In the previous section we saw how to calculate 'm', the slope of a line. In this section we will see it's practical use.
1. Consider two points A and B in a line in fig.34.23 below.
Their coordinates are given to us: (x1,y1) and (x2,y2)
2. Now, a point P' is marked on the x axis.
• This P' is at a distance of 'x' from the y axis.
• The distance 'x' is also given to us
3. In fig.b, a vertical green dashed line is drawn through P'
• This dashed line intersects the line at P
• We want to know 'how high P is'
• That is., we want to know the distance of P from the x axis
4. Let this distance of P from the x axis be y
• So we have:
♦ Distance of P from the y axis = x
♦ Distance of P from the x axis = y
• So coordinates of P are (x,y)
• Out of these, x is given to us
• We have to find y
5. From the known coordinates of A and B, we can calculate the slope 'm'
We have: m = (y2-y1)⁄(x2-x1).
6. We know that, if we know the coordinates of any two points on a line, we can calculate 'm'.
• In step (5) above, we used the known points A and B to find 'm'
• This 'm' will be a unique value for this particular line.
• That is., 'm' is a constant for this particular line
• So if we use A and P, we must get the same 'm'
7. Using A and P, the value of 'm' = (y-y1)⁄(x-x1)
8. But in step (5) we have already calculated 'm' as (y2-y1)⁄(x2-x1)
• So we can equate the two:
[(y-y1)⁄(x-x1)] = [(y2-y1)⁄(x2-x1)]
⟹ (y-y1) = [(y2-y1)⁄(x2-x1)] × (x-x1)
⟹ (y-y1) = [m] × (x-x1)
⟹ y = {[m] × (x-x1)}+ y1
⟹ y = mx - mx1 + y1
⟹ y = mx + (y1 - mx1)
9. The terms 'y1' and 'mx1' are put together because
♦ They are not associated with any of the variables 'x' or 'y
♦ In other words, they are 'constant terms'
♦ Their values can be readily calculated from the known coordinates of the two points A and B
• Since (y1 - mx1) is a constant, we will put it as 'c'. So the result in (8) becomes:
y = mx + c
10. Now, if we put the given value of x at P in the above equation, we will get the corresponding value of y
• Not only the value of x at P, we can put any 'x'. We will get the corresponding 'y'
■ So we can write:
The equation of the line is: y = mx + c
• Where
♦ m = (y2-y1)⁄(x2-x1)
♦ c = (y1 - mx1)
Let us see some examples:
Consider the line in fig.34.21 that we saw in the previous section. It has a lot of points. So we will use that line. It is shown again below:
1. Consider any two points, say C and E
• Then coordinates of C = (x1,y1) = (6,6.5)
• Coordinates of E = (x2,y2) = (10,9.5)
• 'm' of the line = (y2-y1)⁄(x2-x1) = (9.5-6.5)⁄(10-6) = 3⁄4 = 0.75
2. 'c' of the line = (y1-mx1) = [6.5 - (0.75 × 6)] = [6.5 - 4.5] = 2
3. So equation of the line is:
y = mx + c is:
y = 0.75x + 2
4. Once we get the 'equation of the line', we can find the 'y' value for any given 'x'
In our present example, we will check the other points:
■ Consider point A. It's x coordinate is 1
• Put this '1' in the equation. We get:
• y = (0.75 × 1) + 2 = 0.75 + 2 = 2.75
• This is indeed the 'y' coordinate of A in the fig
■ Consider point B. It's x coordinate is 4
• Put this '4' in the equation. We get:
• y = (0.75 × 4) + 2 = 3 + 2 = 5
• This is indeed the 'y' coordinate of B in the fig
■ Consider point D. It's x coordinate is 8
• Put this '8' in the equation. We get:
• y = (0.75 × 8) + 2 = 6 + 2 = 8
• This is indeed the 'y' coordinate of D in the fig
5. An interesting result:
Let us find the 'y' value when x = 0
• Put this '0' in the equation. We get:
• y = (0.75 × 0) + 2 = 0 + 2 = 2
• So when x = 0, y = 2
• Note that, at x = 0, the line cuts the y axis. So '2' is the y intercept of the line
• But this '2' is the value of the constant 'c' in the equation
■ In the equation of a line, the value of 'c' is the 'y intercept' that the line makes
• In the fig., we can see that the line cuts the y axis at exactly 2 units above the origin
• Another line passes through C(3,1) and D(-2,6)
■ We want the coordinates of the point of intersection of the two lines
Solution:
The two lines are shown in fig.34.24 below:
1. The equation of AB:
• 'm' of the line = (y2-y1)⁄(x2-x1) = (4-2)⁄(6-0) = 2⁄6 = 1⁄3
• 'c' of the line = (y1-mx1) = [2 - (1⁄3 × 0)] = [2 - 0] = 2
• So equation of the line is
y = mx + c:
y = (1⁄3)x + 2
2. The equation of CD:
• 'm' of the line = (y2-y1)⁄(x2-x1) = (6-1)⁄(-2-3) = 5⁄-5 = -1
• 'c' of the line = (y1-mx1) = [1 - (-1 × 3)] = [1 + 3] = 4
• So equation of the line is
y = mx + c:
y = - x + 4
4. Any point lying in AB will satisfy the equation in (1)
• Any point lying in CD will satisfy the equation in (2)
• The point of intersection lies in both the lines
• So the point of intersection must satisfy both (1) and (2)
5. Such a point can be found out by solving the two equations (Details here)
Equation (i): y = (1⁄3)x + 2
Equation (ii): y = - x + 4
• From (ii), we get a simple expression for 'y' with out any simplification. We will substitute it in (i)
We get: -x + 4 = (1⁄3)x + 2 `⟹ (1⁄3)x + x = 2 ⟹ (4⁄3)x = 2 ⟹ x = 6⁄4 = 3⁄2 = 11⁄2
• Substituting this value of x in (ii) we get:
y = -(3⁄2) + 4 = 5⁄2 = 21⁄2.
Since it is vertical, it does not have slope. So we cannot write its 'm'
Also it does not have 'y intercept'. So there is no 'c'
So how do we write the equation?
Let the vertical line meet the x axis at (k,0).
Then all points on that vertical line will have x coordinate as 'k'
So the equation of that line is: x = k
For the y axis, 'k' will be zero. So equation of the y axis is: x = 0
Let it meet the y axis at (0,l).
Then all points on that horizontal line will have y coordinate as 'l'
So the equation of that line is: y = l
For the x axis, 'l' will be zero. So equation of the x axis is: y = 0
■ To find the point of intersection of a line
y = mx + c
And a vertical line
x = k,
• We simply put x = k in y = mx + c
■ To find the point of intersection of a line
y = mx + c
And a horizontal line
y = l,
• We simply put y = l in y = mx + c.
Now we will see some solved examples
Solved example 34.12
Prove that the points A(3,1), B(6,4) and C(8,6) are on the same line
Solution:
1. We will write the equation of the line using A and B
• 'm' of the line = (y2-y1)⁄(x2-x1) = (4-1)⁄(6-3) = 3⁄3 = 1
• 'c' of the line = (y1-mx1) = [1 - (1 × 3)] = [1 - 3] = -2
• So equation of the line is
y = mx + c:
y = 1x - 2 ⟹ y = x - 2
• This is shown in fig.34.25(a) below
• We want to know whether C(8,6) also lies on this line
2. In fig.a, the point (8,0) is marked on the x axis
• A vertical dashed line is drawn through it
• There will be infinite number of points on this vertical line
♦ All of them will have x coordinate as 8
♦ C(8,6) is also one among those infinite points
• We want to know whether this C(8,6) lies on AB also. If it does, we have a favourable situation:
■ C(8,6) lies on both the lines. That is., C(8,6) is the point of intersection of the two lines. This is shown in fig.b
3. There will be only one point of intersection between any two lines.
• To find it, we need to solve the two equations:
(i) y = x-2
(ii) x = 8
Substituting x = 8 in (i), we get:
y = 8-2 = 6
• So the point of intersection is (8,6)
• Thus, A, B and C are indeed on the same line
In the next section, we will see a few more solved examples.
1. Consider two points A and B in a line in fig.34.23 below.
Their coordinates are given to us: (x1,y1) and (x2,y2)
 |
| Fig.34.23 |
• This P' is at a distance of 'x' from the y axis.
• The distance 'x' is also given to us
3. In fig.b, a vertical green dashed line is drawn through P'
• This dashed line intersects the line at P
• We want to know 'how high P is'
• That is., we want to know the distance of P from the x axis
4. Let this distance of P from the x axis be y
• So we have:
♦ Distance of P from the y axis = x
♦ Distance of P from the x axis = y
• So coordinates of P are (x,y)
• Out of these, x is given to us
• We have to find y
5. From the known coordinates of A and B, we can calculate the slope 'm'
We have: m = (y2-y1)⁄(x2-x1).
6. We know that, if we know the coordinates of any two points on a line, we can calculate 'm'.
• In step (5) above, we used the known points A and B to find 'm'
• This 'm' will be a unique value for this particular line.
• That is., 'm' is a constant for this particular line
• So if we use A and P, we must get the same 'm'
7. Using A and P, the value of 'm' = (y-y1)⁄(x-x1)
8. But in step (5) we have already calculated 'm' as (y2-y1)⁄(x2-x1)
• So we can equate the two:
[(y-y1)⁄(x-x1)] = [(y2-y1)⁄(x2-x1)]
⟹ (y-y1) = [(y2-y1)⁄(x2-x1)] × (x-x1)
⟹ (y-y1) = [m] × (x-x1)
⟹ y = {[m] × (x-x1)}+ y1
⟹ y = mx - mx1 + y1
⟹ y = mx + (y1 - mx1)
9. The terms 'y1' and 'mx1' are put together because
♦ They are not associated with any of the variables 'x' or 'y
♦ In other words, they are 'constant terms'
♦ Their values can be readily calculated from the known coordinates of the two points A and B
• Since (y1 - mx1) is a constant, we will put it as 'c'. So the result in (8) becomes:
y = mx + c
10. Now, if we put the given value of x at P in the above equation, we will get the corresponding value of y
• Not only the value of x at P, we can put any 'x'. We will get the corresponding 'y'
■ So we can write:
The equation of the line is: y = mx + c
• Where
♦ m = (y2-y1)⁄(x2-x1)
♦ c = (y1 - mx1)
Let us see some examples:
Consider the line in fig.34.21 that we saw in the previous section. It has a lot of points. So we will use that line. It is shown again below:
 |
| Fig.34.21 |
• Then coordinates of C = (x1,y1) = (6,6.5)
• Coordinates of E = (x2,y2) = (10,9.5)
• 'm' of the line = (y2-y1)⁄(x2-x1) = (9.5-6.5)⁄(10-6) = 3⁄4 = 0.75
2. 'c' of the line = (y1-mx1) = [6.5 - (0.75 × 6)] = [6.5 - 4.5] = 2
3. So equation of the line is:
y = mx + c is:
y = 0.75x + 2
4. Once we get the 'equation of the line', we can find the 'y' value for any given 'x'
In our present example, we will check the other points:
■ Consider point A. It's x coordinate is 1
• Put this '1' in the equation. We get:
• y = (0.75 × 1) + 2 = 0.75 + 2 = 2.75
• This is indeed the 'y' coordinate of A in the fig
■ Consider point B. It's x coordinate is 4
• Put this '4' in the equation. We get:
• y = (0.75 × 4) + 2 = 3 + 2 = 5
• This is indeed the 'y' coordinate of B in the fig
■ Consider point D. It's x coordinate is 8
• Put this '8' in the equation. We get:
• y = (0.75 × 8) + 2 = 6 + 2 = 8
• This is indeed the 'y' coordinate of D in the fig
5. An interesting result:
Let us find the 'y' value when x = 0
• Put this '0' in the equation. We get:
• y = (0.75 × 0) + 2 = 0 + 2 = 2
• So when x = 0, y = 2
• Note that, at x = 0, the line cuts the y axis. So '2' is the y intercept of the line
• But this '2' is the value of the constant 'c' in the equation
■ In the equation of a line, the value of 'c' is the 'y intercept' that the line makes
• In the fig., we can see that the line cuts the y axis at exactly 2 units above the origin
Point of intersection of two lines
• A line passes through A(0,2) and B(6,4)• Another line passes through C(3,1) and D(-2,6)
■ We want the coordinates of the point of intersection of the two lines
Solution:
The two lines are shown in fig.34.24 below:
 |
| Fig.34.24 |
• 'm' of the line = (y2-y1)⁄(x2-x1) = (4-2)⁄(6-0) = 2⁄6 = 1⁄3
• 'c' of the line = (y1-mx1) = [2 - (1⁄3 × 0)] = [2 - 0] = 2
• So equation of the line is
y = mx + c:
y = (1⁄3)x + 2
2. The equation of CD:
• 'm' of the line = (y2-y1)⁄(x2-x1) = (6-1)⁄(-2-3) = 5⁄-5 = -1
• 'c' of the line = (y1-mx1) = [1 - (-1 × 3)] = [1 + 3] = 4
• So equation of the line is
y = mx + c:
y = - x + 4
4. Any point lying in AB will satisfy the equation in (1)
• Any point lying in CD will satisfy the equation in (2)
• The point of intersection lies in both the lines
• So the point of intersection must satisfy both (1) and (2)
5. Such a point can be found out by solving the two equations (Details here)
Equation (i): y = (1⁄3)x + 2
Equation (ii): y = - x + 4
• From (ii), we get a simple expression for 'y' with out any simplification. We will substitute it in (i)
We get: -x + 4 = (1⁄3)x + 2 `⟹ (1⁄3)x + x = 2 ⟹ (4⁄3)x = 2 ⟹ x = 6⁄4 = 3⁄2 = 11⁄2
• Substituting this value of x in (ii) we get:
y = -(3⁄2) + 4 = 5⁄2 = 21⁄2.
Equation of a vertical line
Consider the vertical line shown in fig.34.25(a) belowSince it is vertical, it does not have slope. So we cannot write its 'm'
Also it does not have 'y intercept'. So there is no 'c'
So how do we write the equation?
Let the vertical line meet the x axis at (k,0).
Then all points on that vertical line will have x coordinate as 'k'
So the equation of that line is: x = k
For the y axis, 'k' will be zero. So equation of the y axis is: x = 0
Equation of a horizontal line
Consider the horizontal line shown in fig.34.25(b) aboveLet it meet the y axis at (0,l).
Then all points on that horizontal line will have y coordinate as 'l'
So the equation of that line is: y = l
For the x axis, 'l' will be zero. So equation of the x axis is: y = 0
■ To find the point of intersection of a line
y = mx + c
And a vertical line
x = k,
• We simply put x = k in y = mx + c
■ To find the point of intersection of a line
y = mx + c
And a horizontal line
y = l,
• We simply put y = l in y = mx + c.
Now we will see some solved examples
Solved example 34.12
Prove that the points A(3,1), B(6,4) and C(8,6) are on the same line
Solution:
1. We will write the equation of the line using A and B
• 'm' of the line = (y2-y1)⁄(x2-x1) = (4-1)⁄(6-3) = 3⁄3 = 1
• 'c' of the line = (y1-mx1) = [1 - (1 × 3)] = [1 - 3] = -2
• So equation of the line is
y = mx + c:
y = 1x - 2 ⟹ y = x - 2
• This is shown in fig.34.25(a) below
 |
| Fig.34.25 |
2. In fig.a, the point (8,0) is marked on the x axis
• A vertical dashed line is drawn through it
• There will be infinite number of points on this vertical line
♦ All of them will have x coordinate as 8
♦ C(8,6) is also one among those infinite points
• We want to know whether this C(8,6) lies on AB also. If it does, we have a favourable situation:
■ C(8,6) lies on both the lines. That is., C(8,6) is the point of intersection of the two lines. This is shown in fig.b
3. There will be only one point of intersection between any two lines.
• To find it, we need to solve the two equations:
(i) y = x-2
(ii) x = 8
Substituting x = 8 in (i), we get:
y = 8-2 = 6
• So the point of intersection is (8,6)
• Thus, A, B and C are indeed on the same line
No comments:
Post a Comment