Introduction To Graphs

P (−3,5)

Q (0,−8)

R (4,0)

S (−4,−9)

A. Since both X coordinate and Y coordinate are Positive

⇒ (+,+) which is in the first coordinate

B. Since both X coordinate and Y coordinate are Negative

⇒ (-, -) which is in the third coordinate

C. Since both X coordinate is POSITIVE and Y coordinate is Negative

⇒ (+, -) which is in fourth coordinate

In the figure, we have two points marked, P = (-8, 3) and O₁ = (9, 5)

According to question O₁ is the origin for a different coordinate system, X₁’O₁X₁↔ Y₁’O₁Y₁.

For coordinates of P in the new system, we have to find the distance between O₁and point P, measured FROM P.

By measuring distance on the graph above, we get

X₁= -17 and Y₁= -2

⇒ So, the coordinates of Point P in new system is (-17, -2)

OR,

Given, x= -8, y= 3

and a= 9, b=5 (Coordinates of O₁)

As we know, x= a + x’

-8 = 9+ x’

x’ = -8-9

x’ = -17

similarly,

y= b + y₁

3 = 5+ y₁

y₁ = 3-5

y₁ = -2

New coordinate, (x₁, y₁) are (-17,-2)

Given coordinate of P(x, y) = (10,2)

Coordinates of P with respect = (-3,-6)

toX₁’O₁X₁↔ Y₁’O₁Y₁(x’, y’)

to find, a & b

As we know, x = a + x’

10 = a - 3

a = 10+3

a = 13

similarly,

y = b + y’

2 = b - 6

b = 2+6

b = 8

So, Coordinates of O₁ = (13, 8)

⇒ Coordinates of O With respect to O₁ = (-13, -8)

Y = 3 - x

For different values of X and Y

Plotting above values in graph, we get

y = x - 3

For different values of X and Y

Plotting above values in graph, we get

y = 3x - 2

For different values of X and Y

Plotting above values in the graph, we get

y = 5 - 3x

For different values of X and Y

Plotting above values in the graph, we get

4y = −x + 3

For different values of X and Y

Plotting above values in graph, we get

3y = 4x + 1

For different values of X and Y

Plotting above values in graph, we get

x = 4

For different values of X and Y

Plotting above values in graph, we get

General equation for line is ay = bx +c

Here the line is, 0y=1x -4

Since ‘y’ is multiplied by zero every time, the value of ‘y’ does not affect the value of x or the equation.

⇒For every value of y, the value of x will be constant, i.e. 4.

3y =1.

For different values of X and Y

Plotting above values in the graph, we get

General equation for line is ay = bx +c

Here the line is, 3y=0x+ 1

Since ‘x’ is multiplied by zero every time, the value of ‘x’ does not affect the value of y or the equation.

⇒For every value of x, the value of y will be constant, i.e.

_{The given equation is}

Simplifying the equation,

For different values of X and Y

Plotting above values in the graph, we get

Do note that point G (0, ∞) and H (∞, 0) does not lie on the graph. As the Graph is Dis-Continuous, that means graph breaks at point G and H.

The above-mentioned graph is called HYPERBOLA, it has a general equation xy = c.

The general equation of the line is y = mx + c, where m and c are constants.

From the graph, it is visible that the line passes through the point (2, 0) and (0,-3).

Substituting both the values in general equation, y = mx + c, we get

0 = m(2) + c ..(i)

-3 = m(0) + c ..(ii)

From equation (ii), we get c = -3,

Putting c = -3 in equation (i),

2m + (-3) = 0

Putting both m and c in general equation y=mx+c,

So equation of line is 3x-2y = 6

The general equation of the line is y=mx+c, where m and c are constants.

From the graph, it is visible that the line passes through the point (5, 0) and (0,5).

Substituting both the values in general equation, y=mx + c, we get

0 = m(5) + c ..(i)

5 = m(0) + c ..(ii)

From equation (ii), we get c = 5,

Putting c = 5 in equation (i),

5m + (5) = 0

Putting both m and c in general equation y = mx + c,

So equation of line is x + y = 5

Given, Speed of boat = 22 km/h

Speed of stream = 8 km/h

Movement of boat downstream means that the boat is moving with the flow of the river and thus the total speed of boat will be the sum of the speed of boat and speed of the river. Thus,

Net speed = 22 + 8

= 30 km/h

⇒ Distance = Speed × time

Let distance be ‘d’ and time be ‘t’

⇒ d = (30)t

For different values of t and d

Plotting above values in the graph, we get

For equation 3y + 4x = 7

For different values of x and y

For equation 4y + 3x = 7

For different values of x and y

Plotting for both equations in same graph,

From graph it is visible that intersection point is P (1,1).

→ NOTE: You can verify intersection point by substituting it in both equations, and it should satisfy both equations.

Only when we put the point in equation y=0, it satisfies.

Since here the x coordinate is negative and y coordinate is positive, it implies that the point lies in the second quadrant.

Since, Line passes through origin(0,0), this means equation will be of form: y=mx.

And,

⇒ m = 5

⇒ Equation: y = 5x

If a point P has coordinates (x1,y1) in a coordinate system X ′OX ↔ Y′OY, and if O has coordinates (a,b) in another system X₁′O₁X₁↔ Y₁′O₁Y₁ with X ′OX || X ′O₁X₁, then the coordinates of P say (x, y) in the new system X₁′O₁X₁↔ Y₁′O₁Y₁ will be x = x1 + a and y = y1 + a

⇒ x = 3 + 4 and y = 4 + 3

⇒ x = 7 and y = 7

If the coordinates of a point P in a system X ′OX ↔ Y ′OY are (x,y). The coordinates of the same point in the system Y ′OY ↔ XOX′ will be (y,-x)

⇒ co-ordinates will be (8, -5).

In the third quadrant both x and y coordinates are negative.

Firstly, to reach (0,10) it will take 10 steps and afterward 12 steps will be taken to reach (10,12).

Putting x=3 in y=3x+4,

we get,

y=3 × 3 + 4

⇒ y = 13

Only when we put the points in equation y=x, it satisfies.

(i) Since, here both x and y coordinate is positive implies that the point (5, 10) lies in the first quadrant.

(ii) Since, here x coordinate is negative and y coordinate is positive implies that the point (-8, 9) lies in the second quadrant.

(iii) Since, here both x and y coordinate are negative implies that the point (-800, -3000) lies in the third quadrant.

(iv) Since, here x coordinate is positive and y coordinate isneaitive implies that the point (8, -100) lies in first quadrant.

(A)→ (iii)

⇒ On the x-axis, coordinates of a point are of the form (a,0).

And, (B)→ (i)

⇒ In the second quadrant, x coordinate is negative.

And, (C)→ (ii)

⇒ The line y = 3x + 4, cuts the y-axis at (0,4), as it satisfies the equation too.

(a) The y-coordinate of a point on the x-axis is zero.

Explanation:

On x – axis, the values of x keep on increasing and value of y remains zero as there is no movement in vertical direction.

(b) The x-coordinate is called as abscissa.

Explanation: The x coordinate of a coordinate system is called abscissa and y coordinate is called ordinate.

(c) The x-axis and Y-axis intersect at origin (0,0).

At origin values of x and y coordinates are 0 and thus they start increasing or decreasing in value from origin. Thus, they intersect at origin.

(d) If a point (x,y) ≠ (0,0) is in the third quadrant, then x + y has –(negative) sign.

Explanation: In the third quadrant, x and y both are negative, thus x – y = -x + (-y) = - (x + y)

And so its negative.

(e) If a point (x,y) lies above horizontal axis, then y is always +(positive).

(As, in first and second quadrant y is positive.)

(f) The point of intersection of x = y and x = −y is origin(0,0).

(As, after solving both the equations simultaneously by using substitution method, we get- x=0 and y=0).

(g) The line y =4x +5 intersects y-axis at the point (0, 5).

(Since, the point is at y-axis implies that x=0, and after substituting x=0 in equation, we get y=5.

(a) False

The equation of the x-axis is y =0.

(b) True

The y coordinate is 0 and x is constant, so value of y will remain same for this line, so it will be parallel to y axis.

(c) False

Since, x coordinate is 0 and y is constant implies that the line y=8 is parallel to x-axis.

(d) True

Since the multiplication of slope of both the equation is -1, implies that the lines are perpendicular to each other.

(e) True

Since, the one is parallel to horizontal axis and other one is parallel to the vertical axis implies that the both the lines are perpendicular as the axes are perpendicular too.

(f) False

Since, y will always be positive implies that the graph will be a V- shaped graph lying in first and second quadrant.

(g) False

Since, for y=0, and this the point where line intersect x-axis.

(h) True

Since, property of rectangle states that all the angles of a rectangle are 90°.

Let (x₁, y₁) = (0, -8) and (x₂, y₂)=(7, 0)

Now the equation of the line passing through (x₁, y₁) and (x₂, y₂) is given by-

⇒ 7y+56=8x

⇒ 7y=8x-56

(i) We can see in the graph that the graph passes through the points (3, 8) and (-3, -3)

Now, let (x1, y1) = (3, 8) and (x2, y2) = (-3,-3)

Now the equation of the line passing through (x1, y1) and (x2, y2) is given by-

⇒ 6y-48 = 11x-33

⇒ 6y = 11x + 15

(ii) We can see in the graph that the graph passes through the points (-4, 2) and (12, -3)

Now, let (x1, y1)=(-4, 2) and (x2, y2)=(12,-3)

Now the equation of the line passing through (x1,y1) and (x2,y2) is given by-

⇒ 16y – 32 = - 5x - 20

⇒ 16y = - 5x + 12

(If a point P has coordinates (x1,y1) in a coordinate system X ′OX ↔ Y′OY, and if O has coordinates (a,b) in another system X₁′O₁X₁↔ Y₁′O₁Y₁ with X ′OX || X ′O₁X₁, then the coordinates of P(say x,y) in the new system X₁′O₁X₁↔ Y₁′O₁Y₁ will be x = x1 + a and y = y1 + a)

Here, x = 7, y = 10, x1 = 10, and y1 = 7

⇒ a = x - x1

⇒ a = 7 - 10

⇒ a = -3

And, b = y - y1

⇒ b = 10 -7

⇒ b = 3

⇒ the coordinates of O₁ in the system X′OX ↔ Y′OY are (-3, 3).

In the graph the black- lined shaded region is the desired region. (whose left side is covered by the y-axis, downside by the x-axis and remaining by line x+2y=4).

In the above graph, the green line represents the equation 3y=4x-4 and the blue line represents equation 2x=3y+4 and we can see that they are intersecting at (0,-1.33) or

Also, when we’ll solve this two equation simultaneously we’ll get-

x=0 and .

Given, a*b=ab+a+b…….(1)

Now, y=3 * x + 1 * 2

⇒ y = 3x + 3 + x + 2 + 1 + 2 (Using (1))

⇒ y = (3x + x) + (3 + 2 + 1 + 2)

⇒ y = 4x + 8