In the previous section we saw the equation of line. We also saw a solved example. In this section we will see a few more solved examples. Later in this section we will see the slopes of any two lines which are perpendicular to each other.
Solved example 34.13
Find the coordinates of two other points on the line joining (-1,4) and (1,2)
Solution:
1. Fig.34.26 below shows a rough sketch of the line joining A(-1,4) and (1,20)
• Any line can be extended towards the top or bottom indefinitely.
♦ So there will be infinite number of points in any line.
• In our present case, we want any two points other than the given ones
2. A point (x3,0) is marked on the x axis, and a vertical dashed line is drawn through it
• This vertical line intersects the given line at C. Then the x coordinate of C will obviously be x3
• Note that, it is only a rough sketch. The direction of the actual line may be different.
• But what ever be the direction of a line (other than vertical), another vertical line drawn perpendicular to the x axis will surely intersect it at a definite point.
• We are trying to find the coordinates of that point of intersection, which we named as C
3. From the given points A and B, we can find the slope m of the line:
We have: m = (y2-y1)⁄(x2-x1) = (2-4)⁄(1-(-1)) = -2⁄2 = -1
4. Now, if we use C and B, we must get the same slope. So we can write:
m = -1 = (y2-y1)⁄(x2-x1) = (y3-2)⁄(x3-1)
⟹ -1(x3-1) = (y3-2) ⟹ (-x3+1) = (y3-2) ⟹ y3 = 3-x3
5. Thus, if we have a value for x3, we can simply subtract it from 3 to get the corresponding y3
• But we can put any value for x3. Because, what ever be the position of x3, a vertical line drawn through it will surely meet the line through A and B
• So let us put x3 = 2
Then from (4) we get: y3 = (3-2) = 1
• So (2,1) is a point on the line through A and B
6. Like this, we will find one more point:
• If we use D and B, we must get the same slope. So we can write:
m = -1 = (y2-y1)⁄(x2-x1) = (y4-2)⁄(x4-1)
⟹ -1(x4-1) = (y4-2) ⟹ (-x4+1) = (y4-2) ⟹ y3 = 3-x4
• Thus, if we have a value for x4, we can simply subtract it from 3 to get the corresponding y4
• As mentioned above, we can put any value for x4. This time we will put x4 = 5
• Then y3 = (3-5) = -2
• So (5,-2) is a point on the line through A and B
7. Fig.34.26(b) above shows the actual positions of the points
Solved example 34.14
x1, x2, x3, . . . and y1, y2, y3, . . . are arithmetic sequences. Prove that all points with coordinates in the sequence (x1,y1), (x2,y2), (x2,y3), . . . are on the same line
Solution:
1. Given that x1, x2, x3, . . . is an arithmetic sequence
• Let d1 be the common difference of this arithmetic sequence
• Then we can write:
♦ (x2-x1) = d1
♦ (x3-x2) = d1
2. Given that y1, y2, y3, . . . is an arithmetic sequence
• Let d2 be the common difference of this arithmetic sequence
• Then we can write:
♦ (y2-y1) = d2
♦ (y3-y2) = d2
3. We are given one more sequence:
(x1,y1), (x2,y2), (x2,y3), . . .
• But this is not an arithmetic sequence. It is just a sequence of coordinates.
• We have to prove that all members of this sequence lie on a line
4. Let us take the first point and the second point: (x1,y1) and (x2,y2)
They will surely lie on a line. The slope of this line is:
m = (y2-y1)⁄(x2-x1) = d2⁄d1. [∵ (y2-y1) = d2 and (x2-x1) = d1]
5. Let us take the second point and the third point: (x2,y2) and (x3,y3)
They will surely lie on a line. The slope of this line is:
m = (y3-y2)⁄(x3-x2) = d2⁄d1. [∵ (y3-y2) = d2 and (x3-x2) = d1]
6. But this slope obtained in (5) is the same one that we obtained in (4)
• Let us analyse this situation:
(i) Line through points 1 and 2 has a slope m
(ii) Line through points 2 and 3 has the same slope m
(iii) So they are parallel
(iv) But point 2 is common in both lines.
(v) If they are parallel and have one point in common, obviously, the three points lie in one line
7. Let us take the third point and the fourth point: (x3,y3) and (x4,y4)
They will surely lie on a line. The slope of this line is:
m = (y4-y3)⁄(x4-x3) = d2⁄d1. [∵ (y4-y3) = d2 and (x4-x3) = d1]
8. But this slope obtained in (7) is the same one that we obtained in (5)
• Let us analyse this situation:
(i) Line through points 2 and 3 has a slope m
(ii) Line through points 3 and 4 has the same slope m
(iii) So they are parallel
(iv) But point 3 is common in both lines.
(v) If they are parallel and have one point in common, obviously, the three points lie in one line
9. So points 2, 3 and 4 lie in a line.
• But we have proved in (6) that 1, 2 and 3 lie in a line
• In the above two, points 2 and 3 are common. So points 1, 2, 3 and 4 lie in a line
• Continuing like this, we will be able to prove that all points in the sequence (x1,y1), (x2,y2), (x2,y3), . . . lie in a line
Solved example 34.15
Prove that for all points on the line passing through the origin and another point A(4,2), the x coordinate will be double the y coordinate.
Solution:
1. The line passes through two points:
O(0,0) and A(4,2)
• So slope of the line =
m = (y2-y1)⁄(x2-x1) = (2-0)⁄(4-0) = 2⁄4 = 1⁄2
• 'c' of the line = (y1-mx1) = [0 - (1⁄2 × 0)] = [0 - 0] = 0
• So equation of the line is
y = mx + c:
y = 1⁄2 × x + 0 ⟹ y = x⁄2 ⟹ x = 2y
2. So what ever value we put for 'y', the value of 'x' will be double that.
• Let the slopes of AB and CD be m1 and m2 respectively
2. Let the two lines meet at P(x,y).
• A vertical green dashed line is drawn through P
• It meets the x axis at P'
• Obviously, the coordinates of P' will be (x,0)
• This is shown in fig.34.27(a) below
3. Next, we want a point on AB
• P is already a point on AB. We want another point
• How can we obtain it?
4. On the x axis, mark a point 5 units to the left of P'
• Let it be called Q'. It's coordinates will be (x-5,0)
• This is shown in fig.b
• Note that '5 units' is taken arbitrarily. we can take any convenient units towards the left
5. Draw a vertical green dashed line through this Q'
• Let it meet AB at Q. Then x coordinate of Q will be x-5
• Let the y coordinate be y1
• So the coordinates of Q can be written as: (x-5,y1)
• We have to find the value of this y1
6. The slope of line AB will help us to find y1
• The slope of AB is given as m1
• Using P and Q, we can write:
m1 = (y2-y1)⁄(x2-x1) = (y1-y)⁄(x-5-x) = (y1-y)⁄(-5) = (y-y1)⁄5
⟹ 5m1 = y-y1 ⟹ y1 = y-5m1
7. Let the vertical green dashed line through Q' meet CD at R. Then x coordinate of R will be x-5
• Let the y coordinate be y2
• So the coordinates of R can be written as: (x-5,y2)
• We have to find the value of this y2
8. The slope of line CD will help us to find y2
• The slope of CD is given as m2
• Using P and R, we can write:
m2 = (y2-y1)⁄(x2-x1) = (y2-y)⁄(x-5-x) = (y2-y)⁄(-5) = (y-y2)⁄5
⟹ 5m2 = y-y2 ⟹ y2 = y-5m2
9. Thus we get the coordinates of all the three vertices of the ΔPQR:
P(x,y), Q[(x-5),(y-5m1)], R[(x-5),(y-5m2)]
10. Now we can calculate the length of sides of the ΔPQR using the distance formula:
• PQ2 = {[x2-x1]2 + [y2-y1]2} = {[(x-5)-x]2 + [(y-5m1)-y]2} = {[-5]2 + [-5m1]2}
= {25 + 25(m1)2} = 25[1+(m1)2]
• PR2 = {[x2-x1]2 + [y2-y1]2} = {[(x-5)-x]2 + [(y-5m2)-y]2} = {[-5]2 + [-5m2]2}
= {25 + 25(m2)2} = 25[1+(m2)2]
• QR2 = {[x2-x1]2 + [y2-y1]2} = {[(x-5)-(x-5)]2 + [(y-5m2)-(y-5m1)]2}
= {[0]2 + [y-5m2-y+5m1)]2} = {[0]2 + [5m1-5m2]2} = { [5(m1-m2)]2} = 25[m1-m2]2.
11. But PQR is a right triangle. Applying pythagoras theorem, we get:
QR2 = PR2 + PQ2.
• Substituting the values, we get:
25[m1-m2]2 = {25[1+(m1)2] + 25[1+(m2)2]} = 25{[1+(m1)2] + [1+(m2)2]}
⟹ [m1-m2]2 = {[1+(m1)2] + [1+(m2)2]}
⟹ (m1)2 - 2 × m1 × m2 + (m2)2 = 1 + (m1)2 + 1 + (m2)2
⟹ -2 × m1 × m2 = 2
⟹ -1 × m1 × m2 = 1
⟹ m1 × m2 = -1
⟹ m1 = -1⁄m2 OR m2 = -1⁄m1.
Thus we can write:
■ If two lines are perpendicular to each other, the slope of one will be the negative reciprocal of the other
An example:
Prove that the line passing through the points (5,6) and (1,-1) is perpendicular to the line passing through the points (-5,1) and (2,-3)
Solution:
1. Slope of the line passing through the points (5,6) and (1,-1):
m1 = (y2-y1)⁄(x2-x1) = (-1-6)⁄(1-5) = (-7)⁄(-4) = 7⁄4
2. Slope of the line passing through the points (-5,1) and (2,-3):
m2 = (y2-y1)⁄(x2-x1) = (-3-1)⁄(2-(-5)) = -4⁄7
3. Negative reciprocal of m1 = -1 × (reciprocal of m1) = -1×(reciprocal of 7⁄4) = -1 × (4⁄7) = -4⁄7
4. But from (2), we have: -4⁄7 = m2
• Negative reciprocal of m1 = m2
• So the lines are perpendicular to each other
Another example:
A line passes through A(4,3) and B(8,4). A perpendicular line passes through B. Write the coordinates of any one point (other than B) on the perpendicular line
Solution:
1. Slope of the line passing through A and B:
m1 = (y2-y1)⁄(x2-x1) = (4-3)⁄(8-4) = 1⁄4
2. Negative reciprocal of m1 = -1 × (reciprocal of m1) = -1×(reciprocal of 1⁄4) = -1 × (4⁄1) = -4
• So slope of the perpendicular line = m2 = -4
3. On the x axis, mark a point 2 units to the left of B'. See rough sketch in fig.34.28 below:
• Let it be called P'. It's coordinates will be (6,0)
• Note that '2 units' is taken arbitrarily. we can take any convenient units towards the left
4. Draw a vertical green dashed line through this P'
• Let it meet the perpendicular line at P. Then x coordinate of P will be 6
• Let the y coordinate be y1
• So the coordinates of P can be written as: (6,y1)
• We have to find the value of this y1
5. The slope of the perpendicular line PB will help us to find y1
• The slope of PB is obtained as m2 = -4
• Using P and B, we can write:
m2 = -4 = (y2-y1)⁄(x2-x1) = (4-y1)⁄(8-6) = (4-y1)⁄(2) = (4-y1)⁄2
⟹ -8 = 4-y1 ⟹ y1 = 12
• So the coordinates of P are (6,12)
In the next section, we will see circles.
Solved example 34.13
Find the coordinates of two other points on the line joining (-1,4) and (1,2)
Solution:
1. Fig.34.26 below shows a rough sketch of the line joining A(-1,4) and (1,20)
 |
| Fig.34.26 |
♦ So there will be infinite number of points in any line.
• In our present case, we want any two points other than the given ones
2. A point (x3,0) is marked on the x axis, and a vertical dashed line is drawn through it
• This vertical line intersects the given line at C. Then the x coordinate of C will obviously be x3
• Note that, it is only a rough sketch. The direction of the actual line may be different.
• But what ever be the direction of a line (other than vertical), another vertical line drawn perpendicular to the x axis will surely intersect it at a definite point.
• We are trying to find the coordinates of that point of intersection, which we named as C
3. From the given points A and B, we can find the slope m of the line:
We have: m = (y2-y1)⁄(x2-x1) = (2-4)⁄(1-(-1)) = -2⁄2 = -1
4. Now, if we use C and B, we must get the same slope. So we can write:
m = -1 = (y2-y1)⁄(x2-x1) = (y3-2)⁄(x3-1)
⟹ -1(x3-1) = (y3-2) ⟹ (-x3+1) = (y3-2) ⟹ y3 = 3-x3
5. Thus, if we have a value for x3, we can simply subtract it from 3 to get the corresponding y3
• But we can put any value for x3. Because, what ever be the position of x3, a vertical line drawn through it will surely meet the line through A and B
• So let us put x3 = 2
Then from (4) we get: y3 = (3-2) = 1
• So (2,1) is a point on the line through A and B
6. Like this, we will find one more point:
• If we use D and B, we must get the same slope. So we can write:
m = -1 = (y2-y1)⁄(x2-x1) = (y4-2)⁄(x4-1)
⟹ -1(x4-1) = (y4-2) ⟹ (-x4+1) = (y4-2) ⟹ y3 = 3-x4
• Thus, if we have a value for x4, we can simply subtract it from 3 to get the corresponding y4
• As mentioned above, we can put any value for x4. This time we will put x4 = 5
• Then y3 = (3-5) = -2
• So (5,-2) is a point on the line through A and B
7. Fig.34.26(b) above shows the actual positions of the points
Solved example 34.14
x1, x2, x3, . . . and y1, y2, y3, . . . are arithmetic sequences. Prove that all points with coordinates in the sequence (x1,y1), (x2,y2), (x2,y3), . . . are on the same line
Solution:
1. Given that x1, x2, x3, . . . is an arithmetic sequence
• Let d1 be the common difference of this arithmetic sequence
• Then we can write:
♦ (x2-x1) = d1
♦ (x3-x2) = d1
2. Given that y1, y2, y3, . . . is an arithmetic sequence
• Let d2 be the common difference of this arithmetic sequence
• Then we can write:
♦ (y2-y1) = d2
♦ (y3-y2) = d2
3. We are given one more sequence:
(x1,y1), (x2,y2), (x2,y3), . . .
• But this is not an arithmetic sequence. It is just a sequence of coordinates.
• We have to prove that all members of this sequence lie on a line
4. Let us take the first point and the second point: (x1,y1) and (x2,y2)
They will surely lie on a line. The slope of this line is:
m = (y2-y1)⁄(x2-x1) = d2⁄d1. [∵ (y2-y1) = d2 and (x2-x1) = d1]
5. Let us take the second point and the third point: (x2,y2) and (x3,y3)
They will surely lie on a line. The slope of this line is:
m = (y3-y2)⁄(x3-x2) = d2⁄d1. [∵ (y3-y2) = d2 and (x3-x2) = d1]
6. But this slope obtained in (5) is the same one that we obtained in (4)
• Let us analyse this situation:
(i) Line through points 1 and 2 has a slope m
(ii) Line through points 2 and 3 has the same slope m
(iii) So they are parallel
(iv) But point 2 is common in both lines.
(v) If they are parallel and have one point in common, obviously, the three points lie in one line
7. Let us take the third point and the fourth point: (x3,y3) and (x4,y4)
They will surely lie on a line. The slope of this line is:
m = (y4-y3)⁄(x4-x3) = d2⁄d1. [∵ (y4-y3) = d2 and (x4-x3) = d1]
8. But this slope obtained in (7) is the same one that we obtained in (5)
• Let us analyse this situation:
(i) Line through points 2 and 3 has a slope m
(ii) Line through points 3 and 4 has the same slope m
(iii) So they are parallel
(iv) But point 3 is common in both lines.
(v) If they are parallel and have one point in common, obviously, the three points lie in one line
9. So points 2, 3 and 4 lie in a line.
• But we have proved in (6) that 1, 2 and 3 lie in a line
• In the above two, points 2 and 3 are common. So points 1, 2, 3 and 4 lie in a line
• Continuing like this, we will be able to prove that all points in the sequence (x1,y1), (x2,y2), (x2,y3), . . . lie in a line
Solved example 34.15
Prove that for all points on the line passing through the origin and another point A(4,2), the x coordinate will be double the y coordinate.
Solution:
1. The line passes through two points:
O(0,0) and A(4,2)
• So slope of the line =
m = (y2-y1)⁄(x2-x1) = (2-0)⁄(4-0) = 2⁄4 = 1⁄2
• 'c' of the line = (y1-mx1) = [0 - (1⁄2 × 0)] = [0 - 0] = 0
• So equation of the line is
y = mx + c:
y = 1⁄2 × x + 0 ⟹ y = x⁄2 ⟹ x = 2y
2. So what ever value we put for 'y', the value of 'x' will be double that.
Slopes of Perpendicular lines
1. Consider any two lines AB and CD which are perpendicular to each other• Let the slopes of AB and CD be m1 and m2 respectively
2. Let the two lines meet at P(x,y).
• A vertical green dashed line is drawn through P
• It meets the x axis at P'
• Obviously, the coordinates of P' will be (x,0)
• This is shown in fig.34.27(a) below
 |
| Fig.34.27 |
• P is already a point on AB. We want another point
• How can we obtain it?
4. On the x axis, mark a point 5 units to the left of P'
• Let it be called Q'. It's coordinates will be (x-5,0)
• This is shown in fig.b
• Note that '5 units' is taken arbitrarily. we can take any convenient units towards the left
5. Draw a vertical green dashed line through this Q'
• Let it meet AB at Q. Then x coordinate of Q will be x-5
• Let the y coordinate be y1
• So the coordinates of Q can be written as: (x-5,y1)
• We have to find the value of this y1
6. The slope of line AB will help us to find y1
• The slope of AB is given as m1
• Using P and Q, we can write:
m1 = (y2-y1)⁄(x2-x1) = (y1-y)⁄(x-5-x) = (y1-y)⁄(-5) = (y-y1)⁄5
⟹ 5m1 = y-y1 ⟹ y1 = y-5m1
7. Let the vertical green dashed line through Q' meet CD at R. Then x coordinate of R will be x-5
• Let the y coordinate be y2
• So the coordinates of R can be written as: (x-5,y2)
• We have to find the value of this y2
8. The slope of line CD will help us to find y2
• The slope of CD is given as m2
• Using P and R, we can write:
m2 = (y2-y1)⁄(x2-x1) = (y2-y)⁄(x-5-x) = (y2-y)⁄(-5) = (y-y2)⁄5
⟹ 5m2 = y-y2 ⟹ y2 = y-5m2
9. Thus we get the coordinates of all the three vertices of the ΔPQR:
P(x,y), Q[(x-5),(y-5m1)], R[(x-5),(y-5m2)]
10. Now we can calculate the length of sides of the ΔPQR using the distance formula:
• PQ2 = {[x2-x1]2 + [y2-y1]2} = {[(x-5)-x]2 + [(y-5m1)-y]2} = {[-5]2 + [-5m1]2}
= {25 + 25(m1)2} = 25[1+(m1)2]
• PR2 = {[x2-x1]2 + [y2-y1]2} = {[(x-5)-x]2 + [(y-5m2)-y]2} = {[-5]2 + [-5m2]2}
= {25 + 25(m2)2} = 25[1+(m2)2]
• QR2 = {[x2-x1]2 + [y2-y1]2} = {[(x-5)-(x-5)]2 + [(y-5m2)-(y-5m1)]2}
= {[0]2 + [y-5m2-y+5m1)]2} = {[0]2 + [5m1-5m2]2} = { [5(m1-m2)]2} = 25[m1-m2]2.
11. But PQR is a right triangle. Applying pythagoras theorem, we get:
QR2 = PR2 + PQ2.
• Substituting the values, we get:
25[m1-m2]2 = {25[1+(m1)2] + 25[1+(m2)2]} = 25{[1+(m1)2] + [1+(m2)2]}
⟹ [m1-m2]2 = {[1+(m1)2] + [1+(m2)2]}
⟹ (m1)2 - 2 × m1 × m2 + (m2)2 = 1 + (m1)2 + 1 + (m2)2
⟹ -2 × m1 × m2 = 2
⟹ -1 × m1 × m2 = 1
⟹ m1 × m2 = -1
⟹ m1 = -1⁄m2 OR m2 = -1⁄m1.
Thus we can write:
■ If two lines are perpendicular to each other, the slope of one will be the negative reciprocal of the other
An example:
Prove that the line passing through the points (5,6) and (1,-1) is perpendicular to the line passing through the points (-5,1) and (2,-3)
Solution:
1. Slope of the line passing through the points (5,6) and (1,-1):
m1 = (y2-y1)⁄(x2-x1) = (-1-6)⁄(1-5) = (-7)⁄(-4) = 7⁄4
2. Slope of the line passing through the points (-5,1) and (2,-3):
m2 = (y2-y1)⁄(x2-x1) = (-3-1)⁄(2-(-5)) = -4⁄7
3. Negative reciprocal of m1 = -1 × (reciprocal of m1) = -1×(reciprocal of 7⁄4) = -1 × (4⁄7) = -4⁄7
4. But from (2), we have: -4⁄7 = m2
• Negative reciprocal of m1 = m2
• So the lines are perpendicular to each other
Another example:
A line passes through A(4,3) and B(8,4). A perpendicular line passes through B. Write the coordinates of any one point (other than B) on the perpendicular line
Solution:
1. Slope of the line passing through A and B:
m1 = (y2-y1)⁄(x2-x1) = (4-3)⁄(8-4) = 1⁄4
2. Negative reciprocal of m1 = -1 × (reciprocal of m1) = -1×(reciprocal of 1⁄4) = -1 × (4⁄1) = -4
• So slope of the perpendicular line = m2 = -4
3. On the x axis, mark a point 2 units to the left of B'. See rough sketch in fig.34.28 below:
 |
| Fig.34.28 |
• Note that '2 units' is taken arbitrarily. we can take any convenient units towards the left
4. Draw a vertical green dashed line through this P'
• Let it meet the perpendicular line at P. Then x coordinate of P will be 6
• Let the y coordinate be y1
• So the coordinates of P can be written as: (6,y1)
• We have to find the value of this y1
5. The slope of the perpendicular line PB will help us to find y1
• The slope of PB is obtained as m2 = -4
• Using P and B, we can write:
m2 = -4 = (y2-y1)⁄(x2-x1) = (4-y1)⁄(8-6) = (4-y1)⁄(2) = (4-y1)⁄2
⟹ -8 = 4-y1 ⟹ y1 = 12
• So the coordinates of P are (6,12)
