8th Maths : Chapter 3 : Algebra : Graph : Exercise 3.9 : Text Book Back Exercises Questions with Answers, Solution

**Exercise 3.9**

** **

**1. Fill in the blanks:**

(i) *y *=*
p x *where* p *∈*z *always passes through the__________. **[Answer: Origin (0, 0)]**

**Solution:**

[When we substitute *x* = 0 in equation, *y* also
becomes zero. ∴ (0, 0) is a solution]

(ii) The
intersecting point of the line *x* = 4 and
*y* = − 4 is__________. **[Answer: 4, − 4]**

**Solution:**

*x* = 4 is a line parallel to the *y*
− axis and

*y* = − 4 is a line parallel to the *x*
− axis. The point of intersection is a point that lies on both lines &
which should satisfy both the equations. Therefore, that point is (4, −4)

(iii) Scale
for the given graph ,

On the x-axis
1 *cm* = ------- units

y-axis 1
*cm* = ------- units

**[Answer: 3 units, 25 units]**

**Solution:**

With reference to given
graph,

On the *x* − axis, 1 cm = 3 units

*y* axis, 1cm =25 units

** **

**2. Say True or False.**

(i) The points
(1,1) (2,2) (3,3) lie on a same straight line. **[Answer: True]**

**Solution:**

The points (1, 1), (2, 2), (3, 3) all satisfy the equation *y
= x* which is straight line. Hence, it is** true**

(ii) *y* =−9 *x* not passes through the origin. **[Answer: False]**

**Solution:**

*y *= −9*x* substituting for *x*
as zero, we get *y* = − 9 × 0 = 0

∴ for *x *= 0, *y*
= 0. Which means line passes through (0, 0), hence statement is **false.**

**3. Will a line pass through (2, 2) if
it intersects the axes at (2, 0) and (0, 2).**

**Solution:**

Given a line intersects the axis at (2, 0) & (0, 2)

Let line intercept form be expressed as

*ax* + *by* = 1 Where *a*
& *b* are the *x* & *y* intercept respectively.

Since the intercept points are (2, 0) & (0, 2)

*a* = 2, *b* = 2

∴ 2*x* + 2*y* = 1

When the point (2, 2) is considered & substituted in the
equation

2*x* + 2*y* = 1 , we get

2 × 2 + 2 × 2 = 4 ≠ 1

∴ the point (2, 2) does not satisfy the equation. Therefore the
line does not pass through (2, 2)

**Alternatively graphical method**

as we can see the line doesn’t pass through (2, 2)

** **

**4. A line passing through (4, −2) and
intersects the Y-axis at (0, 2). Find a point on the line ****in the second quadrant.**

**Solution:**

Line passes through (4, −2)

y – axis intercept point – (0, 2)

using 2 point formula,

∴ *x* + *y* = 2 is the equation of the line.

Any point in II quadrant will have *x* as negative & *y*
as positive.

So let us take *x* value as − 2

∴ −2 + *y* = 2

∴ *y* = 2 + 2 = 4

∴ Point in II
Quadrant is (−2, 4)

** **

**5. If the points P(5, 3) Q(−3, 3) R(−3,
−4) and S form a rectangle, then find the coordinate of S.**

**Solution:**

Plotting the points on a graph (approximately)

**Steps :**

1. Plot P, Q, R approximately on a graph.

2. As it is a rectangle, RS should be parallel to PQ & QR
should be parallel to PS

3. S should lie on the straight line from R parallel to x−axis
& straight line from P parallel to y−axis

4. Therefore, we get S to be (5, −4)

[Note: We don’t need graph sheet for approximate plotting. This
is just for graphical understanding]

** **

**6. A line passes through (6, 0) and (0,
6) and an another line passes through (−3,0) and (0, −3). What are the points to
be joined to get a trapezium?**

**Solution:**

In a trapezium, there are 2 opposite sides that are parallel.
The other opposite sides are non−parallel.

Now, let us approximately plot the points for our understanding

[no need of graph sheet]

1. Plot the points (0, 6), (6, 0), (−3, 0) & (0, − 3)

2. Join (0, 6) & (6, 0)

3. Join (−3, 0) & (0, −3)

4. We find that the lines formed by joining the points are
parallel lines.

5. So, for forming a trapezium, we should join (0, 6), (−3, 0)
& (0, −3), (6, 0)

** **

**7. Find the point of intersection of
the line joining points (−3, 7) (2, −4) and (4,6) (−5,−7). Also find the point of
intersection of these lines and also their intersection with the axis.**

**Solution:**

Line 1: Joining points (*x*_{1,} *y*_{1})_{
}: (−3, 7) & ( *x*_{2,} *y*_{2}) : (2, −4)

Equation of line joining 2 points by 2 point formula is given by

[*y* – *y*_{1} ] / [ *x* – *x*_{1} ] =
[ *y*_{2 }– *y*_{1} ] / [*x*_{2} – *x*_{1}] ∴ [*y* – 7 ] / [*x*
– (−3) ] = [−4 −7] / [2 – (−3)] ∴ [*y* – 7] / [*x*
+ 3] = −11 / [2 + 3] ∴[*y* – 7] / [*x*
+ 3] = −11 / [2 + 3] ∴[ *y* – 7] / [*x* + 3] = −11 / 5

Cross multiplying, we get [*y* – 7] / [*x*
+ 3] = −11 / 5 ∴ 5(*y* − 7) = −11 (*x* + 3) ∴5*y* − 35 = − 11*x* – 33

Transposing the variables, we get

11*x* + 5*y* = 35 − 33 = 2

**11 x + 5y = 2** Line 1

Similarly, we should find out equation of second line

Joining the points *x*_{1} *y*_{2} : (4,
6) & *x*_{2} *y*_{2} : (−5, −7)

[*y* – *y*_{1}] /
[*x* – *x*_{1}]
= [*y*_{2
}– *y*_{1}] / [*x*_{2}
– *x*_{1}] ; ∴* *[*y* − 6] / [*x*
– 4] = [−7 – 6] / [− 5 − 4]

∴ [*y* − 6] / [*x*
– 4] = −13 / −9 = 13/9

∴ 9*y* − 54 = 13*x* − 52

∴ **9 y −13x = 2** Line 2

For finding point of intersection, we need to solve the 2 line,
equation to find a point that will satisfy both the line equations.

∴ Solving for *x* & *y* from line 1 & line 2 as
below

11*x* + 5*y* = 2 ⇒ multiply both sides by 13,

11 × l3*x* + 5 × l3*y* = 26 ....(3)

Line 2: 9*y* − 13*x* = 2 ⇒ multiply both sides by 11

9 × 11*y* – 13 × 11*x* = 22 ....(4)

i.e 65*y* + 143*x*
= 26 ....(3)

+ 99*y* − 143*x* = 22 ....(4)

Adding (3) & (4)

__65 y + 99y + 0 = 26 + 22__

∴ 164*y* = 48

∴* y* = 48 / 164 = 12 / 41

Substituting this value of *y* in line 1 we get

11*x* + 5*y* = 2

11*x* + 5*y* × [12/41] = 2

11*x *= 2 – 60/41 = [82 – 60] / 41 = 22/41

∴* x *= 22/41

**∴ **

To find point of intersection of the lines with the axis, we
should substitute values & check

Line 1: 11*x* +
5*y* = 2

Point of intersection
of line with *x*− axis, i.e *y* coordinate is ‘0’

∴ Put *y* = 0 in above equation

∴ 11*x* – 5 × 0 = 2

∴ 11*x* + 0 = 2

∴ *x* = 2 / 11

**∴ **

Similarly, point of intersection of line with *y* − axis is
when *x* −coordinate becomes ‘0’

∴ put *x* = 0 in above equation

∴ 11 × 0 + 5*y* = 2

∴ 0 + 5*y* = 2

*y *= 2/5

**∴ **

Similarly for line 2,

9*y* −13*x* = 2

For finding *x* intercept, i.e point where line meets *x*
axis, we know thaty coordinate becomes ‘0’

∴ Substituting* y* = 0 in above eqn. we get

9 × 0 −13*x* = 2

∴ 0 − 13*x* = 2

∴ *x* = −2 / 13

**∴ Point : (− 2/13, 0) **

Similarly for *y* – intercept,* x* − coordinate
becomes ‘0’

∴ Substituting* *for *x* = 0 in above equation. we get

9*y* – 13 × 0 = 2

9*y* − 0 = 2

9*y* = 2

*y* = 2 / 9

**Point (0, 2/9) **

** **

**8. Draw the graph of the following equations,
(i) x = −7 (ii) y = 6**

**Solution: **

** **

**9. Draw the graph of (i) y = −3 x ii) y = x−4 iii) y = 2x+5**

**Solution: **

To draw graph, we need to find out some points.

**(i) y = − 3x**

for *y* = − 3*x*, let us first substituting values
& check

put *x* = 0

*y* = −3 × 0 = 0 ∴
(0, 0) is a point

put *x *= 1

*y* = −3 × l = −3 ∴
(1, −3) is a point

If join these 2 points, we will get the line

**ii) y = x
− 4**

for *y* = *x* − 4

put *x* = 0

*y* = 0 – 4 = − 4 ∴ (0, − 4) is a point

*x *= 4

*y* = 4 – 4 = 0 ∴ (4, 0) is a point

**iii) y = 2x
+ 5**

for *y* = 2*x* + 5

put *x* = − 1

*y* = 2 (−1) + 5 = −2 + 5 = 3 ∴ (− 1, 3) is a point

put *x* = −2

*y* = 2 (−2) + 5 = − 4 + 5 = 1 ∴ (− 2, 1) is a point

Now let us plot the points & join them on graph

** **

**10. Find the values.**

**Solution:**

**(a) Let y = x + 3**

(i) if *x* = 0, *y* = 0 + 3 = 3,

∴ *y* = 3

(ii) *y* = 0, 0 = *x* + 3,

∴ *x* = − 3

(iii) *x* = − 2, *y* = −2 + 3

∴ *y* = 1

(iv) *y* = −3, − 3 = *x* + 3,

∴ *x* = − 6

**(b)**** Let 2 x + y − 6 = 0**

(i) *x* = 0, 2 × 0 + *y* − 6 = 0, *y* = 6

(ii) *y* = 0, 2*x* + 0 − 6 = 0, ∴ 2*x* = 6

*x* = 3

(iii) *x* = − 1, 2 × (− l) + *y* − 6 = 0, − 8 + *y*
= 0

*y* = 8

(iv) *y* = − 2, 2*x* − 2 − 6 = 0, 2*x* = 8

*x* = 4

**(c)**** Let y = 3x + 1**

(i) *x* = −l, *y* = 3 (−1)+1 = 0, ∴ *y* = −2

(ii) *x* = 0, *y* = 3 (0) + 1 = 0, . ∴ *y* = l

(iii) *x* = 1, *y* = 3 (1) + 1 = 0, ∴ *y* = 4

(iv) *x* = 2, *y* = 3 (2) + 1 = 0, ∴ *y* = 7

**Exercise 3.9 **

**1. (i) Origin (ii) (4,-4)
(iii) x-axis 1cm=3 units, y-axis 1cm=25
units**

**2. (i) True (ii) False**

** **

Tags : Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra

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