Question 1 Whole-number lengths
Find the length of the line segment joining A ( 2 , 3 ) A(2, 3) A ( 2 , 3 ) and B ( − 1 , − 1 ) B(-1, -1) B ( − 1 , − 1 ) .
Show the worked solution Step 1: Put the coordinates into the length formula A B = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} A B = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 A B = ( − 1 − 2 ) 2 + ( − 1 − 3 ) 2 AB = \sqrt{(-1 - 2)^2 + (-1 - 3)^2} A B = ( − 1 − 2 ) 2 + ( − 1 − 3 ) 2 = 25 = \sqrt{25} = 25 Step 2: Work out the square root A B = 5 AB = 5 A B = 5 units Question 2 To 3 significant figures
Find the length of the line segment joining A ( − 1 , 1 ) A(-1, 1) A ( − 1 , 1 ) and B ( 0 , 2 ) B(0, 2) B ( 0 , 2 ) , correct to 3 significant figures.
Show the worked solution Step 1: Put the coordinates into the length formula A B = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} A B = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 A B = ( 0 − ( − 1 ) ) 2 + ( 2 − 1 ) 2 AB = \sqrt{(0 - (-1))^2 + (2 - 1)^2} A B = ( 0 − ( − 1 ) ) 2 + ( 2 − 1 ) 2 = 2 = \sqrt{2} = 2 Step 2: Work out the square root, then round A B = 1.4142 … AB = 1.4142\ldots A B = 1.4142 … A B = 1.41 \phantom{AB} = 1.41 A B = 1.41 units (3 s.f.) Question 3 Find a missing coordinate
The distance between A ( − 2 , k ) A(-2, k) A ( − 2 , k ) and B ( 2 , 5 ) B(2, 5) B ( 2 , 5 ) is 5 5 5 units. Find the two possible values of k k k .
Show the worked solution Step 1: Use Pythagoras with k ( 2 − ( − 2 ) ) 2 + ( 5 − k ) 2 = 5 2 (2 - (-2))^2 + (5 - k)^2 = 5^2 ( 2 − ( − 2 ) ) 2 + ( 5 − k ) 2 = 5 2 ( 5 − k ) 2 = 25 − 16 (5 - k)^2 = 25 - 16 ( 5 − k ) 2 = 25 − 16 = 9 = 9 = 9 Step 2: Take ± the square root 5 − k = ± 3 5 - k = \pm 3 5 − k = ± 3 k = 2 k = 2 k = 2 or k = 8 k = 8 k = 8 Question 4 Perimeter of a triangle
Triangle A B C ABC A B C has vertices A ( − 3 , − 1 ) A(-3, -1) A ( − 3 , − 1 ) , B ( 6 , 3 ) B(6, 3) B ( 6 , 3 ) and C ( 2 , − 3 ) C(2, -3) C ( 2 , − 3 ) . Find the perimeter of the triangle, correct to 3 significant figures.
x y −4 −3 −2 −1 1 2 3 4 5 6 −4 −3 −2 −1 1 2 3 0 A B C Show the worked solution Step 1: Find each side with Pythagoras A B = 81 + 16 AB = \sqrt{81 + 16} A B = 81 + 16 A B = 9.8488 … \phantom{AB} = 9.8488\ldots A B = 9.8488 … B C = 16 + 36 BC = \sqrt{16 + 36} B C = 16 + 36 B C = 7.2111 … \phantom{BC} = 7.2111\ldots B C = 7.2111 … C A = 25 + 4 CA = \sqrt{25 + 4} C A = 25 + 4 C A = 5.3851 … \phantom{CA} = 5.3851\ldots C A = 5.3851 … Step 2: Add, then round Perimeter = 22.4451 … = 22.4451\ldots = 22.4451 … = 22.4 = 22.4 = 22.4 units (3 s.f.)