Look at triangle ADM in the figure. Triangle ADM is an acute angled triangle in the first quadrant. How do we find the lengths of its sides AD, DM and MA?
Step-by-step solution
To find: The lengths AD, DM and MA
Idea: None of the sides is parallel to an axis. But the move along the x-axis and the move along the y-axis between two points form the two legs of a right-angled triangle, and the side we want is its hypotenuse.
- AD. Take C (3, 1), straight below A and level with D. Then ACD is right-angled at C. CD = 7 − 3 = 4 and AC = 4 − 1 = 3.½ mark
- By the Baudhāyana–Pythagoras theorem, AD2 = CD2 + AC2 = 42 + 32 = 16 + 9 = 25, so AD = √25 = 5 units.1 mark
- DM. From D (7, 1) to M (9, 6): across 9 − 7 = 2, up 6 − 1 = 5. DM = √(22 + 52) = √(4 + 25) = √29 units (about 5.39).½ mark
- MA. From A (3, 4) to M (9, 6): across 9 − 3 = 6, up 6 − 4 = 2. MA = √(62 + 22) = √(36 + 4) = √40 units (= 2√10, about 6.32).½ mark
- In general, the distance between (x1, y1) and (x2, y2) is √((x2 − x1)2 + (y2 − y1)2).½ mark
Check: The triangle is acute, as stated: the longest side squared, MA2 = 40, is less than AD2 + DM2 = 25 + 29 = 54 ✓.
Answer to write in the exam
Distance = √((x2 − x1)2 + (y2 − y1)2) (Baudhāyana–Pythagoras theorem)
AD = √((7 − 3)2 + (1 − 4)2) = √(16 + 9) = √25 = 5 units
DM = √((9 − 7)2 + (6 − 1)2) = √(4 + 25) = √29 units
MA = √((3 − 9)2 + (4 − 6)2) = √(36 + 4) = √40 units
∴ AD = 5 units, DM = √29 units, MA = √40 units
Common mistakes that cost marks
- Adding the shifts instead of using the theorem: AD = 4 + 3 = 7. The straight path is shorter than going across and then down.
- Forgetting the square root: writing AD = 25 instead of √25 = 5.
- Writing (1 − 4)2 = −9. A square is never negative: (−3)2 = 9.
How this can come in the exam
The distance between the points (2, −1) and (7, 11) is
- 13 units
- 17 units
- √17 units
- 169 units
Show answer
(A) 13 units
Shifts 7 − 2 = 5 and 11 − (−1) = 12; √(25 + 144) = √169 = 13.
Find the perimeter of the triangle with vertices O (0, 0), P (6, 0) and Q (0, 8).
Show answer
OP = 6, OQ = 8 (along the axes) (½ mark). PQ = √(62 + 82) = √100 = 10 (1 mark). Perimeter = 6 + 8 + 10 = 24 units (½ mark).Try one yourself
Find the lengths of the sides of the triangle with vertices K (1, 1), L (4, 2) and N (2, 5).
Show answer
KL = √(32 + 12) = √10, LN = √(22 + 32) = √13, NK = √(12 + 42) = √17 units.
More questions like this
- 1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?
2. Can these distances help you find the distance AD? - What if, x1, x2, y1, y2 take negative values? In the figure, triangle AMD is reflected in the y-axis. What are the coordinates of the images of points A, M, and D?
- 1. What has remained the same and what has changed with this reflection?
2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)? - What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
- Point W has x-coordinate equal to −5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?