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 =
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 = 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
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
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
3k + 2 = 0
3k =– 2
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
= 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
⟹ 2 α = x + x1; 2 β = y + y1; 2 γ = z + z1
x =2 α – x1; x =2 β – y1; 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
⟹ ∆ ABC is an equilateral triangle
We know that, in an equilateral triangle orthocenter, centroid, circumcenter and incentre are same
= (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
∴ 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
= (–3, –4, –2)
Q (–3, –4, –2) is the harmonic conjugate of P (5, 4, – 6)
QUESTION 19
If (3, 2, – 1), (4, 1,1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of the 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 (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is
13 + x = 16, 5 + y = 8, 5 + z = 2
x = 3, y = 3, z = 3
∴ the fourth vertex D = (3, 3, 3)
QUESTION 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 = = 1:2
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