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Question
If the length of the segment joining point L(x, 7) and point M(1, 15) is 10 cm, then the value of x is ______.
Options
7
7 or –5
–1
1
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Solution
If the length of the segment joining point L(x, 7) and point M(1, 15) is 10 cm, then the value of x is 7 or –5.
Explanation:
Here, x1 = x, y1 = 7, x2 = 1, y2 = 15
By distance formula,
d(L, M) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
∴ d(L, M) = `sqrt((1 - x)^2 + (15 - 7)^2)`
∴ `10 = sqrt((1 - x)^2 + 8^2)`
∴ 100 = (1 – x)2 + 64 ...[Squaring both sides]
∴ (1 – x)2 = 100 – 64
∴ (1 – x)2 = 36
∴ `1 - x = ± sqrt(36)` ...[Taking square root of both sides]
∴ 1 – x = ± 6
∴ 1 – x = 6 or 1 – x = – 6
∴ x = –5 or x = 7
∴ The value of x is –5 or 7.
RELATED QUESTIONS
If two vertices of an equilateral triangle be (0, 0), (3, `sqrt3`), find the third vertex using distance formula.
Find the coordinates of the centre of the circle passing through the points (0, 0), (–2, 1) and (–3, 2). Also, find its radius.
Show that the points A (1, –2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.
An equilateral triangle has two vertices at the points (3, 4) and (−2, 3), find the coordinates of the third vertex.
Find all possible values of y for which the distance between the points A(2, –3) and B(10, y) is 10 units.
Find the distance between the following pair of point.
P(–5, 7), Q(–1, 3)
Find the distance between the following pair of points.
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Find the distance between the following pairs of point in the coordinate plane :
(4 , 1) and (-4 , 5)
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(13 , 7) and (4 , -5)
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ABC is an equilateral triangle . If the coordinates of A and B are (1 , 1) and (- 1 , -1) , find the coordinates of C.
A(2, 5), B(-2, 4) and C(-2, 6) are the vertices of a triangle ABC. Prove that ABC is an isosceles triangle.
Find the point on y-axis whose distances from the points A (6, 7) and B (4, -3) are in the ratio 1: 2.
Find distance between point A(–3, 4) and origin O.
Find distance between point Q(3, –7) and point R(3, 3)
Solution: Suppose Q(x1, y1) and point R(x2, y2)
x1 = 3, y1 = –7 and x2 = 3, y2 = 3
Using distance formula,
d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `sqrt(square - 100)`
∴ d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `square`
Find distance between point A(–1, 1) and point B(5, –7):
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = –7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
Show that the point (11, –2) is equidistant from (4, –3) and (6, 3).
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Case Study -2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
Each team plays with 11 players on the field during the game including the goalie. Positions you might play include -
- Forward: As shown by players A, B, C and D.
- Midfielders: As shown by players E, F and G.
- Fullbacks: As shown by players H, I and J.
- Goalie: As shown by player K.
Using the picture of a hockey field below, answer the questions that follow:
If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by ______.
