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Therefore the distance between (2,3) and (4,1) is given by
l =
=
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= √4+4 = √8 = 2√2
(ii)Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
l =
=
= √16+16 = √32 = 4√2
(iii)Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
l =
=
=
2. Find the distance between the points (0, 0) and (36, 15). Also, find the distance between towns A and B if town B is located at 36 km east and15 km north of town A.
= √1296 + 225 = √1521 = 39
Yes, we can find the distance between the given towns A and B.
Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39km.
3. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
BC =
CA =
Since AB+BC ≠ CA
Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.
4. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
BC =
CA =
Since AB = BC.
Therefore, A, B and C are vertices of an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli. “Don’t you think ABCD is a square?”Chameli disagrees. Using distance formula, find which of them is correct.
BC =
CD =
AD =
Therefore, All the sides of ABCD are equal here
Now, we will check the length of its diagonals.
AC =
BD =
So, Diagonals of ABCD are also equal.
we can definitely say that ABCD is a square.
Therefore, Champa is correct.
6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
BC =
CD =
AD =
.png)
Therefore, all four sides of quadrilateral are equal.
Now, we will check the length of diagonals.
AC =
.png)
BD =
Therefore, diagonals of quadrilateral ABCD are also equal.
we can say that ABCD is a square.
(ii)Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB =
BC =
CD =
DA =
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the quadrilateral ABCD.
(iii)Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB =
BC =
CD =
DA =
Here opposite sides of quadrilateral ABCD are equal.
We can now find out the lengths of diagonals.
AC =
BD =
Here diagonals of ABCD are not equal.
we can say that ABCD is not a rectangle therefore it is a parallelogram.
7. Find the point on the x–axis which is equidistant from (2, –5) and (–2, 9).
⇒
Squaring both sides, we get
⇒
(x-2)² + 25 = (x+2)² + 81
x² + 4 - 4x + 25 = x² + 4 + 4x + 81
8x = - 25 - 81
8x = -56
x = - 7
Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)
8. Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.
⇒
⇒ 64 + (y +3)² = 100
⇒ (y+3)² = 100-64 = 36
⇒ y+3 = ± 6
⇒ y+3=6 or y+3 = - 6
Therefore y = 3 or -9
9. If, Q (0, 1) is equidistant from P (5, –3) and R (x, 6), find the values of x. Also, find the distances QR and PR.
⇒√25+16 = √x² + 25
⇒41 = x² + 25
16 = x²
x = ± 4
Thus, R is (4, 6) or (–4, 6).
When point R is (4,6)
PR =
QR =
When point R is (- 4,6)
PR =
QR =
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).
⇒
⇒ (x-3)² + (y-6)² = (x+3)² + (y-4)²
⇒ x² + 9 -6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y
⇒36- 16 = 6x + 6x + 12y - 8y
⇒20 = 12x + 4y
⇒3x + y = 5
⇒3x + y - 5 = 0
