Advertisements
Advertisements
Question
Co-ordinate of point P on a number line is - 7. Find the co-ordinates of points on the number line which are at a distance of 8 units from point P.
Solution
Let point Q be at a distance of 8 units from P and on left side of P
Let point R be at a distance of 8 units from P and on right side of P.
(i) Let the co-ordinate of point Q be x.
Co-ordinate of point P is -7.
Since, point Q is to the left of point P.
∴ -7 > x
∴ d(P, Q) = -7 - x
∴ 8 = -7 - x
∴ x = -7 - 8
∴ x = -15
(ii) Let the co-ordinate of point R be y.
Co-ordinate of point P is -7.
Since, point R is to the right of point P.
∴ y > -7
∴ d(P, R) = 7 - (-7)
∴ 8 = y + 7
∴ 8 - 7 = 7
∴ y = 1
∴ The co-ordinates of the points at a distance of 8 units from P are -15 and 1.
APPEARS IN
RELATED QUESTIONS
If P - Q - R and d(P, Q) = 2, d(P, R) = 10, then find d(Q, R).
On a number line, the co-ordinates of P, Q, R are 3, -5 and 6 respectively. State with reason whether the following statement is true or false.
d(R, P) + d(P, Q) = d(R, Q)
Co-ordinates of the pair of a point is given below. Hence find the distance between the pair.
3, 6
Co-ordinates of the pair of points are given below. Hence find the distance between the pair.
0, - 2
Determine whether the given set of points are collinear or not
(a, −2), (a, 3), (a, 0)
Show that the following points taken in order to form an equilateral triangle
`"A"(sqrt(3), 2), "B"(0, 1), "C"(0, 3)`
Show that the following points taken in order to form the vertices of a parallelogram
A(−7, −3), B(5, 10), C(15, 8) and D(3, −5)
The point (x, y) is equidistant from the points (3, 4) and (−5, 6). Find a relation between x and y
Let A(2, 3) and B(2, −4) be two points. If P lies on the x-axis, such that AP = `3/7` AB, find the coordinates of P.
Show that the point (11, 2) is the centre of the circle passing through the points (1, 2), (3, −4) and (5, −6)