# SELINA Solutions for Class 9 Maths Chapter 28 - Distance Formula

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

## Chapter 28 - Distance Formula Exercise Ex. 28

Find the distance between the following pairs of points:

(i) (-3, 6) and (2, -6)

(ii) (-a, -b) and (a, b)

(iii) and

(iv) and

(i) (-3, 6) and (2, -6)

Distance between the given points

(ii) (-a, -b) and (a, b)

Distance between the given points

(iii) and

Distance between the given points

(iv) and

Distance between the given points

Find the distance between the origin and the point:

(i) (-8, 6) (ii) (-5, -12) (iii) (8, -15)

Coordinates of origin are O (0, 0).

(i) A (-8, 6)

AO =

(ii) B (-5, -12)

BO =

(iii) C (8, -15)

CO =

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.

Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).

Let the coordinates of the point on x-axis be (x, 0).

From the given information, we have:

Thus, the required co-ordinates of the points on x-axis are (26, 0) and (-4, 0).

Find the co-ordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).

Let the coordinates of the point on y-axis be (0, y).

From the given information, we have:

Thus, the required co-ordinates of the points on y-axis are (0, 10) and (0, -2).

A point A is at a distance of unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.

It is given that the co-ordinates of point A are such that its ordinate is twice its abscissa.

So, let the co-ordinates of point A be (x, 2x).

We have:

Thus, the co-ordinates of the point A are (1, 2) and (3, 6).

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

Given that the point P (2, -1) is equidistant from the points A (a, 7) and B (-3, a).

What point on the x-axis is equidistant from the points (7, 6) and (-3, 4)?

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

Thus, the required point is (3, 0).

Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).

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

Thus, the required point is (0, -2).

A point P lies on the x-axis and another point Q lies on the y-axis.

(i) Write the ordinate of point P.

(ii) Write the abscissa of point Q.

(iii) If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.

(i) Since, the point P lies on the x-axis, its ordinate is 0.

(ii) Since, the point Q lies on the y-axis, its abscissa is 0.

(iii) The co-ordinates of P and Q are (-12, 0) and (0, -16) respectively.

PQ =

Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.

Since, PQ = QR, PQR is an isosceles triangle.

Prove that the points P(0, -4), Q(6, 2), R(3, 5) and S(-3, -1) are the vertices of a rectangle PQRS.

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.

Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

Since, AB = BC = CD = DA and AC = BD,

A, B, C and D are the vertices of a square.

Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

Let the given points be A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4).

Since, AB = BC = CD = DA and AC BD

The given vertices are the vertices of a rhombus.

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.

AB = CD

AB^{2} = CD^{2}

(-6 + 3)^{2} + (a + 2)^{2} = (0 + 3)^{2} + (-1 + 4)^{2}

9 + a^{2} + 4 + 4a = 9 + 9

a^{2} + 4a - 5 = 0

a^{2} - 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.

The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.

Let the circumcentre be P (x, y).

Then, PA = PB

PA^{2} = PB^{2}

(x - 5)^{2} + (y - 1)^{2} = (x - 11)^{2} + (y - 1)^{2}

x^{2} + 25 - 10x = x^{2} + 121 - 22x

12x = 96

x = 8

Also, PA = PC

PA^{2} = PC^{2}

(x - 5)^{2} + (y - 1)^{2} = (x - 11)^{2} + (y - 9)^{2}

x^{2} + 25 - 10x + y^{2} + 1 - 2y = x^{2} + 121 - 22x + y^{2} + 81 - 18y

12x + 16y = 176

3x + 4y = 44

24 + 4y = 44

4y = 20

y = 5

Thus, the co-ordinates of the circumcentre of the triangle are (8, 5).

Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.

AB = 5

AB^{2} = 25

(0 - 3)^{2} + (y - 1 - 1)^{2} = 25

9 + y^{2} + 4 - 4y = 25

y^{2} - 4y - 12 = 0

y^{2} - 6y + 2y - 12 = 0

y(y - 6) + 2(y - 6) = 0

(y - 6) (y + 2) = 0

y = 6, -2

Given A = (x + 2, -2) and B (11, 6). Find x if AB = 17.

AB = 17

AB^{2} = 289

(11 - x - 2)^{2} + (6 + 2)^{2} = 289

x^{2} + 81 - 18x + 64 = 289

x^{2} - 18x - 144 = 0

x^{2} - 24x + 6x - 144 = 0

x(x - 24) + 6(x - 24) = 0

(x - 24) (x + 6) = 0

x = 24, -6

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.

Distance between the points A (2x - 1, 3x + 1) and B (-3, -1) = Radius of circle

AB = 10 (Since, diameter = 20 units, given)

AB^{2} = 100

(-3 - 2x + 1)^{2} + (-1 - 3x - 1)^{2} = 100

(-2 - 2x)^{2} + (-2 - 3x)^{2} = 100

4 + 4x^{2} + 8x + 4 + 9x^{2} + 12x = 100

13x^{2} + 20x - 92 = 0

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.

Let the co-ordinates of point Q be (10, y).

PQ = 10

PQ^{2} = 100

(10 - 2)^{2} + (y + 3)^{2} = 100

64 + y^{2} + 9 + 6y = 100

y^{2} + 6y - 27 = 0

y^{2} + 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).

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:

(i) AT (ii) AB.

(i) Given, radius = 13 units

PA = PB = 13 units

Using distance formula,

Using Pythagoras theorem in PAT,

AT^{2} = PA^{2} - PT^{2} = 169 - 25 = 144

AT = 12 units

(ii) We know that the perpendicular from the centre of a circle to a chord bisects the chord.

AB = 2AT = 2 12 units = 24 units

Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.

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).

Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.

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).

Find the point on y-axis whose distances from the points A (6, 7) and B (4, -3) are in the ratio 1: 2.

Let the required point on y-axis be P (0, y).

From the given information, we have:

Thus, the required points on y-axis are (0, 9) and.

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: 5x^{2} + 5y^{2} - 34x + 70y + 58 = 0.

It is given that PA: PB = 2: 3

Hence, proved.

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.

Since, triangle ABC is a right-angled at A, we have:

AB^{2} + AC^{2} = BC^{2}

a^{2} - 6a + 13 + 2 = a^{2} - 8a + 17

2a = 2

a = 1

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