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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

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Question 1.
The equation of the locus of the point whose distance from y-axis is half the distance from origin is ……..
(a) x2 + 3y2 = 0
(b) x2 – 3y2 = 0
(c) 3x2 + y2 = 0
(d) 3x2 – y2 = 0
Solution:
(c) 3x2 + y2 = 0
Hint:
Given that PA = \([\frac{1}{2}/latex]OP
2PA = OP
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 1
4PA2 = OP2
4(x)2 = x2 + y2 ⇒ 3x2 – y2 = 0

Question 2.
Which of the following equation is the locus of (at2, 2at) ……
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 2
Solution:
(d) y2 = 4ax
Hint:
y2 = 4ax ⇒ Equation that satisfies the given point (at2, 2at)

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Question 3.
Which of the following point lie on the locus of 3x2 + 3y2 – 8x – 12y + 17 = 0?
(a) (0, 0)
(b) (-2, 3)
(c) (1, 2)
(d) (0, -1)
Solution:
(c) (1, 2)
Hint:
The point that satisfies the given equations (0, 0) ⇒ 17 ≠ 0
(-2, 3) ⇒ 3 (4) + 3 (9) + 16 – 36 + 17 ≠ 0
(1, 2) ⇒ 3 + 3 (4) – 8 (1) – 12 (2) + 17
32 – 32 = 0, 0 = 0

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 382
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 20

Question 5.
Straight line joining the points (2, 3) and (-1, 4) passes through the point (α, β) if
(a) α + 2β = 7
(b) 3α + β = 9
(c) α + 3β = 11
(d) 3α + 3β = 11
Solution:
(c) α + 3β = 11
Hint:
Equation joining (2, 3), (-1, 4)
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 25
3y – 12 = – x -1 ⇒ x + 3y – 11 = 0, (α, β) lies on it ⇒ α + 3β – 11 = 0.

Question 6.
The slope of the line which makes an angle 45° with the line 3x – y = – 5 are
(a) 1, -1
(b) [latex]\frac{1}{2},-2\)
(c) \(1, \frac{1}{2}\)
(d) \(2,-\frac{1}{2}\)
Solution:
(c) \(1, \frac{1}{2}\)
Hint:
Equation of line 3x – y = -5, y = 3x + 5, m1 = 3
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 26
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 27

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Question 7.
Equation of the straight line that forms an isosceles triangle with coordinate axes in the I-quadrant with perimeter \(4+2 \sqrt{2}\) is
(a)x + y + 2 = 0
(b) x + y – 2 = 0
(c) x + y – \(\sqrt{2}\) = 0
(d) x + y + \(\sqrt{2}\) = 0
Solution:
(b) x + y – 2 = 0
Hint.
Let the sides be x, x
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 28
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 288

Question 8.
The coordinates of the four vertices of a quadrilateral are (-2, 4), (-1, 2), (1, 2) and (2, 4) taken in order. The equation of the line passing through the vertex (-1, 2) and dividing the quadrilateral in the equal areas is ………
(a) x + 1 = 0
(b) x + y = 1
(c) x + y + 3 = 0
(d) x – y + 3 = 0
Solution:
(b) x + y = 1
Hint:
This equation passes through (-1, 2)
-1 + 2 = 1 ⇒ 1 = 1
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 29

Question 9.
The intercepts of the perpendicular bisector of the line segment joining (1, 2) and (3, 4) with coordinate axes are ……….
(a) 5, -5
(b) 5, 5
(c) 5, 3
(d) 5, -4
Solution:
(b) 5, 5
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 30

Question 10.
The equation of the line with slope 2 and the length of the perpendicular from the origin equal to \(\sqrt{5}\) is ……
(a) x + 2y = \(\sqrt{5}\)
(b) 2x + y = \(\sqrt{5}\)
(c) 2x + y = 5
(d) x + 2y – 5 = 0
Solution:
(c) 2x + y = 5
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 31
The required line is y = 2x + 5 ⇒ 2x – y + 5 = 0

Question 11.
A line perpendicular to the line 5x – y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5 sq. units, then its equation is …….
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 32
Solution:
(a) x + 5y ± 5\(\sqrt{2}\) = 0
Hint:
Equation of a line perpendicular to 5x – y = 0 is
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 33

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Question 12.
Equation of the straight line perpendicular to the line x – y + 5 = o, through the point of intersection the y-axis and the given line …….
(a) x – y – 5 = 0
(b) x + y – 5 = 0
(c) x + y + 5 = 0
(d) x + y + 10 = 0
Solution:
(b) x + y – 5 = 0
Hint:
x – y + 5 = 0 ⇒ put x = 0, y = 5
The point is (0, 5)
Equation of a line perpendicular to x – y + 5 = 0 is x + y + k = 0
This passes through (0, 5)
k = -5
x + 7 – 5 = 0

Question 13.
If the equation of the base opposite to the vertex (2, 3) of an equilateral triangle is x + y = 2, then the length of a side is ………
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 289
Solution:
\(\sqrt{6}\)
Hint:
In an equilateral ∆ the perpendicular will bisects the base in to two equal parts. Length of the perpendicular drawn from (2, 3) to the line x + 7 – 2 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 299

Question 14.
The line (p + 2q) x + (p – 3q)y = p – q for different values of p and q passes through the point ……
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 300
Solution:
(d) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
Hint:
(p + 2 q)x + (p – 3q)y = p – q
px + 2qx + py – 3qy = p – q
P(x + y) + q (2x – 3y) = p – q
The fourth option x = 2/5, y = 3/5
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 311
= p – q = RHS

Question 15.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …
(a) (7, 3)
(b) (4, 1)
(c) (1, -1)
(d) (-2, 3)
Solution:
(b) (4, 1)
Hint:
Let (a, b) be on 2x – 3y = 5 ⇒ 2a – 3b = 5
It is equidistance from (1, 2) and (3, 4)
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 322
(a – 1)2 + (b – 2)2 = (a – 3)2 + (6 – 4)2
a2 – 2a + 1 + b2 – 4b + 4 = a2 – 6a + 9 + b2 – 8b + 16
4a + 4b = 20
2a+ 2b = 10
2a – 3b = 5
5b = 5
b = 1 ∴ a = 4
∴ The point is (4, 1)

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Question 16.
The image of the point (2, 3) in the line y = – x is ………
(a) (-3, -2)
(b) (-3, 2)
(c) (-2, -3)
(d) (3, 2)
Solution:
(a) (-3, -2)
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 333
x – 2 = -5, y – 3 = -5
x = -3, y = -2
(-3,-2)

Question 17.
The length of ⊥ from the origin to the line \(\frac{x}{3}-\frac{y}{4}=1\) is ……
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 34
Solution:
(c) \(\frac{12}{5}\)
Hint:
4x – 3y = 12 ⇒ 4x – 3y – 12 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 35

Question 18.
The y-intercept of the straight line passing through (1, 3) and perpendicular to 2x – 3y + 1 = 0 is ……..
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 355
Solution:
(b) \(\frac{9}{2}\)
Hint:
Equation of a line perpendicular to 2x – 3y + 1 = 0 is 3x + 2y = k. It passes through (1, 3).
3 + 6 = k ⇒ k = 9, 3x + 2y = 9
To find y-intercept x = 0, 2y = 9, y = 9/2

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Question 19.
If the two straight lines x + (2k – 7)y + 3 = 0 and 3kx + 9y – 5 = 0 are perpendicular then the value of k is ……
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 356
Solution:
(a) k = 3
Hint.
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 37
Since the lines are perpendicular m1m2 = – 1
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 377

Question 20.
If a vertex of a square is at the origin and its one side lies along the line 4x + 3y – 20 = 0, then the area of the square is ……..
(a) 20 sq. units
(b) 16 sq. units
(c) 25 sq. units
(d) 4 sq.units
Solution:
(b) 16 sq. units
Hint:
One side of a square = Length of the perpendicular from (0, 0) to the line.
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 38

Question 21.
If the lines represented by the equation 6x2 + 41xy – 7y2 = 0 make angles α and β with
x – axis, then tan α tan β =
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 39
Solution:
(a) \(-\frac{6}{7}\)
Hint.
6x2 + 41xy – 7y2 = 0 ⇒ 6x2 – xy + 42xy – 7y2 = 0 ⇒ x (6x – y) + 7y (6x – y) = 0
(x + 7y) (6x – y) = 0 ⇒ x + 7y = 0, 6x – y = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 40

Question 22.
The area of the triangle formed by the lines x2 – 4y2 = 0 and x = a is …….
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 378
Solution:
(c) \(\frac{1}{2} a^{2}\)
Hint:
x2 – 4y2 = 0 , (x – 2y) (x + 2y) = 0 ⇒ x – 2y = 0, x + 2y = 0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 42
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 379

Question 23.
If one of the lines given by 6x2 – x + 4x2 = 0 is 3x + 4y = 0, then c equals to ……
(a) -3
(b) -1
(c) 3
(d) 1
Solution:
(a) -3
Hint.
6x2 – xy + 4cy2 = 0, 3x + 4y = 0
The other line may be (2x + by)
(3x + 4y) (2x + by) = 6x2 – xy + 4cy2
6x2 + 3xby + 8xy + 4by2 = 6x2 – xy + 4cy2
6x2 + xy (3b + 8) + 4by2 = 6x2 – xy + 4cy2
paring, 3b + 8 = -1
3b = -9 ⇒ b = -3
4b = 4c ⇒ 4(-3) = 4c
-12 = 4c ⇒ c = -3

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Question 24.
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 43
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 380
Solution:
(c) \(\frac{5}{9}\)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 381

Question 25.
The equation of one the line represented by the equation x2 + 2xy cot θ – y2 = 0 is ………
(a) x – y cotθ = 0
(b) x + y tan θ = 0
(e) x cos θ + y(sin θ + 1) = 0
(d) x sin θ + y(cos θ + 1) = 0
Solution:
(d) x sin θ + y(cos θ + 1)=0
Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 56

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