# These solutions 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.

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

**Sol:**

Distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is PQ =

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

**2.Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x ^{2} + 9y^{2} + 8 z^{2} – 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

PA^{2} = (3PB)^{2}

PA^{2}= 9 PB^{2 }

⟹ x^{2} + z^{2} = 9[(x – 1)^{2} + (y – 2)^{2} + (z + 1)^{2}]

x^{2} + z^{2} = 9[x^{2 }– 2x + 1 + y^{2} – 4y + 4 + z^{2} + 2z + 1]

x^{2} + z^{2} = 9x^{2 }– 18x + 9 + 9y^{2} – 36y + 36 +9 z^{2} + 18z + 9

∴ 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 = 0

**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

⟹ AP^{2} =OP^{2}

(x – a )^{2 }+ y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

x^{2} + a^{2} – 2ax + y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

a^{2} – 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)

**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)

**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

⟹ AB^{2} = 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

**6.If the point (1, 2, 3) is changes to the point (2, 3, 1) through translation of axes. Find 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)

**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 (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) in the ratio, then

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

**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

**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

2 – 4k = 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

**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 (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) is

**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 tetrahedron whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

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

**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 int eh ratio k:1

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

**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

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)

**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

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

⟹ D divides BC in the ratio 5:3

**15. If M (α, β, γ) is the midpoint of the line joining the points (x _{1}, y_{1}, z_{1}) 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 + x_{1}; 2 β = y + y_{1}; 2 γ = z + z_{1}

x =2 α – x_{1}; x =2 β – y_{1;} x =2 γ – z_{1}

_{ }∴ B = (2 α – x_{1}, 2 β – y_{1}, 2 γ –)

**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)

⟹ ∆ 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)

**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)

∴ I = (1, 1,0)

**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 hormonic conjugate of P**

**Sol:**

Hormonic Conjugate: If P divides AB in the ratio m : n, then the Hormonic 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 x_{2} – x : x – x_{2}

= 3 – 5 : 5 – 9

= 1 : 2 (internally)

Let Q be the hormonic conjugate of P

⟹ Q divides AB in the ratio –1 : 2

Q =

= (–3, –4, –2)

Q (–3, –4, –2) is the hormonic conjugate of P (5, 4, – 6)

**19. If (3, 2, – 1), (4, 1,1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of tetrahedron, find the fourth Vertex.**

**Sol:**

Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)

Let the fourth vertex is D = (x, y, z)

The centroid of the tetrahedron whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

13 + x = 16, 5 + y = 8, 5 + z = 2

x = 3, y = 3, z = 3

∴ the fourth vertex D = (3, 3, 3)

**20. Show that the points A(3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear and find the ratio in which B divides AC.**

**Sol:**

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

∴ the points A (3, 2, –4), (5, 4, –6) and C (9, 8, –10) are collinear

B divides AB in the ratio is AB:BC =

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