Co-Ordinate Geometry A-1

CLASS : XIth SUBJECT : MATHS
DATE : Solutions DPP NO. :1
Topic :- CO-ORDINATE GEOMETRY
1 (a)
𝑎 cos 𝐵 ― cos2 𝐶) + 𝑏2(cos2 𝐶 ― cos2 𝐴) + 𝑐2 ( cos2 𝐴 ― cos2 𝐵 )
2( 2
= 𝑎2(1 ― sin2 𝐵 ― 1 + sin2 𝐶) + 𝑏2(1 ― sin2 𝐶 ― 1 + sin2 𝐴)

+ 𝑐2 (1 ― sin2 𝐴 ― 1 + sin2 𝐵 )
= 𝑎2(sin2 𝐶 ― sin2 𝐵) + 𝑏2(sin2 𝐴 ― sin2 𝐶) + 𝑐2 (sin2 𝐵 ― sin2 𝐴)
= 𝑘2𝑎2(𝑐2 ― 𝑏2) + 𝑘2𝑏2(𝑎2 ― 𝑐2) + 𝑘2𝑐2 (𝑏2 ― 𝑐2)
=0
2 (a)
Let sin 𝐴 = 3𝑘,sin 𝐵 = 4𝑘,sin 𝐶 = 5𝑘
sin 𝐴 sin 𝐵 sin 𝐶
∵ 𝑎
=
= 𝑐 = 𝑝 [say]
𝑏
3𝑘 4𝑘 5𝑘
⇒ = = =𝑝
𝑎 𝑏 𝑐
𝑘 𝑘
⇒ 𝑎=3 ,𝑏=4 ,𝑐=5
𝑝 𝑝 () 𝑘
𝑝 () ()
⇒ 𝑎 = 3𝑙, 𝑏 = 4𝑙, 𝑐 = 5𝑙 [let 𝑙 = 𝑝𝑘]
𝑏2 + 𝑐2 ― 𝑎2
∴ cos 𝐴 =
2𝑏𝑐
16 + 25 ― 9 32 4
= = =
2×4×5 40 5
𝑐2 + 𝑎2 ― 𝑏2
∴ cos 𝐵 =
2𝑎𝑐
25 + 9 ― 16 18 3
= = =
2×3×5 30 5
43
Now, cos 𝐴:cos 𝐵 = 5:5 = 4:3
4 (b)
Slope of perpendicular to the line joining the points
cos α ― cos β
(𝑎cos α, 𝑎sin α) and (𝑎 cos β , 𝑎 sin β) = ― sin α ― sin β
α+β
= tan
2
Hence, equation of perpendicular is
𝑦 = tan ( )𝑥
α+β
…(i)
2
Now, on solving the equation of line with Eq. (i), we get
𝑎 𝑎
[2
(cos α + cos β), ( sin α + sin β )
2
]
5 (a)
∆𝑃𝐵𝐶
[
Area of ∆ 𝐴𝐵𝐶 = { {6(5 ]
― 3( ―2 ― 𝑦) + 4(𝑦 ― 5) + 𝑥(5 + 2)}
+ 2) ― 3( ―2 ― 3) + 4(3 ― 5)}
7𝑥 + 7𝑦 ― 14 𝑥+𝑦―2
= | 49
= | |
7 |
6 (b)
Let 𝑃𝑄 and 𝑅𝑆 be the poles of height 20 m and 80 m subtending angles α and β at 𝑅 and 𝑃
respectively. Let ℎ be the height of the point 𝑇, the intersection of 𝑄𝑅 and 𝑃𝑆
S
Q
80 m T
20 m
h
R V P
Then, 𝑃𝑅 = ℎcot α +ℎcot β
= 20 cot α = 80 cot β
⇒ cot 𝛼 = 4 cot β
cot α
⇒ =4
cot β
Again, ℎcot 𝛼 +ℎcot β = 20cot 𝛼
⇒ (ℎ ― 20)cot 𝛼 = ―ℎ cot β
cot α ℎ
⇒ = =4
cot β 20 ― ℎ
⇒ ℎ = 80 ― 4ℎ
⇒ ℎ = 16 m
8 (a)
Since, α, β, γ are the roots of the equation
𝑥3 ― 3𝑝𝑥2 + 3𝑞𝑥 ― 1 = 0
∴ α + β + γ = 3𝑝
αβ + βγ + γα = 3𝑞
and αβγ = 1
Let 𝐺(𝑥, 𝑦) be the centroid of the given triangle
α+β+γ
∴ 𝑥= =𝑝
3
1 1 1
+ +
and 𝑦 = α β γ
3
βγ + γα + αβ
= =𝑞
3αβγ
Hence, coordinates of the centroid of triangle are (𝑝, 𝑞)
9 (d)
Let 𝑂(0, 0) be the orthocenter, 𝐴(ℎ, 𝑘) be the third vertex and 𝐵( ― 2, 3) and 𝐶(5, ― 1) the other
𝑘
two vertices. Then, the slope of the line through 𝐴 and 𝑂 is ℎ, while the line through 𝐵 and 𝐶 has the
( ― 1 ― 3) 4
slope (5 + 2)
= ― 7. By the property of the orthocenter, these two lines must be perpendicular, so
we have
(ℎ𝑘)( ― 47) = ―1 ⇒ℎ𝑘 = 74 …(i)
5―2+ℎ ―1 + 3 + 𝑘
Also, 3
+ 3
=7
⇒ ℎ + 𝑘 = 16 …(ii)
Which is not satisfied by the points given in the options (a), (b) or (c)
10 (b)
Let (ℎ, 𝑘) be the point
According to question,
4 (ℎ ― ℎ)2 + 𝑘 = ℎ2 + 𝑘2
⇒ 4|𝑘| = ℎ2 + 𝑘2
Locus of the point is
4|𝑦| = 𝑥2 + 𝑦2 ⇒ 𝑥2 + 𝑦2 ― 4|𝑦| = 0
12 (a)
Given points are 𝑃(4, ― 2),𝐴(2, ― 4) and 𝐵(7,1)
Suppose 𝑃 divides 𝐴𝐵 in the ratio 𝜆 :1. Then,
7𝜆 + 2 2
= 4⇒𝜆 =
𝜆+1 3
Thus, 𝑃 divides 𝐴𝐵 internally in the ratio 2 :3
The coordinates of the point dividing 𝐴𝐵 externally in the ratio 2 :3 are
2 × 7 ― 3 × 2 2 × 1 ― 3 × ―4
( 2―3
,
2―3 )
= ( ― 8, ― 14)
Hence, the harmonic conjugate of 𝑅 with respect to 𝐴 and 𝐵 is ( ― 8, ― 14)

13 (c)
If 𝑂 is the origin and 𝑃(𝑥1,𝑦1),𝑄(𝑥2,𝑦2) are two points, then
𝑂𝑃 × 𝑂𝑄 cos ∠𝑃𝑂𝑄 = 𝑥1𝑥2 + 𝑦1𝑦2
∴ 𝑂𝑃 × 𝑂𝑄 × sin ∠𝑃𝑂𝑄
= 𝑂𝑃2 × 𝑂𝑄2 ― 𝑂𝑃2 × 𝑂𝑄2 × cos2 ∠𝑃𝑂𝑄
= (𝑥21 + 𝑦21)(𝑥22 + 𝑦22) ― (𝑥1𝑥2 + 𝑦1𝑦2)2
= (𝑥1𝑦2 ― 𝑥2𝑦1)2 = |𝑥1𝑦2 ― 𝑥2𝑦1|
14 (c)
(3)2 + (5)2 ― (4)2 3
cos 𝐵 = =
2×3×5 5
9 4
⇒ sin 𝐵 = 1 ― =
25 5
∴ sin 2𝐵 = 2 sin 𝐵 cos 𝐵
4 3 24
=2× × =
5 5 25
16 (d)
Given that, ∠𝐴 = 45°, ∠𝐵 = 75°
∠𝑐 = 180° ― 45° ― 75° = 60°
∴ 𝑎 + 𝑐 2 = 𝑘( sin 𝐴 + 2 sin 𝐶 )
= 𝑘( sin 45° + 2 sin 60 °)
=𝑘 ( 1
2
+ 2
2
3
) = 𝑘( 1+ 3
2
) …(i)
𝑏
And 𝑘 = sin 𝐵
𝑏 2 2𝑏
= =
sin 75° 3+1
On putting the value of 𝑘 in Eq. (i), we get
𝑎 + 𝑐 2 = 2𝑏
18 (d)
From figure 𝐴𝐵𝐶𝐷 is s square
Whose diagonals 𝐴𝐶 and 𝐵𝐷 are of length 2 unit

1
Hence, required area = 2 𝐴𝐶 × 𝐵𝐷
1
= 2 × 2 × 2 = 2 sq units
19 (c)
In ∆𝐴𝑃𝐷,
𝑎
tan 45° = ⇒𝐴𝑃 = 𝑎
𝐴𝑃
C
D E
b
a
45o 45o
A P B
and in ∆ 𝐵𝑃𝐶,
𝑏
tan 45° =
𝑃𝐵
⇒𝑃𝐵 = 𝑏
∴ 𝐷𝐸 = 𝑎 + 𝑏 and 𝐶𝐸 = 𝑏 ― 𝑎
In ∆𝐷𝐸𝐶,
𝐷𝐶2 = 𝐷𝐸2 + 𝐸𝐶2
= (𝑎 + 𝑏2) + (𝑏 ― 𝑎2)
= 2(𝑎2 + 𝑏2)
20 (b)
If the axes are rotated through 30°, we have
3𝑋 ― 4
𝑥 = 𝑋 cos 30° ― 𝑌 sin 30° =
2
𝑋 + 3𝑌
and, 𝑦 = 𝑋 sin 30° + 𝑌 cos 30° =
2
Substituting these values in 𝑥 +2 𝑥𝑦 ― 𝑦2 = 2𝑎2, we get
2 3
( 3𝑋 ― 𝑌)2 + 2 3( 3𝑋 ― 𝑌)(𝑋 + 3𝑌) ― (𝑋 + 3𝑌)2 = 8𝑎2
⇒𝑋2 ― 𝑌2 = 𝑎2
ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. A A D B A B D A D B
Q. 11 12 13 14 15 16 17 18 19 20
A. B A C C B D C D C B

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CLASS : XIth SUBJECT : MATHS

DATE : Solutions DPP NO. :1

Topic :- CO-ORDINATE GEOMETRY

= 𝑎2(1 ― sin2 𝐵 ― 1 + sin2 𝐶) + 𝑏2(1 ― sin2 𝐶 ― 1 + sin2 𝐴)

Hence, the harmonic conjugate of 𝑅 with respect to 𝐴 and 𝐵 is ( ― 8, ― 14)

Whose diagonals 𝐴𝐶 and 𝐵𝐷 are of length 2 unit

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