Step: 1

Distance, d = √[(x _{2} - x _{1})^{2} + (y _{2} - y _{1})^{2}]

[Use the distance formula.]

Step: 2

[Replace (x _{2}, y _{2}) with (-8, -5) and (x _{1}, y _{1}) with (-5, -4).]

Step: 3

[Subtract.]

Step: 4

[Simplify.]

Step: 5

[Find the positive square root.]

Step: 6

So, distance between L and M = 3.16 units.

Correct Answer is : 3.16 units

Step: 1

The distance between A and B is 10 units.

Step: 2

Distance d = √[(x _{2} - x _{1})^{2} + (y _{2} - y _{1})^{2}]

[Use the distance formula.]

Step: 3

Distance between (-5, 2) and (6, 2) is √[(6 - (-5))^{2} + (2 - 2)^{2}] = √(11^{2} + 0^{2}) = √(121 + 0) =√121 ≠ 10

Step: 4

Distance between (-5, 2) and (5, 3) is √[(5 - (-5))^{2} + (3 - 2)^{2}] = √(10^{2} + (1)^{2}) = √(100 + 1) =√101 ≠ 10

Step: 5

Distance between (-5, 2) and (5, 2) is √[(5 - (-5))^{2} + (2 - 2)^{2}] = √(10^{2} + 0^{2}) = √(100 + 0) =√100 = 10.

Step: 6

The coordinates in choices B and C are at a distance other than 10 units from the point A.

Step: 7

The point (5, 2) is at a distance of 10 units from A.

Step: 8

So, the coordinates of B are (5, 2).

Correct Answer is : (5, 2)

Step: 1

The distance between (x _{1}, y _{1}) and (x _{2}, y _{2}) = d = √[(x _{2} - x _{1})^{2} + (y _{2} - y _{1})^{2}]

[Use the distance formula.]

Step: 2

Distance between AB, d = √[(3 - 5)^{2} + (x - 6)^{2})]

[Replace (x _{1}, y _{1}) with (5, 6) and (x _{2}, y _{2}) with (3, x ).]

Step: 3

[Subtract.]

Step: 4

[Simplify.]

Step: 5

[Apply exponents.]

Step: 6

= √(x ^{2} - 12x + 40)

Step: 7

The distance between A and B is 2√2 units.

Step: 8

√(x ^{2} - 12x + 40) = 2√2

[Equate distances.]

Step: 9

[Squaring on both sides.]

Step: 10

[Write in general form.]

Step: 11

(x - 4)(x - 8) = 0

[Factorize.]

Step: 12

Step: 13

The coordinates of B are either (3, 4) or (3, 8)

[Substitute x values.]

Correct Answer is : Either A or B

Step: 1

Distance d = √[(x _{2} - x _{1}) ^{2} + (y _{2} - y _{1}) ^{2}]

[Use the distance formula.]

Step: 2

[Replace (x _{2}, y _{2}) with (5, 5) and (x _{1}, y _{1}) with (3, 4).]

Step: 3

[Subtract.]

Step: 4

[Simplify.]

Step: 5

[Find the positive square root.]

Step: 6

The distance between, P (3, 4) and Q (5, 5) is 2.23 units.

Correct Answer is : 2.23 units

Step: 1

Distance between two points d = √[(x _{2} - x _{1}) ^{2} + (y _{2} - y _{1}) ^{2}]

[Use the distance formula.]

Step: 2

[Replace (x _{2}, y _{2}) with (8, -5) and (x _{1}, y _{1}) with (5, -3).]

Step: 3

[Subtract.]

Step: 4

[Simplify.]

Step: 5

[Find the positive square root.]

Step: 6

The distance between, A and B is 3.60 units.

Correct Answer is : 3.60 units

Step: 1

The distance between (x _{1}, y _{1}) and (x _{2}, y _{2}) = d = √[(x _{2} - x _{1})^{2} + (y _{2} - y _{1})^{2}]

[Use the distance formula.]

Step: 2

[Replace (x _{1}, y _{1}) with (-3, 6) and (x _{2}, y _{2}) with (-4, 5).]

Step: 3

[Subtract.]

Step: 4

Step: 5

Step: 6

The distance between A and B is 1.4 units.

[Round the distance to the nearest tenth.]

Correct Answer is : 1.4 units

Step: 1

Distance d = √[(x _{2} - x _{1}) ^{2} + (y _{2} - y _{1}) ^{2}]

[Use the distance formula.]

Step: 2

Consider choice A, the distance between (3, 4) and (8, 4),

Step: 3

[Replace (x _{1}, y _{1}) with (3, 4) and (x _{2}, y _{2}) with (8, 4).]

Step: 4

[Subtract.]

Step: 5

[Simplify.]

Step: 6

[Find the positive square root.]

Step: 7

Checking with the choices B, C and D, we find that those are not at a distance of 5 units from (3, 4).

Step: 8

thus the point (8, 4) is at a distance of 5 units from (3, 4).

Correct Answer is : (8, 4)

Step: 1

The coordinates of A are (2, 2).

Step: 2

The first coordinate of B is 4 and the second coordinate is twice the second coordinate of A.

Step: 3

The second coordinate of B = 2 x second coordinate of A = 2 x 2 = 4

[Substitute the second coordinate of A = 2.]

Step: 4

The coordinates of B are (4, 4)

Step: 5

Distance between the line-segment with endpoints (x _{1}, y _{1}) and (x _{2}, y _{2}) is
√[(x _{2} - x _{1})^{2} + (y _{2} - y _{1})^{2}]

Step: 6

Distance between A and B = √[(4 - 2)^{2} + (4 - 2)^{2}]

[Replace (x _{1}, y _{1}) with (2, 2) and (x _{2}, y _{2}) with (4, 4).]

Step: 7

= √(2^{2} + 2^{2})

[Subtract.]

Step: 8

= √(4 + 4)

[Evaluate powers.]

Step: 9

= √8

Step: 10

= 2√2

[Simplify.]

Step: 11

The distance between A and B is 2√2 units.

Correct Answer is : 2√2 units

Step: 1

Plot the points R(- 3, 2), S(- 3, - 2) and T(6, - 2) in a Cartesian plane and join them.

Step: 2

Distance between the point R and the point S = 2 - (-2) = 4 units

[Difference between the y -coordinates.]

Step: 3

Distance between the point S and the point T = 6 - (-3) = 9 units

[Difference between the x -coordinates.]

Step: 4

Distance between Andrew and Peter = Distance between the point R and the point T

Step: 5

Since ΔRST is a right-angled triangle,

RT^{2} = RS^{2} + ST^{2}

RT

[Pythagoras′ theorem.]

Step: 6

RT^{2} = 4^{2} + 9^{2}

[Substitute the values.]

Step: 7

RT^{2} = 16 + 81 = 97

[Add.]

Step: 8

RT = 9 7 = 9.85 units

[Take square root on both sides.]

Step: 9

Therefore, the distance between Andrew and Peter is 9.85 units.

Correct Answer is : 9.85 units

Step: 1

Distance between Mary′s house P and the park R = 5 - 2 = 3 units

[Distance between the y -coordinates.]

Step: 2

Distance between Mary′s house P and the library Q = 5 - 1 = 4 units

[Distance between the x -coordinates.]

Step: 3

The distance between the park R and the library Q = Length of RQ

Step: 4

Since ΔRPQ is a right-angled triangle,

RQ^{2} = PQ^{2} + PR^{2}

RQ

[Pythagoras′ theorem.]

Step: 5

RQ^{2} = 4^{2} + 3^{2}

[Substitute the values.]

Step: 6

RQ^{2} = 16 + 9 = 25

[Add.]

Step: 7

RQ = 2 5 = 5 units

[Take square root on both sides.]

Step: 8

Therefore, the distance between the park and the library is 5 units.

Correct Answer is : 5 units

Step: 1

Distance between the point A and the point B = 4 - (-2) = 4 + 2 = 6 units

[Difference between the x -coordinates.]

Step: 2

Distance between the point B and the point C = 3 - (- 3) = 3 + 3 = 6 units

[Difference between the y -coordinates.]

Step: 3

Distance between Michael′s house A and the point C = Length of AC

Step: 4

Since ΔABC is a right-angled triangle, AC^{2} = AB^{2} + BC^{2}

[Pythagoras′ theorem.]

Step: 5

AC^{2} = 6^{2} + 6^{2}

[Substitute the values.]

Step: 6

AC^{2} = 36 + 36 = 72

[Add.]

Step: 7

AC = 7 2 = 8.49 units

[Take square root on both sides.]

Step: 8

So, length of AC = 8.49 units

Step: 9

Therefore, Michael is at a distance of 8.49 units from his house now.

Correct Answer is : 8.49 units

Step: 1

Plot the points A(5, 3), B(5, - 2) and C(0, 3) in a Cartesian plane and join them.

Step: 2

Distance between school A and C = 5 - 0 = 5 units

[Difference between the x -coordinates.]

Step: 3

Distance between the school A and school B = 3 - (- 2) = 3 + 2 = 5 units

[Difference between the y -coordinates.]

Step: 4

Since ΔBAC is a right-angled triangle,

BC^{2} = AB^{2} + AC^{2}

BC

[Pythagoras′ theorem.]

Step: 5

BC^{2} = 5^{2} + 5^{2}

[Substitute the values.]

Step: 6

BC^{2} = 25 + 25 = 50

[Add.]

Step: 7

BC = 5 0 = 7.07 units

[Take square root on both sides.]

Step: 8

Therefore, school B is 7.07 units far from school C.

Correct Answer is : 7.07 units

Step: 1

From the figure, AB = 2 - (- 3) = 2 + 3 = 5 units.

Step: 2

AC = 4 - (- 2) = 4 + 2 = 6 units.

Step: 3

From ΔABC, BC^{2} = AB^{2} + AC^{2}

[Apply Pythagorean theorem.]

Step: 4

BC^{2} = 5^{2} + 6^{2}

[Replace AB with 5 and AC with 6.]

Step: 5

BC^{2} = 25 + 36 = 61

[Simplify.]

Step: 6

BC = 6 1

[Take square root on both sides.]

Step: 7

The length of BC = 6 1 units.

Correct Answer is : 6 1 units

- Parallel Lines and Transversals-Gr 8-Solved Examples
- Angle Sum Theorem of a Triangle-Gr 8-Solved Examples
- Exterior Angle Theorem-Gr 8-Solved Examples
- Pythagorean Theorem-Gr 8-Solved Examples
- Transformations-Gr 8-Solved Examples
- Volume of Cylinders-Gr 8-Solved Examples
- Volume of Cones-Gr 8-Solved Examples
- Volume of a Sphere-Gr 8-Solved Examples

- Distance Formula