of BC = 3/5.
AB || CD and DA || CD.
(a) So, DA || CB
Slope of AD = Slope of BC = 3/5
(b) If CD || AD
Slope of CD = Slope of AD = -2.
(c) Slope of altitude of AD = [-1/slope of AD]
= -5/3
(d) Slope of altitude of CD = [-1/slope of CD]
= -1/-2 = 1/2.
Que-11: The vertices of a ΔABC are A(1,1), B(7,3) and C(3,6). State the slope of the altitude to (a) AB (b) BC (c) AC
Sol: The vertices of a ΔABC are A(1,1), B(7,3) and C(3,6).
(a) Slope of AB = {(y2−y1)/(x2−x1)}
= {(3-1)/(7-1)}
= 2/6 = 1/3
and slope of its altitude = -1/slope of AB
= -3.
(b) Slope of BC = {(y2−y1)/(x2−x1)}
= {(6-3)/(3-7)}
= 3/-4 = -3/4
and slope of its altitude = -1/slope of BC
= 4/3.
(c) Slope of AC = {(y2−y1)/(x2−x1)}
= {(6-1)/(3-1)}
= 5/2
and slope of its altitude = -1/slope of AB
= -2/5.
Que-12: The line joining (-5,7) and (0,-2) is perpendicular to the line joining (1,-3) and (4,x). Find x.
Sol: Slope of line joining the lines (-5,7) and (0,-2) is
Slope m1 = {(y2−y1)/(x2−x1)}
m1 = (-2-7)/(-0+5)
m1 = -9/5
Slope of line joining the lines (1,-3) and (4,x) is
Slope m2 = {(y2−y1)/(x2−x1)}
m2 = (x+3)/(4-1) = (x+3)/3
If these lines are perpendicular to each other then,
m1 × m2 = -1
(-9/5) × {(x+3)/3} = -1
{-3(x+3)}/5 = -1
(x+3) = -1 × (-5/3) = 5/3
x = (5/3) – 3
x = (5-9)/3, x = -4/3
Que-13: The vertices of a quad. PMQS are P(0,0), M(3,2), Q(7,7) and S(4,5). Show that PMQS is a parallelogram.
Sol: The vertices of a quad. PMQS are P(0,0), M(3,2), Q(7,7) and S(4,5)
Slope of PM m1 = (y2-y1)/(x2-x1)
(2-0)/(3-0) = 2/3
Slope MQ m2 = (7-2)/(7-3)
= 5/4
Slope of QS m3 = (5-7)/(4-7)
= 2/3
Slope of SP m4 = (5-0)/(4-0)
= 5/4.
If slope of PM = Slope of QS = 2/3
So, PM || QS
Slope of MQ = Slope of SP = 5/4
So, MQ || SP
Therefore, PMQS is a parallelogram.
Que-14: Show that P(a,b), Q(a+3,b+4), R(a-1,b+7), S(a-4,b+3) are the vertices of a square. What is the area of the square.
Sol: Let the points be A(a, b), B(a + 3, b + 4), C(a – 1, b + 7) and D(a – 4, b + 3).
We know that,
Distance formula = √{(x2−x1)²+(y2−y1)²}By distance formula,
AB = √{(a+3−a)²+(b+4−b)² = √(3²+4²)
= √(9+16) = √25 = 5 units.
BC = √{[(a−1)−(a+3)]²+[(b+7)−(b+4)]²} = √{[a−a−1−3]²+[b−b+7−4]²}
= √{[−4]²+[3]²} = √(16+90) = √25 = 5 units.
CD = √{[(a−4)−(a−1)]²+[(b+3)−(b+7)]²} = √{[a−a−4+1]²+[b−b+3−7]²}
= √{[−3]²+[−4]² = √(9+16) = √25 = 5 units.
DA = √{[a−(a−4)]²+[b−(b+3)]²} = √(4²+(−3)²)
= √(16+9) = √25 = 5 units.
Since, AB = BC = CD = DA.
So, PQRS is a square.
Now area of square PQRS is = (side)
= (PR)
Que-15: Without using Pythagoras theorem, show that the points A(0,4) ,B(1,2) and C(3,3) are the vertices of a right angled triangle.
Sol: We have, A (0, 4), B (1, 2) and C (3, 3)
Now, Slope of m1 = Slope of AB = (2−4)/(1−0) = −2
Slope of BC m2 = Slope of BC = (3−2)/(3−1) = 1/2
Slope of CA m3 = Slope of CA = (4−3)/(0−3) = −1/3
∴ m1m2 = −2 × (1/2) = −1
Therefore, AB is perpendicular to BC, i.e.
∠ABC = 90∘.
Thus, the given points are the vertices of a right angled triangle.
Que-16: If the points (x,1), (1,2) and (0,y+1) are collinear, show that (1/x)+(1/y) = 1.
Sol: Slope of line joining (x, 1), (1, 2) = (2−1)/(1−x)
Slope of line joining (1, 2), (0, y + 1) = (y+1−2)/(0−1)
⇒ {1/(1−x)} = {(y−1)−1} = (1 – x) (y – 1)
⇒ –1 = y – xy – 1 + x
⇒ x + y = xy
= (1/x) + (1/y) = 1.
–: End of Coordinate Geometry Class 10 OP Malhotra Exe-11B ICSE Maths Solutions Ch-11 :–
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