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Math Class VIII 1 Question Bank
1. Plot the following points on a graph paper :
(i) A (3, 5) (ii) C (0, 3) (iii) D (5, 0)
(iv) E (2, – 4) (v) F (– 3, 5) (vi) H (– 2, – 3)
(vii) J (– 3, 0) (viii) K (0, – 4)
Ans.
1 2 3 4 5 6 7–1–2–3–4–5–6–7x′ x
O
1
–1
–2
–3
–4
–5
–6
–7
2
3
4
5
6
y
y′
C (0, 3)
A(3, 5)F(–3, 5)
J(–3, 0)
H(–2, –3)
K(0, –4)
E(2, –4)
D(5, 0)
2. Write down the coordinates of each of the following points P, Q,
R, S and T shown on the graph paper.
14COORDINATE SYSTEM
AND GRAPHS
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Math Class VIII 2 Question Bank
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
–6
2
3
4
5
6
y
y′
T
S
R
HQ
P 4
Ans.From graph the coordinates of
P are (3, 0)
Q are (4, – 3)
R are (– 4, – 2 )
S are (– 6, 3) and T are (– 3, 2)
3. Plot the points A (4, 1), B (– 2, 1), C (– 3, –2) and D (3, – 2). Name
the figure ABCD.
Ans. We have plotted the points A (4, 1), B (– 2, 1), C (– 3, –2) and
D (3, – 2) on the graph paper and join AB, BC, CD and DA.
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Math Class VIII 3 Question Bank
1 2 3 4 5 6 7–1–2–3–4–5–6–7x′ x
O
1
–1
–2
–3
–4
–5
–6
–7
–8
2
3
4
5
6
7
y′
(–2, 1)B A(4, 1)
D(3, –2)(–3, –2)C
4. Plot the points A (5, 1), B (– 2, 2) and C (4, 2). Name the figure
ABC.
Ans.We have plotted the points A (5, 1), B (– 2, 2) and C (4, 2) on the
graph and join AB, BC and CA. The figure formed is triangle ABC
1 2 3 4 5 6 7–1–2–3–4–5–6–7x′ x
O
1
–1
–2
–3
–4
–5
–6
–7
–8
2
3
4
5
6
7
y
y′
(–2, 2)BC (4, 2)
A (5, 1)
© EDULA
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Math Class VIII 4 Question Bank
5. Draw the graph of 2x – 3y = 6 and 2x – 3y = 3 on the same paper.What do you observe?
Ans. The given equations can be written as
2x = 6 + 3y ⇒6 3
2
yx
+= ... (i)
and 2x = 3 + 3y ⇒3 3
2
yx
+= ... (ii)
Table of values for (i) we have
0 1 2
2 1.33 0.66
x
y − − −
Table of values for (ii), we have
0 1 2
1 0.33 0.33
x
y − −
Plot these points on the graph paper as under :
1 2 3–1–2–3x′ xO
1
–1
–2
–3
2
3y
y′
(–1, –1) C
From the graph we see that the graph of two equations are parallelto each other.
6. A(5, 3), B(– 1, 3) and C(– 1, – 1) are three vertices of a rectangleABCD. Plot the given points on a graph paper and then use thisgraph to find the co-ordinates of the fourth vertex D.
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Math Class VIII 5 Question Bank
Ans.
1 2 3 4 5–1–2–3–4–5x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–1, 3) B
(–1, –1) C
A (5, 3)
D (5, –1)
Co-ordinates of the fourth vertex D = (5, – 1)
7. A(1, 0), B(–2, 0) and D(1, –3) are the vertices of a square ABCD. Byplotting the given points on a graph paper, find the co-ordinates ofthe unknown vertex C.
Ans.
1 2 3 4 5–1–2–3–4–5x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–2, –3) CD (1, –3)
A (1, 0)B (–2, 0)
Co-ordinates of the unknown vertex C = (–2, –3)
8. Plot the points A(4, 5), B(–1, 5), C (–1, –2) and D(4, –2) on a
graph paper. Join AB, BC, CD and DA. Give a special name to the
quadrilateral ABCD obtained. Also, find its area.
© EDULA
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Math Class VIII 6 Question Bank
Ans.
1 2 3 4 5–1–2–3–4–5x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–1, –2) C D (4, –2)
A (1, 0)
(–1, 5) B A (4, 5)
Take each small square = 1 unit
A(4, 5), B(–1, 5), C(–1, –2) and D(4, –2) are plotted or on thegraph paper.
Join AB, BC, CD, DA. Thus, ABCD is a rectangle.
Here AD = 5 + 2 = 7, AB = 4 + 1 = 5
Area of rectangle ABCD = AB × AD = 5 × 7 = 35 sq . units.
9. Plot the points A(1, –1), B(–1, 4) and C(–3, –1) on a graph paperto obtain the triangle ABC. Give a special name to the triangleABC and, if possible, find its area.
Ans. As Plot the points
A(1, –1), B(–1, 4), C(–3, –1)
1 2 3 4 5–1–2–3–4–5x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–3, –1) CL (–1, –1)
A (1, –1)
(–1, 4) B
© EDULA
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Math Class VIII 7 Question Bank
Join AB, BC, CA. Clearly AC = 1 + 3 = 4 units
Draw BL perpendicular to AC. Then L will be (–1, –1)
[∵ ordinate on each points of AC = –1]
∵ AL = 1 + 1 = 2 units ∴ L is the mid point of AC
∆ABC is an isosceles triangle.
Area of triangle ABC = 1
2Base × height
= 1
2 (4) (5) = 10 sq. units.
10. Plot the points A(2, 0), B(0, 5) and C(–2, 0), what kind of triangleis ∆ ABC ? Find its area.
Ans. Marked the points A(2, 0), B(0, 5) and C(–2, 0) on the graphpaper. Joint AB, BC, CA.
From graph ABC is an isosceles triangle.
Area of isosceles triangle = 1
2 × base × height
= 1
2 × AC × OB =
1
2 × 4 × 5 sq units
1 2 3 4 5–1–2–3–4–5x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y ′
B(0, 5)
C(–2, 0) A(2, 0)
= 2 × 5 sq units
= 10 sq units.
© EDULA
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Math Class VIII 8 Question Bank
11. Plot a rectangle which lies in first quadrant, has origin as one vertex, is6 units long along x-axis and 4 units long along y-axis. Give the co-ordinates of its vertices.
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y ′
A
BC
Ans. The rectangle which lies in first quadrant has origin as vertex, is 6unit long along x-axis and 4 units long along y-axis is follows :
Co-ordinates of its vertices are O (0, 0), A(6, 0), B(6, 4) andC(0, 4).
12. Draw the graphs of the following equations:
(i) x + 2y = 0 (ii) 3x – 2y = 9
(iii) 2x + 3y = 12 (iv) x + 2y + 1 = 0
(v) 5x – 3y = 9 (vi) 7x + 3y = 13
(vii)1
13
y x= + (viii)1
32
x y= −
(ix)5
32
x y= − (x) x = – 2 – y
Ans. (i) x + 2y = 0 ⇒ x = – 2y
Put y = 1, then x = – 2 × 1 = – 2
Put y = 2, then x = – 2 × 2 = – 4
Put y = 3, then x = – 2 × 3 = – 6
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Math Class VIII 9 Question Bank
Put y = – 1, then x = – 2 × (– 1) = 2
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–6, 3)
(–4, 2)
(–2, 1)
(2, –1)
(ii) 3x – 2y = 9
⇒ 3x = 9 + 2y
⇒ 9 2
3
yx
+=
Put y = 0, then9 2 0
3x
+ ×= =
9 0 93
3 3
+= =
Put y = 3, then 9 2 3
3x
+ ×=
9 6 155
3 3
+= = =
Put y = – 3 then 9 2 ( 3)
3x
+ × −=
9 6 31
3 3
−= = =
Put y = – 6 then 9 2 ( 6)
3x
+ × −=
9 12 31
3 3
− −= = = −
Table of values 0 3 3 6
3 1 5 1
y
x
− −
−
1 2 3 1
2 4 6 2
−
− − −
y
x
© EDULA
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Math Class VIII 10 Question Bank
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
–6(–1, –6)
(1, –3)
(3, 0)
(5, 3)
(iii) 2x + 3y = 12 ⇒ 2x = 12 – 3y
⇒ 12 3
2
yx
−=
Put, y = 0, then 12 3 0
2x
− ×=
12 0 126
2 2
−= = =
Put, y = 2, then 12 3 2
2x
− ×=
12 6 63
2 2
−= = =
Put, y = 3, then 12 3 3
2x
− ×=
12 9 31.5
2 2
−= = =
Put, y = 4, then 12 3 4
2x
− ×=
12 12 00
2 2
−= = =
Table of values0 2 3 4
6 3 1.5 0
y
x
© EDULA
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Math Class VIII 11 Question Bank
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(0, 4)
(1.5, 3)
(3, 2)
(6, 0)
(iv) x + 2y + 1 = 0
⇒ x = –2y – 1⇒ x = – (2y + 1)
Put y = 0 then x = 1 – (2 × 0 + 1) = – (0 + 1) = – 1
Put y = 1, then x = – (2 × 1 + 1) = – (2 + 1) = – 3
Put y = – 2, then x = – {2 × (– 2) + 1} = – (– 4 +1) = 3
Put y = –1 then x = – (2 × (–1) + 1 ) = – ( – 2 + 1) = 1
Table of values 0 1 2 1
1 3 3 1
y
x
− −
− −
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–3, 1)
(–1, 0)
(1, –1)(3, –2)
© EDULA
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Math Class VIII 12 Question Bank
(v) 5x – 3y = 9
⇒ 5x = 9 + 3y ⇒ 9 3
5
yx
+=
Put y = 2, then 9 (3)(2) 9 6
35 5
x+ +
= = =
Put y = – 3, then 9 3 ( 3) 9 9
05 5
x+ × − −
= = =
Put y = 7, then 9 (3) (7) 9 21 30
65 5 5
x+ × +
= = = =
2 3 7
3 0 6
y
x
−
Plot the points
A(3, 2), B(0, –3), C(6, 7) on the graph
1 2 3 4 5 6 7–1–2–3–4–5–6–7x′ x
O
1
–1
–2
–3
–4
–5
–6
–7
–8
2
3
4
5
6
7
y
y′
C (6, 7)
A (3, 2)
B (0, –3)
(vi) 7x + 3y = 13 ⇒ 3y = 13 – 7x ⇒ 13 7
3
xy
−=
© EDULA
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Math Class VIII 13 Question Bank
Put x = – 1, then 13 7( 1) 13 7 20
6.63 3 3
y− − +
= = = =
Put x = 1, then 13 7(1) 13 7 6
23 3 3
y− −
= = = =
Put x = 4, then 13 7(4) 13 28 15
53 3 3
y− − −
= = = = −
6.6 2 5
1 1 4
y
x
−
−
Plot the points A(– 1, 6.6), B(1, 2) and C(4, –5) on the graph.
Join them to form a line. then it is the required graph of
7x + 3y = 13
1 2 3 4 5 6 7–1–2–3–4–5–6–7x′ x
O
1
–1
–2
–3
–4
–5
–6
–7
–8
2
3
4
5
6
7
y
y′
C (4, –5)
B (1, 2)
A (–1, 6.6)
(vii) 1
13
y x= +
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Math Class VIII 14 Question Bank
Put y = 2 in 1
13
y x= + , we get
12 1
3x= + ⇒
12 1
3x− =
⇒ 1
13
x= , then x = 3
Put y = 3, we get, 3 – 1 = 1
3x
⇒ 1
23
x= , then x = 6
Put y = 0, we get, 1
0 13
x= +
⇒ 1
13
x = − then x = –3
2 3 0
3 6 3
y
x −
Plot the points on the graph. The line formed by joining the
points is the required graph of 1
13
x= + .
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–3, 0)
(3, 2)
(6, 3)
© EDULA
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Math Class VIII 15 Question Bank
(viii)1
32
x y= −
Put y = 2, then 2
3 1 3 22
x = − = − = −
Put y = – 2, then 2
3 42
x−
= − = −
Put y = 6, then 6
3 02
x = − =
2 2 6
2 4 0
y
x
−
− −
Plot the points on the graph paper. Join them. The line we get
is the required graph of 32
yx = − .
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
6
y
y′
(–4, –2)
(–2, 2)
(0, 6)
(ix)5
32
x y= − ⇒ 5
32
y x= −
⇒ 5y = 6 – 2x ⇒ 6 2
5
xy
−=
Put y = 0, then 6 2
05
x−= ⇒ 0 = 6 – 2x
⇒ – 2x = – 6 ⇒ x = 3
© EDULA
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Math Class VIII 16 Question Bank
Put y = 4, then 6 2
45
x−= ⇒ 20 = 6 – 2x
⇒ – 2x = 14 ⇒ x = – 7
Put y = – 2 then ,6 2
25
x−− = ⇒ –10 = 6 – 2x ⇒ 16 = – 2x
⇒ x = 8
0 4 2
3 7 8
y
x
−
−
Plot the points on the graph paper. Join them, then it is the
graph of 5
32
x y= −
1 2 3 4 5 6 7 8–1–2–3–4–5–6–7x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
6
y
y′
(–7, 4)
(3, 0)
(8, –2)
(x) x = – 2 – y, y = – 2 – x
Put y = – 2 then –2 = – 2 – x ⇒ x = 0
Put y = – 5 then – 5 = – 2 – x ⇒ x = 3
Put y = 1 then 1 = – 2 – x ⇒ – x = 3 ⇒ x = – 3
2 5 1
0 3 3
y
x
− −
−
© EDULA
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Math Class VIII 17 Question Bank
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–3, 1)
(0, –2)
(3, –5)
x = –2 –y
13. Solve the following pairs of equations, graphically :
(i) x = 0 and x + 3y = 6 (ii) x = 4 and 2x – 3y + 1 = 0
(iii) 3x – 4y = 1 and 2y – x = 1 (iii) x + 2y = 11 and 2x – y = 2
(v) 3x – 4y = 1 and x – 2y + 1 = 0 (vi) 32 3
x y− = and x + y = 1
(vii) x = – y – 1 and 2y = 1 – x (viii) 2x – 3y = – 6 and 12
yx − =
Ans.
1 2 3 4 5 6–1–2–3–4–5–6x′ x
0
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(–3, 3)(0, 2)
(3, 1)
x + 3y = 6
x = 0
(i) x = 0, and x + 3y = 6
⇒ 3y = 6 – x
Put x = 0 then y = 2, Put x = 3 then y = 1
© EDULA
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Math Class VIII 18 Question Bank
Put x = – 3 then y = 3 ⇒ 6
3
xy
−=
2 1 3
0 3 3
y
x −
Hence, the required solution is x = 0 and y = 2.
(ii) x = 4, and
2x – 3y + 1 = 0
⇒ 2x + 1 = 3y
⇒2 1
3
xy
+=
Put y = 1 then x = 1
Put y = 3 then x = 4
Put y = – 1 then x = – 2
1 4 2
1 3 1
x
y
−
−
Plot the points on the graph. From the graph the two linesintersect at point (4, 3).
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
2x – 3y + 1 = 0
(1, 1)
(–2, –1)
(4, 3)
x =
4
Hence, the required solution is x = 4, and y = 3.
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Math Class VIII 19 Question Bank
(iii) 3x – 4y = 1
⇒ 3x – 1 = 4y
⇒ 4y = 3x – 1 ⇒ 3 1
4
xy
−=
If x = – 1, then 3( 1) 1 3 1 4
14 4 4
y− − − − −
= = = = −
If x = 1, then 3(1) 1 3 1 2
0.54 4 4
y− −
= = = =
If x = 3, then 3(3) 1 9 1 8
24 4 4
y− −
= = = =
1 0.5 2
1 1 3
y
x
−
−
Plot the points A(–1, –1), B(1, 0.5) and C(3, 2) on the graph.
Now 2y – x = 1 ⇒ 2y = x + 1 ⇒ 1
2
xy
+=
Put x = – 3, then 3 1 2
12 2
y− + −
= = = −
Put x = 1, then 1 1 2
12 2
y+
= = =
Put x = 3, then 3 1 4
22 2
y+
= = =
1 1 2
3 1 3
y
x
−
−
Plot the points P(–3, –1), Q(1, 1) and R(3, 2) on the same graph.
© EDULA
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Math Class VIII 20 Question Bank
1 2 3 4 5 6–1–2–3–4–5–6 O
1
–1
–2
–3
–4
–5
2
3
4
5
P(–3, –1)
A(–
1, –
1)
Q(1,1)C (3,2)
R (3,2)
2y – x
= 1
3x –
4y =
1
B(1, 0.5)
From the graph two lines intersect each other at a point (3, 2)
Hence, x = 3 and y = 2 is the required solution.
(iv) x + 2y = 11 ⇒ 2y = 11 – x ⇒11
2
xy
−=
Put x = 1, then 11 1 10
52 2
y−
= = =
Put x = 3, then 11 3 8
42 2
y−
= = =
Put x = 5, then 11 5 6
32 2
y−
= = =
5 4 3
1 3 5
y
x
Plot the points A(1, 5), B(3, 4) and C(5, 3) on the graph.
Now 2x – y = 2 ⇒ 2x – 2 = y ⇒ y = 2x – 2
Put x = 1, then y = 2(1) – 2 = 2 – 2 = 0
Put x = 2, then y = 2(2) – 2 = 4 – 2 = 2
Put x = 3, then y = 2(3) – 2 = 6 – 2 = 4
0 2 4
1 2 3
y
x
x′
y′
x
y
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Math Class VIII 21 Question Bank
Plot the points P(1, 0), Q(2, 2) and R(3, 4) on the same graph
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
Q (2, 2)
P (1,0)
C (5,3)B (3,4)
R (3,4)A (1, 5)
2x –
y =
2 x + 2y = 11
From the graph two lines intersect each other at a point (3, 4)
Hence, x = 3 and y = 4 is the required solution.
(v) 3x – 4y = 1
x – 2y + 1 = 0
Now, 3x – 4y = 1 ⇒ 3x – 1 = 4y ⇒ 3 1
4
xy
−=
Put y = –1, then x = –1Put y = 2, then x = 3
Put y = – 4, then x = – 5
1 2 4
1 3 5
y
x
− −
− −
x – 2y + 1 = 0
⇒ x + 1 = 2y ⇒ 1
2
xy
+=
Put y = 2, then x = 3
Put y = –1, then x = – 3
Put y = 3, then x = 5
2 1 3
3 3 5
y
x
−
−
© EDULA
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Math Class VIII 22 Question Bank
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(5, 3)(3, 2)
(–1, –1)
(–5, –4)
(–3, –1)
x – 2y + 1 = 0
3y – 4y
= 1
From the graph two lines intersects at the point (3, 2).
Hence, x = 3 and y = 2 is the required solution.
(vi) 32 3
x y− = ⇒ 3
2 3
x y− = ⇒ 3 3
2
xy
= −
⇒
39
2
xy = −
Put y = – 9, then x = 0
Put y = – 6, then x = 2
Put y = – 3, then x = 4
9 6 3
0 2 4
y
x
− − −
x + y = 1 y = 1 – x
Put y = 1, then x = 0
Put y = 4, then x = – 3
Put y = – 3, then x = 4
1 4 3
0 3 4
y
x
−
−
Plot the points on the graph. From the graph, the two linesintersects at (4, –3).
Hence, x = 4 and y = –3 is the required solution.
© EDULA
BZ
INTER
NATIO
NAL
Math Class VIII 23 Question Bank
1 2 3 4 5 6 7–1–2–3–4–5–6–7x′ x
O
1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
2
3
4
5
y
y′
(–3, 4)x + y = 1
(0, 1)
(4, –3)
(2, –6)
(0, –9)
x/2
– y/
3 =
3
(vii) x = – y – 1 ⇒ y = – 1 – x
Put y = – 1, then x = 0
Put y = 3, then x = – 4
Put y = – 3, then x = 2
1 3 3
0 4 2
y
x
− −
−
And, 2y = 1 – x ⇒1
2
xy
−=
Put y = – 1, then x = 3
Put y = 2, then x = – 3
Put y = – 2, then x = 5
3 3 5
1 2 2
x
y
−
− −
Plot the points on the graph paper. From the graph we see that,the two lines intersects at the point (–3, 2).
© EDULA
BZ
INTER
NATIO
NAL
Math Class VIII 24 Question Bank
Hence, x = –3 and y = 2 is the required solution.
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(2, –3)
(0, –1)
(–3, 2)
(–4, 3)
(5, –2)
x = –y –1
2y = 1 – x
(viii) 2x – 3y = –6 ⇒ 2x + 6 = 3y ⇒ 2 6
3
xy
+=
Put y = 2, then x = 0
Put y = 4, then x = 3
Put y = 0, then x = –3
2 4 0
0 3 3
y
x −
And, 12
yx − = ⇒ 1
2
yx = + ⇒
2
2
yx
+=
Put y = 0, then x = 1
Put y = 4, then x = 3
Put y = – 2, then x = 0
0 4 2
1 3 0
y
x
−
Plot the points on the graph. From the graph the two lines inter-sects at the point P(3, 4).
© EDULA
BZ
INTER
NATIO
NAL
Math Class VIII 25 Question Bank
Hence, x = 3 and y = 4 is the required solution.
1 2 3 4 5 6–1–2–3–4–5–6x′ x
O
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
(0, –1)
(–3, 0)
(0, 2)
(3, 4)
(1, 0)
(0, –2)
x –
y/2
= 1
2x – 3y = 6
14. Draw the graph of the linear equations x = – 2, x = 5, y = 0 and
y = 4 on the same graph paper. Hence find the area of the quadri-
lateral enclosed by these lines.
Ans.The graph of the linear equations
x = – 2 ... (i)
x = 5 , ... (ii)
y = 0 ... (iii)
and, y = 4 ... (iv)
are shown in the given graph.
1 2 3 4 5 6–1–2–3–4–5–6x′ x
0
1
–1
–2
–3
–4
–5
2
3
4
5
y
y′
y = 4
y =
4
y = 0
Area of the quadrilateral enclosed by these lines = base × height
= 7 units × 4 units = 28 square units.
