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The correct option is: (A) only one
The correct option is: (C) 1
The correct option is: (C) one or three
Distance between two points A and B is calculated by subtarcting the smaller co-ordinate from the greater co-ordinate.
d (A, B) = 5 − (−2)
∴ d (A, B) = 5 + 2
∴ d (A, B) = 7
∴ The correct option is: (C) 7
P – Q – R
∴ d (P, R) = d (P, Q) + d (Q, R)
∴ 10 = 2 + d (Q, R)
∴ 10 − 2 = d (Q, R)
∴ 8 = d (Q, R)
i.e. d (Q, R) = 8
The correct option is: (B) 8
As seen in the figure:
P: 3, Q: − 5, R: 6
∴ d (R, P) = 6 - 3 = 3 ... (i)
d (P, Q) = 3 − (− 5) = 3 + 5 = 8 ... (ii)
and d (R, Q) = 6 − (− 5) = 6 + 5 = 11 ... (iii)
From (i), (ii) and (iii):
d (R, P) + d (P, Q)
= 8 + 3
= d (R, Q)
Therefore,
The distance between two points is calculated by subtracting the smaller co-ordinate from the greater co-ordinate.
Let, those points be X and Y respectively.
Now, 3 < 6
∴ d (X, Y) = 6 − 3
∴ d (X, Y) = 3
Let, those points be X and Y respectively.
Now, − 9 < − 1
∴ d (X, Y) = − 1 − (− 9)
∴ d (X, Y) = − 1 + 9
∴ d (X, Y) = 8
Let, those points be X and Y respectively.
Now, − 4 < 5
∴ d (X, Y) = 5 − (− 4)
∴ d (X, Y) = 5 + 4
∴ d (X, Y) = 9
Let, those points be X and Y respectively.
Now, − 2 < 0
∴ d (X, Y) = 0 − (− 2)
∴ d (X, Y) = 0 + 2
∴ d (X, Y) = 2
Let, those points be P and Q respectively.
Now, (x − 3) < (x + 3)
∴ d (P, Q) = (x + 3) − (x − 3)
∴ d (P, Q) = x + 3 − x + 3
∴ d (P, Q) = 6
Let, those points be P and Q respectively.
Now, (− 47) < (− 25)
∴ d (P, Q) = (− 25) − (− 47)
∴ d (P, Q) = (− 25) + 47
∴ d (P, Q) = 22
Let, those points be P and Q respectively.
Now, (− 85) < 80
∴ d (P, Q) = 80 − (− 85)
∴ d (P, Q) = 80 + 85
∴ d (P, Q) = 165
There are 2 points which are at a distance of 8 from point P.
Let, those points be R and T.
Let, the co-ordinate of T be x and the co-ordinate of R be y.
Now, x > − 7
∴ d (P, T) = x − (− 7)
∴ 8 = x + 7
∴ 8 − 7 = x
∴ 1 = x
i.e. x = 1
∴ The co-ordinate of T is 1.
Also, y < − 7
∴ d (P, R) = (− 7) − y
∴ 8 = − 7 − y
∴ 8 + 7 = − y
∴ 15 = − y
i.e. y = − 15
∴ The co-ordinate of R is − 15.
∴ The co-ordinates of points which are at a distance of 8 from P are 1 and − 15.
A – B – C ... (Given)
∴ d (A, B) + d (B, C) = d (A, C)
∴ d (A, B) + 6.5 = 17
∴ d (A, B) = 17 − 6.5
∴ d (A, B) = 10.5
P – Q – R ... (Given)
∴ d (P, R) = d (P, Q) + d (Q, R)
∴ d (P, R) = 3.4 + 5.7
∴ d (P, R) = 9.1
There are 2 points which are at a distance of 7 from point A.
Let, those points be P and Q.
Let, the co-ordinate of Q be x and the co-ordinate of P be y.
The distance between two points is calculated by subtracting the smaller co-ordinate from the greater co-ordinate.
Now, x > 1
∴ d (A, Q) = x − 1
∴ 7 = x − 1
∴ 7 + 1 = x
∴ 8 = x
i.e. x = 8
∴ The co-ordinate of Q is 8.
Also, y < 1
∴ d (A, P) = 1 − y
∴ 7 = 1 − y
∴ y = 1 − 7
∴ y = − 6
∴ The co-ordinate of P is − 6.
∴ The co-ordinates of points which are at a distance of 7 from A are 8 and − 6.
If a quadrilateral is a square, then it is a rhombus.
If the angles are in linear pair, then they are supplementary.
If a (closed) figure is formed by three segments, then it is a triangle.
If a number has only two divisors, then it is a prime number.
If a figure is a triangle, then the sum of the measures of angles in it is 180°.
If two angles are complements of each other, then the sum of their measures is 90°.
If two lines are parallel, then the corresponding angles formed by a transversal of the two lines are congruent.
If a number is divisible by 3, then the sum of its digits is divisible by 3.
In \(\triangle\)ABC,
side AB ≅ side BC ≅ side CA
\(\angle\) A ≅ \(\angle\) B ≅ \(\angle\) C
\(\square\)ABCD is a parallelogram.
Its diagonals AC and BD intersect in point M.
AM = CM and BM = DM.
In \(\triangle\)ABC,
side AB ≅ side BC ≅ side CA
And in \(\triangle\)PQR,
side PQ ≅ side QR ≅ side RP
\(\triangle\) ABC ~ \(\triangle\) PQR
\(\angle\)ABC and \(\angle\)ABD form a linear pair
and \(\angle\)ABC ≅ \(\angle\)ABD
\(\angle\)ABC = \(\angle\)ABD = 90°
In \(\triangle\)ABC,
seg BD ⊥ side AC
and seg CE ⊥ side AB
and seg BD ≅ seg CE
side AB ≅ side AC
This page was last modified on
13 April 2026 at 13:11