# ICSE Solutions for Selina Concise Chapter 28 Distance Formula Class 9 Maths

**Exercise 28**

**1 .**Find the distance between the following pairs of points:**(i) (-3, 6) and (2, -6 )****(ii) (-a, -b) and (a, b)****(iii) (3/5 ,2) and (-1/5 , 1 ²/5)****(iv) (√3+1, 1) and (0, √3)**

**Answer**

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

Distance between the given points

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

Distance between the given points

**(iv)**(√3 + 1, 1) and (0, √3)

Distance between the given points

**2.**Find the distance between the origin and the point:**(i) (-8, 6)****(ii) (-5, -12)****(iii) (8, -15)**

**Answer**

Coordinates of origin are O (0, 0). **(i)** A (-8, 6)

**3 .**The distance between the points (3, 1) and (0, x) is 5. Find x.**Answer**

It is given that the distance between the points A(3, 1) and B(0, x) is 5.

∴ AB = 5

AB^{2} = 25

⇒ (0-3)^{2} + (x-1)^{2} = 25

⇒ 9 + x^{2} + 1 – 2x = 25

⇒ x^{2} – 2x – 15 = 15

⇒ x^{2} – 5x + 3x – 15 = 0

⇒ x(x-5) + 3(x-5) = 0

⇒ (x-5) (x+3) = 0

⇒ x = 5, -3

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

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

From the given information, we have :

⇒ x = 26 , -4

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

**5 .**Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).**Answer**

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

From the given information, we have :

⇒ y = 10, -2

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

**6.**A point A is at a distance of √10 unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.**Answer**

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 :

⇒ x = 1, 3

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

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

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

∴ PA = PB

PA^{2} = PB^{2}

⇒ (a-2)^{2} + (7+1)^{2} = (-3-2)^{2} + (a+1)^{2}

⇒ a^{2} + 4 – 4a + 64 = 25 + a^{2} + 1 + 2a

⇒ 42 = 6a

⇒ a = 7

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

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

PA^{2} = PB^{2}

⇒ (x-7)^{2} + (0-6)^{2} = (x+3)^{2} + (0-4)^{2}

⇒ x^{2} + 49 – 14x + 36 = x^{2} + 9 + 6x + 16

⇒ 60 = 20 x

⇒ x = 3

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

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

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

PA^{2} = PB^{2}

⇒ (0-5)^{2} + (y-2)^{2} = (0+4)^{2} + (y-3)^{2}

⇒ 25 + y^{2} + 4 – 4y = 16 + y^{2} + 9 – 6y

⇒ 2y = -4

⇒ y = -2

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

**10.**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.**

**Answer **

**(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.

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

**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**Answer **

**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.**Answer**

Hence, A, B, C are the vertices of a right - angled triangle.

Hence, △ABCD is an isosceles right - angled triangle.

Area of △ABC = (1/2) × AB ×CA

= (1/2) × 5×5

= 12.5 sq. units

**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.**Answer **

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

**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.**Answer **

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 - 1)(a+5) = 0

⇒ a = 1 or -5

It is given that a is negative, thus the value of a is -5.

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

Let the circumcntre 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

⇒ 3x + 4y = 44

⇒ 24 + 4y = 44

⇒ 4y = 20

⇒ y = 5

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

**18.**Given A = (3, 1) and B = (0, y – 1). Find y if AB = 5.**Answer **

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 + 6y - 12 = 0

⇒ (y - 6)(y +2) = 0

⇒ y = 6, -2

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

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 -24)(x + 6) = 0

⇒ x = 24, -6

**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.**Answer **

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

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

**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.**Answer **

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

PQ = 10

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).

**22.**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:**(i) AT**

**(ii) AB**

**Answer **

**(i)** Given, radius = 13 units

∴ PA = PB = 13 units

Using distance formula,

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

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

**24.**Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.**Answer**

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).

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

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).

**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.**Answer **

Let the required point of y - axis be P(0

Show that: 5x

**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: 5x

^{2}+ 5y^{2}– 34x + 70y + 58 = 0.**Answer **

It is given that PA : PB = 2 : 3

**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.**Answer **