Maharashtra Board Class 10 Math Part – 2 Solution Chapter 5 Practice Set 5.1 – Co-ordinate Geometry
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 5: Co-ordinate Geometry. Marathi or English Medium Students of Class 10 get here Co-ordinate Geometry full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Co-ordinate Geometry |
Practice Set |
5.1 |
Practice set 5.1
(1) Find the distance between each of the following pairs of points.
(1)A(2, 3), B(4, 1)
(2) P(-5, 7), Q(-1, 3)
(3) R (0, -3) S (0, 5/2
(4) L (5, -8), (-7, -3)
(5) T (-3, 6), R (9, -10)
(6) W (-7/2, 4), X (11 (11, 4)
Solution:
(1) Given A (2, 3), B (4, 1)
∴ AB = √(4 – 2)2 + (1 – 3)2 [Distance formula]
= √22 + (-2)2
= √4+4
= √8
= 2√2
(2) Given, P (-5, 7), Q (-1, 3)
∴ PQ √{-1 – (-5)}2 + (3 – 7)2 [Distance formula]
= √42 + (-4)2
= √16+16
= √32
= 4√2
(3) Given, R (0, – 3), S (0, 5/2)
∴ RS = √(0 – 0)2 + {5/2 – (3)}2
= √0+ (5+6/2)2
= √(11/2)2
= 11/2
(4) Given, L (5, -8), M (-7, -3)
∴ LM = √(-7 – 5)2 + {-3 – (-8)}2
= √(-12)2+ 52
= √144 + 25
= √169
= 13
(5) Given, T (-3, 6), R (9, – 10)
∴ TR = √{9 – (-3)}2 + (-10, -6)2
= √122 + (-16)2
= √144 + 256
= √400
= 20
(6) Given, W (- 7/2, 4), × (11, 4)
∴ W × = √{11 – (- 7/2)}2 + (4 – 4)2
= √(22+7/2)2 + 82
= √(29/2)2
= 29/2
(2) Determine whether the points are collinear.
(1) A(1, -3), B(2, -5), C(-4, 7)
(2) L(-2, 3), M(1, -3), N(5, 4)
(3) R(0, 3), D(2, 1), S(3, -1)
(4) P(-2, 3), Q(1, 2), R(4, 1)
Solution:
(1) Given, A (1, -3), B (2, -5), C (-4, 7)
∴ AB = √(2 – 1)2 + (-5 – (-3)}2
= √12 + (-2)2
= √1+4
= √5 —- (i)
BC = √(-4 – 2)2 + {7 – (-5)}2
= √(-6)2 + 122
= √36+144
= √180
= 6√5 —- (2)
AC = √(-4-1)2 + (7 – (-3)2
= √(-5) + 102
= √25+100
= √125
= 5√5 — (3)
Now, From 1, 2, 3, clearly BC is the greatest.
∴ AB + AC = √5 + 5√5 = √5 (1+5) = 6√5 = BC [From 2]
∴ AB + AC = BC
∴ Points A, B & C are collinear.
(2) Given, L (-2, 3) M (1, -3), N (5, 4)
∴ LM = √{1 – (-2)}2 + (-3 – 3)2
= √32 + (-6)2
= √9+36
= √45
Or, LM = 3√5 —- (i)
MN = √(5 – 1)2 + {4 – (-3)}2
= √42 + 72
= √16 + 49
= √65
Or, MN = √13 × √5 — (ii)
LN = √{5 – (-2)}2 + (4 – 3)2
= √72 + 12
=49+1
= √50
Or, LN = 5√2 —- (iii)
Comparing equation (i), (ii) & (iii)
Clearly, MN is the greatest ∵ √65 > √50 > √45
∴ LM + LN = 3√5 + 5√2 ≠ MN
∴ Points are not collinear.
(3) Given R (0, 3), D (2, 1), S(3, -1)
∴ RD = √(2 – 0)2 + (1 – 3)2
= √22 + (-2)2
= √4+4
= √8
Or, RD = 2√2 —- (i)
DS = √(3 – 2)2 + (-1 – 1)2
= √12 + (-2)2
Or, DS = √1+4
= √5 —- (ii)
RS = √(3 – 6)2 + (-1 -3)2
= √32 + (-4)2
= √9 + 16
= √25
Or, RS = 5 —- (iii)
From equation (i) (ii) & (iii)
Clearly, RS is greatest
∴ RD + DS = 2√2 + √5 ≠ RS
∴ Points R, D & S are not collinear.
(4) P (-2, 3), Q (1, 2), R (4, 1)
∴ PQ = √{1 – (-2)}2 + (2 – 3)2
= √32 + (-1)2
= √9+1
Or, PQ = √10 —– (1)
QR = √(4 – 1)2 + (1 – 2)2
= √32 + (-1)2
= √9+1
Or, QR = √10 —- (2)
PR = √{4 – (-2)}2 + (1 – 3)2
= √62 + (-2)2
= √36+4
Or, PR = √40 —– (3)
From (1), (2), (3)
Clearly PR > QR & PQ
∴ QR + PQ = √10 + √10
= 2√10
PR = √40 = 2√10
∴ QR + PQ = PR
∴ Points, P, Q & R are collinear
(3) Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).
Solution:
Given points are A (-3-4), B (1, -4)
Let, the point on a X-axis be C (x, O)
∴d (AC) =√{x – (-3)}2 + (0 – 4)2
= √(x + 3)2 + (-4)2
= √(x + 3)2 + 16
d (BC) = √(x – 1)2 + {0 (-4)}2
= √(x – 1)2 + 42
= √(x – 1)2 + 42
= √(x – 1)2 + 16
(ATQ) d (AC) = d (BC)
Or, √(x+3)2 + 16 = √(x-1)2 + 16
Or, (x + 3)2 + 16 = (x – 1)2 + 16 [Squaring both sides]
Or, (x + 3)2 – (x – 1)2 = 0
Or, (x + 3 – x + 7) (x + 3 + x – 1) = 0
Or, 4 (2x + 2) = 0
Or, 2x + 2 = 0
Or, 2x = -2
Or, x = -1
∴ The point equidistance from point A & B. is C (-1, 0)
(4) Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Given, points are P (-2, 2), Q (2, 2), R (2, 7)
∴ PQ = √{2 – (-2)} + (2 – 2)2
= √(2 +2)2 + 0
= √42
Or, PQ = 4 —- (1)
QR = √(2 – 2)2 + (7 – 2)2
= √0 + 52
Or, QR = 5 — (2)
PR = √{2 – (-2)}2 + (7 – 2)2
= √42 + 52
= √16 + 25
Or, PR = √41 —- (3)
If points P, Q, R are to be vertices of aright angled triangle it must satisfy Pythagoras theorem.
From (1), (2), (3), clearly PR is the greatest.
∴ Let, consider PR as hypotenuse.
∴ PR2 = (√41)2 = 41 —- (4)
PQ2 = 42 = 16, QR2 = 52 = 25
∴ PQ2 + QR2 = 16 + 25 = 41 – (5)
∴ PR2 = PQ2 + QR2 [From (4) & (5)]
∴ Points P (-2, 2), Q (2, 2), R (2, 7) satisfices
Pythagoras theorem having PR as hypotenuse.
Hence ∆PQR is a right angled triangle (Proved)
(5) Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram.
Solution:
Given, point are P (2, -2), Q (7, 3), R (11, -1), S (6, -6)
∴ PQ = √(7 – 2)2 + {3 – (-2)}2
= √52 + 52
= √25 + 25
Or, PQ = √50 —- (i)
QR = √(11 – 7)2 + (-1 – 3)2
= √42 + (-4)2
= √16 + 16
Or, QR = √32 —- (ii)
RS = √(6 – 11)2 + {-6 – (-1)}2
= √(-5)2 + (-5)2
= √25+25
Or, RS = √50 — (iii)
PR = √(6 – 2)2 + {-6 – (-2)}2
= √42 + (-4)2
= √16 + 16
Or, RS = √50 —- (iii)
Or, PR = √32 — (iv)
Now, From (i), (ii), (iii) & (iv) we can see that,
PQ = RS & QR = PR
∴ If we consider PQRS we can see that opposite sides are equal.
Therefore, PQRS is a parallelogram
(6) Show that points A(-4, -7), B(-1, 2), C(8, 5) and D (5, -4) are vertices of a rhombus ABCD.
Solution:
The given points are A (-4, -7), B (-1, 2)
C (8, 5) & D (5, -4)
∴ AB = √{-1 – (-4)}2 + (2 – (-7)}2
= √32 + 92
= √9-81
Or, AB = √90 —– (1)
BC = √{8 – (-1)}2 + (5 – 2)2
= √92 + 32
= √81 + 9
Or, BC = √90 —- (2)
CD = √(5 – 8)2 + (-4 – 5)2
= √32 + (9)2
= √9+81
Or, CD = √90 – (3)
AD = √{5 – (-4)}2 + {-4 – (-7)}2
= √92 + 32
= √81 + 9
Or, AD = √90 – (4)
Now, AC = √{8 – (-4)}2 + {5 – (-7)}2
= √122 + 122 = √144+144
= √288
Or, AC = 15
BD = √{5 – (-1)}2 + (-4-2)
= √62 + (-6)2
= √56 + 36
Or, BC = √72
Now, For ABCD to be rhombus it has to satisfy the following conditions.
(1) All sides must be equal
(2) Diagonals must bisect each other at right angles.
From (1), (2), (3) & (4)
AB = BC
CD = AD
∴ Condition 1 is satisfied.
Now let’s check condition 2.
Now, AC & BD are the diagonals for them to be at right angle we have to test them under Pythagoras theorem with any of the sides.
So to prove,
AB2 = (AC/2)2 + (BD/2)2 [∵ diagonals bisect each other]
L.H.S
AB2 = (√90)2
= 90
R.H.S
(AC/2)2 + (BD/2)2
= (√288/2)2 + (√72/2)2
= 288/4 + 72/4
= 360/4
= 90
∴ L.H.S = R.H.S
Or, AB2 = (AC/2)2 + (BD/2)2
∴ Condition 2 is satisfied
Therefore, ABCD is a rhombus.
(7) Find x if distance between points L(x, 7) and M(1, 15) is 10.
Solution:
The given points are L (x, 7), M (1, 15)
Also LM = 10 [Given]
Also LM = √(1 – x)2 + (15 – 7)2
Or, 10 = √(1 – x)2 + 82
Or, 102 = (1 – x)2 + 64
Or, (1-x)2 = 100 – 64
Or, 1 – x = √36
Or, 1 – x = 6
Or, x = -5
(8) Show that the points A (1, 2), B (1, 6), C (1 + 2√3, 4) are vertices of an equilateral triangle.
Solution:
The given points are A (1, 2), B (1, 6), C (1 + 2√3, 4).
∴ AB = √(1 – 1)2 + (6 – 2)2
= √0+42
= 4 —- (1)
BC = √(1 + 2√3 – 1)2 + (4 – 6)2
= √(2√3)2 + (-2)2
= √12 + 4
= √16
BC = 4 —- (2)
AC = √(1 + 2√3 – 1)2 + (4 – 2)2
= √(2√3)2 + (2)2
= √12+4
= √16
Or, AC = 4 – 3
∴ AB = BC = AC [From 1, 2, 3]
∴ ∆ ABC is an equilateral triangle since all sides are equal.
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 5 Co-ordinate Geometry Practice Set 5.1
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