Question 1.1
Find the distance between the following pairs of points:
(-3, 6) and (2, -6)
(-3, 6) and (2, -6)
Distance between the given points
=
=
=
=
= 13
Question 1.2
Find the distance between the following pairs of points:
(-a, -b) and (a, b)
(-a, -b) and (a, b)
Distance between the given points
=
=
=
= 2
Question 1.3
Find the distance between the following pairs of points:
Distance between the given points
=
=
=
=
= 1
Question 1.4
Find the distance between the following pairs of points:
Distance between the given points
=
=
=
= 2
Question 2.1
Find the distance between the origin and the point:
(-8, 6)
Coordinates of origin are O (0, 0).
A (-8, 6)
AO =
=
=
= 10
Question 2.2
Find the distance between the origin and the point:
(-5, -12)
Coordinates of origin are O (0, 0).
B (-5, -12)
BO =
=
=
= 13
Question 2.3
Find the distance between the origin and the point:
(8, -15)
Coordinates of origin are O (0, 0).
C (8, -15)
CO =
=
=
= 17
Question 3
The distance between the points (3, 1) and (0, x) is 5. Find x.
It is given that the distance between the points A (3, 1) and B (0, x) is 5.
∴ AB = 5
AB2 = 25
(0 - 3)2 + (x - 1)2 = 25
9 + x2 + 1 - 2x = 25
x2 - 2x - 15 = 0
x2 - 5x + 3x - 15 = 0
x(x - 5) + 3(x - 5) = 0
(x - 5)(x + 3) = 0
x = 5, -3
Question 4
Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).
Sol;Let the coordinates of the point on x-axis be (x, 0).
From the given information, we have:
(x - 11)2 + (0 + 8)2 = 289
x2 + 121 - 22x + 64 = 289
x2 - 22x - 104 = 0
x2 - 26x + 4x - 104 = 0
x(x - 26) + 4(x - 26) = 0
(x - 26)(x + 4) = 0
x = 26, -4
Thus, the required co-ordinates of the points on x-axis are (26, 0) and (-4, 0).
Question 5
Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).
Sol:Let the coordinates of the point on y-axis be (0, y).
From the given information, we have:
(0 + 8)2 + (y - 4)2 = 100
64 + y2 + 16 - 8y = 100
y2 - 8y - 20 = 0
y2 - 10y + 2y - 20 = 0
y(y - 10) + 2(y - 10) = 0
(y - 10)(y + 2) = 0
y = 10, - 2
Thus, the required co-ordinates of the points on y-axis are (0, 10) and (0, -2).
Question 6
A point A is at a distance of
It is given that the coordinates of point A are such that its ordinate is twice its abscissa.
So, let the coordinates of point A be (x, 2x).
We have:
(x - 4)2 + (2x - 3)2 = 10
x2 + 16 - 8x + 4x2 + 9 - 12x = 10
5x2 - 20x + 15 = 10
x2 - 4x + 3 = 0
x2 - x - 3x + 3 = 0
x(x - 1) -3(x - 1) = 0
(x - 1)(x - 3) = 0
x = 1, 3
Thus, the co-ordinates of the point A are (1, 2) and (3, 6).
Question 7
A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
Sol:We know that the distance between the two points (x1, y1) and (x2, y2) is
d =
Let the given points be A = (a, 7) and B = (−3, a) and the third point given is P(2, −1).
We first find the distance between P(2, −1) and A =(a, 7) as follows:
PA =
=
=
=
=
Similarly, the distance between P(2,−1) and B = (−3, a) is:
PB =
=
=
=
Since the point P(2,−1) is equidistant from the points A(a, 7) and B = (−3, a), therefore, PA = PB that is:
⇒ a2 + 4 − 4a − a2 − 1 − 2a = −39
⇒ −6a + 3 = −39
⇒− 6a =−39 −3
⇒ −6a = −42
⇒ a =
Hence, a = 7.
Question 8
What point on the x-axis is equidistant from the points (7, 6) and (-3, 4)?
Sol:Let the co-ordinates of the required point on x-axis be P (x, 0).
The given points are A (7, 6) and B (-3, 4).
Given, PA = PB
PA2 = PB2
(x - 7)2 + (0 - 6)2 = (x + 3)2 + (0 - 4)2
x2 + 49 - 14x + 36 = x2 + 9 + 6x + 16
60 = 20x
x = 3
Thus, the required point is (3, 0).
Question 9
Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).
Sol:Let the co-ordinates of the required point on y-axis be P (0, y).
The given points are A (5, 2) and B (-4, 3).
Given, PA = PB
PA2 = PB2
(0 -5)2 + (y -2)2 = (0 + 4)2 + (y - 3)2
25 + y2 + 4 - 4y = 16 + y2 + 9 - 6y
2y = -4
y = -2
Thus, the required point is (0, -2).
Question 10.1
A point P lies on the x-axis and another point Q lies on the y-axis.
Write the ordinate of point P.
Since, the point P lies on the x-axis, its ordinate is 0.
Question 10.2
A point P lies on the x-axis and another point Q lies on the y-axis.
Write the abscissa of point Q.
Since, the point Q lies on the y-axis, its abscissa is 0.
Question 10.3
A point P lies on the x-axis and another point Q lies on the y-axis.
If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.
The co-ordinates of P and Q are (-12, 0) and (0, -16) respectively.
PQ =
=
=
= 20
Question 11
Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.
Sol:PQ =
=
=
= 5
QR =
=
=
= 5
RP =
=
=
= 2
Since, PQ = QR, ΔPQR is an isosceles triangle.
Question 12
Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
Sol :PQ =
=
=
= 5
QR =
=
=
= 5
RP =
=
=
= 2
Since, PQ = QR, ΔPQR is an isosceles triangle.
PQ =
QR =
RS =
PS =
PR =
QS =
∵ PQ = RS and QR = PS,
Also PR = QS
∴ PQRS is a rectangle.
Question 13
Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.
Sol:AB =
BC =
CA =
∵ AB = CA
A, B, C are the vertices of an isosceless triangle.
AB2 + CA2 = 25 + 25 = 50
BC2 =
∴ AB2 + CA2 = BC2
Hence, A, B., C are the vertices of a right - angled triangle.
Hence, ΔABC is an isosceles right-angled triangle.
Area of ΔABC =
=
= 12.5 sq.units
Question 14
Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.
Sol:AB =
BC =
CD = =
DA = =
AC = =
BD = =
Since, AB = BC = CD = DA and AC = BD,
A, B, C and D are the vertices of a square.
Question 15
Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
Sol:Let the given points be A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4).
AB =
BC = =
CD= =
DA =
AC =
BD =
Since, AB = BC = CD = DA and AC ≠ BD
The given vertices are the vertices of a rhombus.
Question 16
Points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.
Sol:AB = CD
AB2 = CD2
(- 6 + 3)2 + (a + 2)2 = (0 + 3)2 + (- 1 + 4)2
9 + a2 + 4 + 4a = 9 + 9
a2 + 4a - 5 = 0
a2 - a + 5a - 5 = 0
a(a - 1) + 5 (a - 1) = 0
(a - 1) (a + 5) = 0
a = 1 or - 5
It is given that a is negative, thus the value of a is - 5.
Question 17
The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.
Sol:Let the circumcentre be P (x, y).
Then, PA = PB
PA2 = PB2
(x - 5)2 + (y - 1)2 = (x - 11)2 + (y - 1)2
x2 + 25 - 10x = x2 + 121 - 22x
12x = 96
x = 8
Also, PA = PC
PA2 = PC2
(x - 5)2 + (y - 1)2 = (x - 11)2 + (y - 9)2
x2 + 25 - 10x + y2 + 1 - 2y = x2 + 121 - 22x + y2 + 81 - 18y
12x + 16y = 176
3x + 4y = 44
24 + 4y = 44
4y = 20
y = 5
Thus, the coordinates of the circumcentre of the triangle are (8, 5).
Question 18
Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.
Sol:AB = 5
AB2 = 25
(0 - 3)2 + (y - 1 - 1)2 = 25
9 + y2 + 4 - 4y = 25
y2 - 4y - 12 = 0
y2 - 6y + 2y - 12 = 0
y(y - 6) + 2(y - 6) = 0
(y - 6) (y + 2) = 0
y = 6, -2
Question 19
Given A = (x + 2, -2) and B (11, 6). Find x if AB = 17.
Sol:AB = 17
AB2 = 289
(11 - x - 2)2 + (6 + 2)2 = 289
x2 + 81 - 18x + 64 = 289
x2 - 18x - 144 = 0
x2 - 24x + 6x - 144 = 0
x(x - 24) + 6(x - 24) = 0
(x - 24) (x + 6) = 0
x = 24, -6
Question 20
The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 unit.
Sol:Distance between the points A (2x - 1, 3x + 1) and B (- 3, - 1) = Radius of circle
AB = 10 (Since, diameter = 20 units, given)
AB2 = 100
(-3 - 2x + 1)2 + (-1 - 3x - 1)2 = 100
(-2 - 2x)2 + (-2 - 3x)2 = 100
4 + 4x2 + 8x + 4 + 9x2 + 12x = 100
13x2 + 20x - 92 = 0
x =
x =
x = 2, -
Question 21
The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.
Sol:Let the co-ordinates of point Q be (10, y).
PQ = 10
PQ2 = 100
(10 - 2)2 + (y + 3)2 = 100
64 + y2 + 9 + 6y = 100
y2 + 6y - 27 = 0
y2 + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0
(y + 9) (y - 3) = 0
y = -9, 3
Thus, the required co-ordinates of point Q are (10, -9) and (10, 3).
Question 22.1
Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of: AT
Given, radius = 13 units
PA = PB = 13 units
Using distance formula,
PT =
=
=
= 5
Using Pythagoras theorem in Δ PAT,
AT2 = PA2 - PT2
AT2 = 169 - 25
AT2 = 144
AT = 12 units.
Question 22.2
Point P (2, -7) is the centre of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of AB.
We know that the perpendicular from the center of a circle to a chord bisects the chord.
∴ AB = 2AT
= 2 x 12 units
= 24 units.
Question 23
Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.
Sol:PQ =
=
=
= 3.6055
= 3.61 units
Question 24
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
We know that any point on x-axis has coordinates of the form (x, 0).
Abscissa of point B = 11
Since, B lies of x-axis, so its co-ordinates are (11, 0).
AB =
=
=
= 5 units
Question 25
Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.
' Sol:We know that any point on y-axis has coordinates of the form (0, y).
Ordinate of point B = 9
Since, B lies of y-axis, so its co-ordinates are (0, 9).
AB =
=
=
= 13 units
Question 26
Find the point on y-axis whose distances from the points A (6, 7) and B (4, -3) are in the ratio 1: 2.
Sol:Let the required point on y-axis be P (0, y).
PA =
=
=
PB =
=
=
From the given information, we have:
4y2 - 56y + 340
= y2 + 6y + 25
3y2 - 62y + 315
= 0
y =
y =
y =
Thus, the required points on y-axis are (0, 9) and
Question 27
The distances of point P (x, y) from the points A (1, - 3) and B (- 2, 2) are in the ratio 2: 3.
Show that: 5x2 + 5y2 - 34x + 70y + 58 = 0.
It is given that PA: PB = 2: 3
9(x2 - 2x + y2 + 10 + 6y) = 4(x2 + 4x + y2 + 8 - 4y)
9x2 - 18x + 9y2 + 90 + 54y = 4x2 + 16x + 4y2 + 32 - 16y
5x2 + 5y2 - 34x + 70y + 58 = 0
Hence, proved.
Question 28
The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.
Sol:AB =
=
=
BC =
=
=
CA =
=
=
Since, triangle ABC is a right-angled at A, we have:
AB2 + AC2 = BC2
⇒ a2 - 6a + 13 + 2 = a2 - 8a + 17
⇒ 2a = 2
⇒ a = 1
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