This note is designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations.

QUESTION 1

Show that the points A ( – 4, 9, 6), B (– 1, 6, 6), and C(0, 7, 10) right-angled isosceles triangle.

Sol:

Distance between two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is PQ =

AB = BC =

AB² + BC² = 18 + 18 = 36 = AC²

∴ ∠ B = 90⁰

points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle

QUESTION 2

Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x ² + 9y² + 8 z² – 18x – 36y + 18z + 54 = 0

Sol:

Let P (x, y, z) be the locus of the point

A (0, y, 0) be any point on Y – axis

B = (1, 2, – 1)

Condition is PA = 3PB

PA² = (3PB)²

z² = 9[x²– 2x + 1 + y² – 4y + 4 + z² + 2z + 1]

x² + z² = 9x²– 18x + 9 + 9y² – 36y + 36 +9 z² + 18z + 9

∴ 8x ² + 9y² + 8 z² – 18x – 36y + 18z + 54 =0

QUESTION 3

A, B, C are three points on OX, OY and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C and O

Sol:

Let P (x, y, z) be the required point

O = (0, 0, 0) A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)

Given, AP = BP = CP = OP

AP = OP

⟹ AP² =OP²

(x – a )²+ y² + z² = x² + y² + z²

x² + a² – 2ax + y² + z² = x² + y² + z²

a² – 2ax = 0

a (a – 2x) = 0

a – 2x = 0 (∵ a≠0)

a = 2x ⟹ a/2

Similarly, y = b/2 and z = c/2

∴ P = (a/2, b/2, c/2)

QUESTION 4

Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear

Sol:

Given points are A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2)

AB + BC =

AB + BC = AC

∴ A B and C are collinear

QUESTION 5

Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.

Sol:

Let A = (5, – 1, 7), B = (x, 5, 1)

Given AB = 9

⟹ AB² = 81

(5 – x)² + (– 1 – 5)² + (7 – 1)² = 81

(5 – x)²+ 36 + 36 = 81

(5 – x)² + 72 = 81

(5 – x)²= 81 – 72 = 9

(5 – x) = ± 3

5 – x = 3 or 5 – x = – 3

5 – 3 = x or 5 + 3 = x

x = 2 or x = 8

QUESTION 6

If the point (1, 2, 3) is changed to the point (2, 3, 1) through the translation of axes. Find a new origin.

Sol:

Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)

x = X + h, y = Y + k, z = Z + l

h = x – X, k = y – Y, l = z – Z

h = 1 – 2, k = 2 – 3, l = 3 – 1

h = – 1, k = – 1, l = 2

New origin is (– 1, – 1, 2)

QUESTION 7

By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

Sol:

If a point P divides the line segment joining the points (x₁, y₁, z₁), (x₂, y₂, z₂) in the ratio, then P =

Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

QUESTION 8

Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).

Sol:

Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)

Suppose P divides AB in the ratio k : 1

P =

(1,2,3) =

1 =

7 – 2k = k + 1

7 – 1 = k + 2k

6 = 3k

k = 2

∴ P divides AB in the Ratio 1 : 2

QUESTION 9

Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not.

Sol:

Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)

Let C divides AB in the ratio k : 1

C =

(4, –7, 2) =

4 =

k = 4 (k + 1)

2 – 4k = 4k + 4

– 4k– 4k = 4 – 2

– 8k = 2

K = -1/4

C divides AB in the Ratio 1 : 4 externally

∴ A, B, C are collinear

QUESTION 10

Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)

Sol:

The centroid of the triangle whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is

∴ The centroid of the triangle =

QUESTION 11

Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)

Sol:

The centroid of the triangle whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) (x₃, y₃, z₃), and (x₄, y₄, z₄) is

the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is

QUESTION 12

Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)

Sol:

Let P be any point on the YZ-plane

P = (0, y, z)

Let P divides AB in the ratio k:1

P =

⟹ (0, y, z) =

⟹ = 0

3k + 2 = 0

3k =– 2

k =

YZ-plane divides AB in the ratio – 2:3

QUESTION 13

Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).

Sol:

let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)

ABCD is a parallelogram

The midpoint of AC = Midpoint of BD

⟹

⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0

x = 3, y = 3, z = 1

∴ The fourth vertex D = (3, 3, 1)

QUESTION 14

A (5, 4, 6), B (1, –1, 3), C (3, 3, 1) are three points. Find the coordinates of the point in which the bisector of ∠BAC meets the side BC.

Sol:

We know that the bisector of ∠BAC divides BC in the ratio AB:AC

AB =

AC =

AB:AC = 5 : 3

= 5:3

If D is a point where the bisector of ∠BAC meets BC

⟹ D divides BC in the ratio 5:3

QUESTION 15

If M (α, β, γ) is the midpoint of the line joining the points (x₁, y₁, z₁) and B, then find B

Sol:

Let B (x, y, z) be the required point

M is the midpoint of AB

⟹ (α, β, γ) =

⟹ 2 α = x + x₁; 2 β = y + y₁; 2 γ = z + z₁

x =2 α – x₁; x =2 β – y_1; x =2 γ – z₁

∴ B = (2 α – x₁, 2 β – y₁, 2 γ –)

QUESTION 16

If H, G, S and I respectively denote orthocenter, centroid, circumcenter and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.

Sol:

Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)

AB =

BC =

AC=

AB = BC = AC

⟹ ∆ ABC is an equilateral triangle

We know that, in an equilateral triangle orthocenter, centroid, circumcenter and incentre are same

Centroid G =

= (2, 2, 2)

∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)

QUESTION 17

Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).

Sol:

Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)

a = 5

b = 4

c = 3

incentre =

⟹ I =

∴ I = (1, 1,0)

QUESTION 18

Find the ratio in which the point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also, find the harmonic conjugate of P

Sol:

Harmonic Conjugate: If P divides AB in the ratio m: n, then the Harmonic Conjugate of P (i.e., Q) divides AB in the ratio –m: n.

Given points are A (3, 2, – 4), B (9, 8, –10) and P (5, 4, – 6)

P divides AB in the ratio is x₂ – x : x – x₂

= 3 – 5 : 5 – 9

= 1 : 2 (internally)

Let Q be the harmonic conjugate of P

⟹ Q divides AB in the ratio –1: 2

Q =