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

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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

4PA² = OP²
4(x)² = x² + y² ⇒ 3x² – y² = 0

Question 2.
Which of the following equation is the locus of (at², 2at) ……

Solution:
(d) y² = 4ax
Hint:
Given x = at², y = 2at
y = 2at ⇒ t = [latex]\frac{y}{2 a}\)
x = at2 ⇒ x = a × \(\left(\frac{y}{2 a}\right)^{2}\)
x = a × \(\frac{y^{2}}{4 a^{2}}\) = \(\frac{y^{2}}{4 a^{2}}\)
y² = 4ax

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 equation of the given locus is
3x² + 3y² – 8x – 12y + 17 = 0
(0, 0) does not lie on the locus since the locus contains constant term. Substituting (-2, 3) in the locus
3(- 2)² + 3(3)² – 8 × – 2 – 12 × 3 + 17
= 3 × 4 + 3 × 9 + 16 – 36 + 17
= 12 + 27 + 16 – 36 + 17 ≠ 0
∴ (- 2, 3) does not lie on the locus
Substituting (1, 2) on the locus
3(1)² + 3(2)² – 8 × 1 – 12 × 2 + 17
= 3 + 12 – 8 – 24 + 17
= 32 – 32
= 0
∴ (1, 2 ) lies on the locus

Question 4.

(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3

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)

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) \(\frac{1}{2},-2\)
(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

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

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 into 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

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

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

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 …….

Solution:
(a) x + 5y ± 5\(\sqrt{2}\) = 0
Hint:
Equation of a line perpendicular to 5x – y = 0 is

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 ………

Solution:
\(\sqrt{6}\)
Hint:
In an equilateral, ∆ the perpendicular wall bisects the base into two equal parts. Length of the perpendicular drawn from (2, 3) to the line x + 7 – 2 = 0

Question 14.
The line (p + 2q) x + (p – 3q)y = p – q for different values of p and q passes through the point ……

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

= 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)

(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)

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)

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 ……

Solution:
(c) \(\frac{12}{5}\)
Hint:
4x – 3y = 12 ⇒ 4x – 3y – 12 = 0

Question 18.
The y-intercept of the straight line passing through (1, 3) and perpendicular to 2x – 3y + 1 = 0 is ……..

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

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 ……

Solution:
(a) k = 3
Hint.

Since the lines are perpendicular m₁m₂ = – 1

Question 20.
If a vertex of a square is at the origin and it’s 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.

Question 21.
If the lines represented by the equation 6x² + 41xy – 7y² = 0 make angles α and β with the x-axis, then tan α tan β =

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

Question 22.
The area of the triangle formed by the lines x² – 4y² = 0 and x = a is …….

Solution:
(c) \(\frac{1}{2} a^{2}\)
Hint:
x² – 4y² = 0 , (x – 2y) (x + 2y) = 0 ⇒ x – 2y = 0, x + 2y = 0

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

Question 24.


Solution:
(c) \(\frac{5}{9}\)
Hint:

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