In the previous section we completed a discussion on surface area and volume of some solids. In this chapter we will see Geometry and Algebra.
• This chapter is in fact a continuation of chapter 31 - Coordinate Geometry.
• The last section of that chapter can be seen here. In that chapter 31, we saw an interesting case:
1. We were given the coordinates of two points
• Those were the end points of a diagonal of a rectangle or a square
2. We were asked to find the coordinates of the other two corners of the rectangle
• We solved such problems very easily. (Details here)
• Now we will see a similar case.
♦ This time, instead of rectangles or squares, we look at parallelograms.
• Coordinates of any three of the four vertices of a parallelogram will be given.
• We will be asked to find the coordinates of the fourth vertex.
Let us see how it is done:
1. Consider the parallelogram OABC in fig.34.1(a) below
• Coordinates of three vertices O, A and C are known. We have to find the coordinates of the vertex B
2. In fig.b, green lines are drawn.
• These green lines have a special property: They are all parallel to the axes
♦ The horizontal green lines are parallel to the x axis
♦ The vertical green lines are parallel to the y axis
3. Consider the two green lines attached to the side CB
• They intersect at B'.
• We know that the angle between the axes will always be 90o
• Since the green lines are parallel to the axes, the angle at B' will also be 90o
• So triangle BB'C is a right triangle
4. In the same way, the bottom triangle AA'O is also a right triangle
5. Now we will see the relation between the two right triangles:
• OA and CB are the opposite sides of a parallelogram.
♦ So OA and CB are parallel to each other
♦ and they will be inclined at the same angle with the x axis
• CB' and OA' are both parallel to the x axis. So we get: ∠BCB' = ∠AOA'
• These two equal angles are denoted as α in fig.c
6. Again, OA and CB are parallel to each other and they will be inclined at the same angle with the y axis also
• BB' and AA' are both parallel to the y axis. So we get: ∠CBB' = ∠OAA'
• These two equal angles are denoted as β in fig.c
7. We have a situation:
(i) Angles at the ends of the hypotenuse BC are α and β
(ii) Angles at the ends of the hypotenuse OA are also α and β
(iii) The two triangles are right angled.
(iv) The two hypotenuses, being opposite sides of a parallelogram, have the same lengths
■ In this situation, we can apply the RHS criterion. The two triangles are equal. (Details here)
Let us write the correspondence:
• The 90o angle is at A' and B'. So A'↔B'
• The αo angle is at O and C. So O↔C
• The βo angle is at A and B. So A↔B
• So the correspondence is A'↔B', O↔C and A↔B. This is same as A'OA↔B'CB
• So we can write:
The two triangles are congruent to one another under the correspondence: A'OA↔B'CB
• This is same as ΔA'OA ≅ ΔB'CB
Now we can write the corresponding sides. We get:
♦ A'O↔B'C
♦ OA↔CB
♦ A'A↔B'B
8. Among the above three correspondences, the middle one is already known to us.
• Because, they are the opposite sides of a parallelogram.
• The first and the last are of great use to us.
■ The first one indicates this:
The bases of the two triangles are of equal length
■ The last one indicates this:
The altitudes of the two triangles are of equal length
9. Now consider fig.34.2(a) below:
• AA' is parallel to the y axis. So coordinates of A' are (6,0)
• Thus length of OA' is 6 units. This is shown in fig.34.2(b)
10. We have seen that the bases are equal.
• So CB' must also be 6 units in length
• Then coordinates of B' will be (3+6, 5). That is., (9,5)
11. Now consider the coordinates of A. They are (6,2)
• So AA' will be 2 units in length
12. We have seen that the altitudes are equal
• So BB' will also be 2 units in length
13. We already wrote the coordinates of B' as (9,5).
• So the coordinates of B will be (9,5+2). That is., (9,7)
• This is shown in fig.c. Thus we determined the unknown coordinates at B. The problem is solved.
• We were able to solve it because, the two bases as well as the two altitudes are equal. We proved it using the RHS criterion.
• There is an easier method to prove those equalities. In that method we use trigonometric ratios. We have already seen the basics of trigonometry here.
1. In fig.34.2(a) above, consider the right triangle BCB'. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = BB'⁄CB.
cos α = adjacent side⁄hypotenuse = B'C⁄CB.
2. Again in fig.34.2(a) above, consider the right triangle AOA'. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = AA'⁄OA.
cos α = adjacent side⁄hypotenuse = A'O⁄OA.
3. Now, the sine in ⊿BCB' can be equated to the sine in ⊿AOA'. Because, both are taken for the same angle α.
• For example, suppose α be 36o. Then from the tables, we have: sin 36 = 0.5878
• So in ⊿BCB', sin α = sin 36 = BB'⁄CB = 0.5878
• In ⊿AOA', AA'⁄OA will also be equal to 0.5878. Because:
sin α = sin 36 = AA'⁄OA= 0.5878
4. So we can write: BB'⁄CB = AA'⁄OA.
• But CB = OA. So we get: BB' = AA'
■ That is., altitudes are equal
5. The same is applicable to cosine also:
• Equating the cosines, we get:
B'C⁄CB = A'O⁄OA.
• But CB = OA. So we get: B'C = A'O
■ That is., bases are equal
• So now we are in a position to find the unknown vertex of any parallelogram.
• The method can be applied very quickly just by doing some metal calculations.
Let us analyse:
1. In fig.34.2(b), imagine a person standing at O
• To reach A, he must first travel 6 units horizontally to the right
♦ This '6 units' is determined from the x coordinate of A
• Then he must travel 2 units vertically upwards
♦ This '2 units' is determined from the y coordinate of A
2. Another person at C must follow the same procedure to reach B. That is:
• He must first travel 6 units horizontally to the right
• Then he must travel 2 units vertically upwards
■ The direction of each travel must be clearly specified:
• Horizontal - Towards left OR Towards right
• Vertical - Upwards OR Downwards
3. We are able to use this easy method because:
♦ bases are equal
♦ altitudes are also equal
• This method will be more clear when we see a solved example
Solved example 34.1:
What are the coordinates of the vertex C of the parallelogram shown in the fig.34.3 below:
Solution:
1. Group the vertices into two: [A,B] and [C,D]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach B from A:
• First travel 4 units horizontally to the right [∵ (5-1) = 4]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
3. The same procedure of travel must be followed for the travel from D to C
• First travel 4 units horizontally to the right
♦ At the end of this travel, the coordinates will be (6,5) [∵ (2+4) = 6]
• Then travel 1 unit vertically upwards
♦ At the end of this final lap, the coordinates will be (6,6) [∵ (5+1) = 6]
So the coordinates of C are (6,6)
■ In the above steps, we chose [A,B] and [C,D]
Let us choose [A,D] and [B,C]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach A from D:
• First travel 1 unit horizontally to the right [∵ (2-1) = 1]
• Then travel 2 units vertically upwards [∵ (5-3) = 2]
3. The same procedure of travel must be followed for the travel from B to C
• First travel 1 unit horizontally to the right
♦ At the end of this travel, the coordinates will be (6,4) [∵ (5+1) = 6]
• Then travel 2 units vertically upwards
♦ At the end of this final lap, the coordinates will be (6,6) [∵ (4+2) = 6]
So the coordinates of C are (6,6)
• We get the same answer. So we can choose groups in two different ways
■ But it is important to note that, the members of a group should not be diagonally opposite. Then the method will not work
In the next section, we will see a few more solved examples.
• This chapter is in fact a continuation of chapter 31 - Coordinate Geometry.
• The last section of that chapter can be seen here. In that chapter 31, we saw an interesting case:
1. We were given the coordinates of two points
• Those were the end points of a diagonal of a rectangle or a square
2. We were asked to find the coordinates of the other two corners of the rectangle
• We solved such problems very easily. (Details here)
• Now we will see a similar case.
♦ This time, instead of rectangles or squares, we look at parallelograms.
• Coordinates of any three of the four vertices of a parallelogram will be given.
• We will be asked to find the coordinates of the fourth vertex.
Let us see how it is done:
1. Consider the parallelogram OABC in fig.34.1(a) below
 |
| Fig.34.1 |
2. In fig.b, green lines are drawn.
• These green lines have a special property: They are all parallel to the axes
♦ The horizontal green lines are parallel to the x axis
♦ The vertical green lines are parallel to the y axis
3. Consider the two green lines attached to the side CB
• They intersect at B'.
• We know that the angle between the axes will always be 90o
• Since the green lines are parallel to the axes, the angle at B' will also be 90o
• So triangle BB'C is a right triangle
4. In the same way, the bottom triangle AA'O is also a right triangle
5. Now we will see the relation between the two right triangles:
• OA and CB are the opposite sides of a parallelogram.
♦ So OA and CB are parallel to each other
♦ and they will be inclined at the same angle with the x axis
• CB' and OA' are both parallel to the x axis. So we get: ∠BCB' = ∠AOA'
• These two equal angles are denoted as α in fig.c
6. Again, OA and CB are parallel to each other and they will be inclined at the same angle with the y axis also
• BB' and AA' are both parallel to the y axis. So we get: ∠CBB' = ∠OAA'
• These two equal angles are denoted as β in fig.c
7. We have a situation:
(i) Angles at the ends of the hypotenuse BC are α and β
(ii) Angles at the ends of the hypotenuse OA are also α and β
(iii) The two triangles are right angled.
(iv) The two hypotenuses, being opposite sides of a parallelogram, have the same lengths
■ In this situation, we can apply the RHS criterion. The two triangles are equal. (Details here)
Let us write the correspondence:
• The 90o angle is at A' and B'. So A'↔B'
• The αo angle is at O and C. So O↔C
• The βo angle is at A and B. So A↔B
• So the correspondence is A'↔B', O↔C and A↔B. This is same as A'OA↔B'CB
• So we can write:
The two triangles are congruent to one another under the correspondence: A'OA↔B'CB
• This is same as ΔA'OA ≅ ΔB'CB
Now we can write the corresponding sides. We get:
♦ A'O↔B'C
♦ OA↔CB
♦ A'A↔B'B
8. Among the above three correspondences, the middle one is already known to us.
• Because, they are the opposite sides of a parallelogram.
• The first and the last are of great use to us.
■ The first one indicates this:
The bases of the two triangles are of equal length
■ The last one indicates this:
The altitudes of the two triangles are of equal length
9. Now consider fig.34.2(a) below:
 |
| Fig.34.2 |
• Thus length of OA' is 6 units. This is shown in fig.34.2(b)
10. We have seen that the bases are equal.
• So CB' must also be 6 units in length
• Then coordinates of B' will be (3+6, 5). That is., (9,5)
11. Now consider the coordinates of A. They are (6,2)
• So AA' will be 2 units in length
12. We have seen that the altitudes are equal
• So BB' will also be 2 units in length
13. We already wrote the coordinates of B' as (9,5).
• So the coordinates of B will be (9,5+2). That is., (9,7)
• This is shown in fig.c. Thus we determined the unknown coordinates at B. The problem is solved.
• We were able to solve it because, the two bases as well as the two altitudes are equal. We proved it using the RHS criterion.
• There is an easier method to prove those equalities. In that method we use trigonometric ratios. We have already seen the basics of trigonometry here.
1. In fig.34.2(a) above, consider the right triangle BCB'. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = BB'⁄CB.
cos α = adjacent side⁄hypotenuse = B'C⁄CB.
2. Again in fig.34.2(a) above, consider the right triangle AOA'. Taking trigonometric ratios, we will get:
sin α = opposite side⁄hypotenuse = AA'⁄OA.
cos α = adjacent side⁄hypotenuse = A'O⁄OA.
3. Now, the sine in ⊿BCB' can be equated to the sine in ⊿AOA'. Because, both are taken for the same angle α.
• For example, suppose α be 36o. Then from the tables, we have: sin 36 = 0.5878
• So in ⊿BCB', sin α = sin 36 = BB'⁄CB = 0.5878
• In ⊿AOA', AA'⁄OA will also be equal to 0.5878. Because:
sin α = sin 36 = AA'⁄OA= 0.5878
4. So we can write: BB'⁄CB = AA'⁄OA.
• But CB = OA. So we get: BB' = AA'
■ That is., altitudes are equal
5. The same is applicable to cosine also:
• Equating the cosines, we get:
B'C⁄CB = A'O⁄OA.
• But CB = OA. So we get: B'C = A'O
■ That is., bases are equal
• So now we are in a position to find the unknown vertex of any parallelogram.
• The method can be applied very quickly just by doing some metal calculations.
Let us analyse:
1. In fig.34.2(b), imagine a person standing at O
• To reach A, he must first travel 6 units horizontally to the right
♦ This '6 units' is determined from the x coordinate of A
• Then he must travel 2 units vertically upwards
♦ This '2 units' is determined from the y coordinate of A
2. Another person at C must follow the same procedure to reach B. That is:
• He must first travel 6 units horizontally to the right
• Then he must travel 2 units vertically upwards
■ The direction of each travel must be clearly specified:
• Horizontal - Towards left OR Towards right
• Vertical - Upwards OR Downwards
3. We are able to use this easy method because:
♦ bases are equal
♦ altitudes are also equal
• This method will be more clear when we see a solved example
Solved example 34.1:
What are the coordinates of the vertex C of the parallelogram shown in the fig.34.3 below:
 |
| Fig.34.3 |
1. Group the vertices into two: [A,B] and [C,D]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach B from A:
• First travel 4 units horizontally to the right [∵ (5-1) = 4]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
3. The same procedure of travel must be followed for the travel from D to C
• First travel 4 units horizontally to the right
♦ At the end of this travel, the coordinates will be (6,5) [∵ (2+4) = 6]
• Then travel 1 unit vertically upwards
♦ At the end of this final lap, the coordinates will be (6,6) [∵ (5+1) = 6]
So the coordinates of C are (6,6)
■ In the above steps, we chose [A,B] and [C,D]
Let us choose [A,D] and [B,C]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach A from D:
• First travel 1 unit horizontally to the right [∵ (2-1) = 1]
• Then travel 2 units vertically upwards [∵ (5-3) = 2]
3. The same procedure of travel must be followed for the travel from B to C
• First travel 1 unit horizontally to the right
♦ At the end of this travel, the coordinates will be (6,4) [∵ (5+1) = 6]
• Then travel 2 units vertically upwards
♦ At the end of this final lap, the coordinates will be (6,6) [∵ (4+2) = 6]
So the coordinates of C are (6,6)
• We get the same answer. So we can choose groups in two different ways
■ But it is important to note that, the members of a group should not be diagonally opposite. Then the method will not work
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