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Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1
Question 1.
Find the distance between the following pairs of points :
i) (2, 3) (4, 1)
ii) (- 5, 7). (- 1, 3)
iii) (a, b), (- a, – b)
Answer:
i) If A (x1, y1) = A (2, 3)
B (x2, y2) = B (4, 1), then
distance between pairs = 2√2
ii) Let P(x1, y1) = P(- 5, 7)
9(x2, y2) = Q(- 1, 3)
Distance = 4√2
iii) A (a, b), B (-a, -b)
Question 2.
Find the distance between the points (0, 0) and (36, 15). can you now find the distance between the two towns A and B discussed in Section 7.2.
Answer:
The distance between the points A(0, 0) and B(36, 15):
Distence = 39
∴ Distance between two towns A and B discussed in section 7.2 are A(O, O) & B
(36, 15) ∴ Distance = 39 km
Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Answer:
Whether the points A (1, 5), B (2, 3) and C (- 2, – 11) are collinear.
The longest distance = CA = √265
But, AB + BC ≠ AC
AC + CB ≠ AB
BA + AC ≠ BC
∴ Hence, the given points are not collinear.
Question 4.
Check whether (5, -2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Answer:
A = (5, – 2), B = (6, 4) and C = (7, – 2)
Distence =
AB = BC
Hence, the given points are vertices of a isosceles triangle.
Question 5.
In a classroom, 4 friends are seated at the points, A, B, C and D as shown in the following figure. 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.
Answer:
A → (3, 4), B → (6, 7), C → (9, 4) & D → (6, 1)
∴ AB = BC = CD = DA
i,e all the sides are equal ∴ The figure is either a square (or) Rhombus.
BD = 6
∴ diagonal are equal & sides are equal
∴ ABCD is a square
Hence, Champa iš correct.
Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons 1 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)
Answer:
(i) A → (- 1, – 2), B → (1, 0), C → (- 1, 2) and D →(- 3, 0)
AC = 4
BD = 4
∴ AC = BD
In a quadrilateral all the sides are equal & diagonal are equal
∴ Quadrilateral is a square.
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
As per Distance formula,
AC = √13
From (1), (2) & (3)
BC + AC = √13 + √13 = 2√13
BC + AC = √52
i.e.,AC + BC = AB
Hence, the points A, B & C are collinear so, ABCD is not even a quadrilateral
(iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)
As per Distance formula =
AC ≠ BD
In a Quadrilateral ABCD, opposite sides are equal and diagonals are not equal
∴ It is a parallelogram.
Question 7.
Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Answer:
Let the required point be (x, 0)
Since (x, 0) is equidistance from the point (2, -5) & (-2, 9).
p → (x, 0), A → (2, – 5) & B → (- 2, 9)
(2 – x )2 + (- 5)2 = (- 2 – x )2 + (9)2
4 – 4x + x2 + 25 = 4 + 4x + x2 + 81
x2 – 4x + 29 = x2 + 4x + 85 – 29
x2 – x2 – 4x – 4x = 85 – 29
– 8x = 56
x = – 7
Hence, the required point is (x, 0) = (- 7, 0)
Question 8.
Find the values of ‘y’ for which the distance between the points P(2, – 3) and Q (10, y) is 10 units.
Answer:
If the distance between P and Q, PQ = 10 units, then y =?
PQ = Distance
Squaring both sides
100 = 64 + y2 + 6y + 9
y2 + 6y + 73 – 100 = 0
y2 + 6y – 27 = 0
y2 + 9y – 3y – 27 = 0
y(y + 9) – 3(y + 9) = 0
(y + 9)(y – 3) = 0
y + 9 = 0 or y – 3 = 0
y = – 9 or y = 3
Hence required value of y is – 9 or 3
Question 9.
If Q (0, 1) is equidistant from P (5, -3) and R(x, 6), find the value of x. Also, find the distances QR and PR.
Answer:
Since, Point Q → (0, 1) is equidistant from P → (5, – 3) and R(x, 6)
Question 10.
Find a relation between ‘x’ and ‘y’ such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Answer:
Let P (x, y) be equidistant from the points, A(3, 6) and B(- 3, 4)
i.e., PA = PB
(x – 3)2 + (y – 6)2
= (x + 3)2 + (y – 4)2
x2 – 6x + 9 + y2 – 12y + 36
= x2 + 6x + 9 + y2 – 8y + 16
x2 – x2 – 6x – 6x + y2
+ 9 – 9 – 12y + 8y + 36 – 16 = 0
– 12 – 4y + 20 = 0
divide by – 4
3x + y – 5 = 0