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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) x^{2} + 3y^{2} = 0

(b) x^{2} – 3y^{2} = 0

(c) 3x^{2} + y^{2} = 0

(d) 3x^{2} – y^{2} = 0

Solution:

(c) 3x^{2} + y^{2} = 0

Hint:

Given that PA = \([\frac{1}{2}/latex]OP

2PA = OP

4PA^{2} = OP^{2}

4(x)^{2} = x^{2} + y^{2} ⇒ 3x^{2} – y^{2} = 0

Question 2.

Which of the following equation is the locus of (at^{2}, 2at) ……

Solution:

(d) y^{2} = 4ax

Hint:

y^{2} = 4ax ⇒ Equation that satisfies the given point (at^{2}, 2at)

Question 3.

Which of the following point lie on the locus of 3x^{2} + 3y^{2} – 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.

(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) [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, m_{1} = 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 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

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 will bisects the base in to 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}

a^{2} – 2a + 1 + b^{2} – 4b + 4 = a^{2} – 6a + 9 + b^{2} – 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 m_{1}m_{2} = – 1

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.

Question 21.

If the lines represented by the equation 6x^{2} + 41xy – 7y^{2} = 0 make angles α and β with

x – axis, then tan α tan β =

Solution:

(a) \(-\frac{6}{7}\)

Hint.

6x^{2} + 41xy – 7y^{2} = 0 ⇒ 6x^{2} – xy + 42xy – 7y^{2} = 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 x^{2} – 4y^{2} = 0 and x = a is …….

Solution:

(c) \(\frac{1}{2} a^{2}\)

Hint:

x^{2} – 4y^{2} = 0 , (x – 2y) (x + 2y) = 0 ⇒ x – 2y = 0, x + 2y = 0

Question 23.

If one of the lines given by 6x^{2} – x + 4x^{2} = 0 is 3x + 4y = 0, then c equals to ……

(a) -3

(b) -1

(c) 3

(d) 1

Solution:

(a) -3

Hint.

6x^{2} – xy + 4cy^{2} = 0, 3x + 4y = 0

The other line may be (2x + by)

(3x + 4y) (2x + by) = 6x^{2} – xy + 4cy^{2}

6x^{2} + 3xby + 8xy + 4by^{2} = 6x^{2} – xy + 4cy^{2}

6x^{2} + xy (3b + 8) + 4by^{2} = 6x^{2} – xy + 4cy^{2}

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 x^{2} + 2xy cot θ – y^{2} = 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