Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.5 Text Book Back Questions and Answers, Notes.

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

Choose the correct or more suitable answer.

Question 1.

The equation of the locus of the point whose distance from y – axis is half the distance from origin is

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

(2) x^{2} – 3y^{2} = 0

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

(4) 3x^{2} – y^{2} = 0

Answer:

(4) 3x^{2} – y^{2} = 0

Explaination:

Let the point be (h,k)

Equation of y axis is x = 0

Given the distance of the point from y – axis is

\(\frac{1}{2}\) × distance of the point from the origin Distance of the point (h, k) from the y- axis is

= \(\frac{\mathrm{h}}{\sqrt{1^{2}}}\) = h

The distance of the point P (h , k) from the origin

Squaring on both sides

4h^{2} = h^{2} + k^{2}

4h^{2} – h^{2} – k^{2} = 0

3h^{2} – k^{2} = o

The locus of ( h, k ) is 3x^{2} – y^{2} = 0

Question 2.

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

(1) \(\frac{x^{2}}{\mathbf{a}^{2}}-\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}\) = 1

(2) \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1

(3) x^{2} + y^{2} = a^{2}

(4) y^{2} = 4ax

Answer:

(4) y^{2} = 4ax

Explaination:

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

Question 3.

Which of the following points lie on the locus of 3x^{2} + 3y^{2} – 8x – 12y + 17 = 0

(1) (0, 0)

(2) (-2, 3)

(3) (1, 2)

(4) (0, – 1)

Answer:

(3) (1, 2)

Explaination:

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.

If the point (8, – 5) lies on the focus \(\frac{x^{2}}{16}-\frac{y^{2}}{25}\) = k, then the value of k is

(1) 0

(2) 1

(3) 2

(4) 3

Answer:

(4) 3

Explaination:

Given the point (8, – 5) lies on the locus

Question 5.

Straight line joining the points (2, 3) and (- 1, 4) passes through the point (α, β) if

(1) α + 2β = 7

(2) 3α + β = 9

(3) α + 3β = 11

(4) 3α + β = 11

Answer:

(3) α + 3β = 11

Explaination:

The equation of the straight line joining the points (2, 3) and (-1, 4) is

x – 2 = – 3(y – 3)

x – 2 = – 3y + 9

x + 3y – 2 – 9 = 0

x + 3y – 11 = 0

Given (α, P) lies on this line

α + 3β – 11 = 0

Question 6.

The slope of the line which males an angle 45°with the line 3x – y = – 5 are

(1) 1, – 1

(2) \(\frac{1}{2}\), – 2

(3) 1, \(\frac{1}{2}\)

(4) 2, –\(\frac{1}{2}\)

Answer:

(2) \(\frac{1}{2}\), – 2

Explaination:

The equation of the given line is

3x – y + 5 = 0 ………… (1)

the slope of the line (1) is m = tan θ

Angle made by the required line = θ + 45°

where tan θ = 3

Required slope m_{1} = tan (θ + 45° )

Question 7.

Equation of the straight line that forms an isosceles triangle with coordinate axes in the

1-quadrant with perimeter 4 + 2√2 is

(1) x + y + 2 = 0

(2) x + y – 2 = 0

(3) x + y – √2 = 0

(4) x + y + √2 = 0

Answer:

(2) x + y – 2 = 0

Explaination:

Let OAB be the isosceles triangle formed in the first quadrant. Let OA = OB = a

In the right angle A OAB

AB^{2} = OA^{2} + OB^{2}

AB^{2} = a^{2} + a^{2} = 2a^{2}

AB = √2a

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

(1) x + 1 = 0

(2) x + y = 1

(3) x + y + 3 = 0

(4) x – y + 3 = 0

Answer:

(4) x – y + 3 = 0

Explaination:

ABCD is a quadrilateral in which the sides AD and BC are parallel. Draw AE and DF perpendicular to BC. ADFE is a square with sides

AD = AE = EF = DF = 2

Area of ADFE = 2 × 2 = 4

In ∆ AEB , BE = 1, AE = 2

∴ Area of ∆ AEB = \(\frac{1}{2}\) × 1 × 2 = 1

Similarly Area of ∆ DFC = 1

∆ Area of the quadrilateral A B C D = 4 + 1 + 1 = 6

Given the line through the vertex (- 1, 2 ) divides the quadrilateral ABCD into two half.

Let E (x , 4) is the point on the line BC such that the line DE divides the area of the quadrilateral ABCD into two half.

∴ Area of ∆ EDC = 3

\(\frac{1}{2}\) × EC × height = 3

\(\frac{1}{2}\) × (x + 2) × 2 = 3

x + 2 = 3 ⇒ x = 1

∴ The coordinates of E are (1, 4)

The equation of the line joining the points (- 1, 2) and (1, 4) is

x – 1 = y – 4

x – y + 3 = 0

Question 9.

The intercepts of the perpendicular bisector of the line segment joining (1,2) and (3,4) with coordinate axes are

(1) 5, – 5

(2) 5, 5

(3) 5, 3

(4) 5, – 4

Answer:

(2) 5, 5

Explaination:

Let the given points be A (1, 2) and B (3, 4) P( h,k )be a point in the plane such that PA = PB

Squaring on both the sides

(h – 1)^{2} + (k – 2)^{2} = (3 – h)^{2} + (k – 4)^{2}

h^{2} – 2h + 1 + k^{2} – 4k + 4 = 9 – 6h + h^{2} + k^{2} – 8k + 16

– 2h – 4k + 5 = – 6h – 8k + 25

6h + 8k – 2h – 4k + 5 – 25 = 0

4h + 4k – 20 = 0

h + k – 5 = 0

The locus of (h, k)is x + y – 5 = 0 which is the perpendicular bisector of A and B

x + y = 5

\(\frac{x}{5}+\frac{y}{5}\) = 1

x – intercept = 5 , y – intercept = 5

Question 10.

The equation of the line with slope 2 and the length of the perpendicular from the origin equal to √5 is

(1) x + 2y = √5

(2) 2x + y = √5

(3) 2x + y = 5

(4) x + 2y – 5 = 0

Answer:

Let the equation of the required line be

y = mx + c ……….. (1)

Given Slope m = 2

(1) ⇒ y = 2x + c

2x – y + c = 0 ……….. (2)

The length of the perpendicular from (0, 0) to line (2)

Substituting for c in equation (2), we have

2x – y + 5 = 0

Question 11.

If a line is perpendicular to the line Sx-y = 0 and forms a triangle with the coordinate axes of area 5 sq. units, then its equation is

(1) x + 5y ± 5√2 = 0

(2) x – 5y ± 5√2 = 0

(3) 5x + y ± 5√2 = 0

(4) 5x – y ± 5√2 = 0

Answer:

(1) x + 5y ± 5√2 = 0

Explaination:

The equation of the given line is 5x – y = 0 —— (1)

The equation of any perpendicular to line (1) is

-x – 5y + k = 0

x + 5y – k = 0 —– (2)

x + 5y = k

x- intercept = k , y – intercept = \(\frac{\mathrm{k}}{5}\)

The line (2) intersect the x-axis at A and y-axis at B

∴ A is (k, 0) and B is \(\left(0, \frac{k}{5}\right)\)

OA = k and OB = \(\frac{\mathrm{k}}{5}\)

Area of ∆ OBA = \(\frac{1}{2}\) × OA × OB

∴ The required equation is x + 5y = ± 5√2

Question 12.

Equation of the straight line perpendicular to the line x – y + 5 = 0 through the point of intersection y-axis and the given line

(1) x – y – 5 = 0

(2) x + y – 5 = 0

(3) x + y + 5 = 0

(4) x + y + 10 = 0

Answer:

(2) x + y – 5 = 0

Explaination:

The equation of the given line is

x – y + 5 = 0 ………. (1)

To find the point of intersection with y – axis, Put x = 0 in (1)

0 – y + 5 = 0 ⇒ y = 5

∴ The point of intersection is ( 0, 5)

The equation any line perpendicular to line (1) is

– x – y + k = 0

x + y – k = 0

This line passes through the point (0,5)

0 + 5 – k = 0 ⇒ k = 5

∴ The equation of the required line is

x + y – 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

(1) \(\frac{\sqrt{3}}{2}\)

(2) 6

(3) √6

(4) 3√2

Answer:

(3) √6

Explaination:

∆ ABC is an equilateral triangle. Vertex A is (2, 3)

Equation of the base BC is x + y = 2

Draw AD ⊥ BC. By the property of the equilateral triangle, D is the midpoint of BC

BD = DC

Let AB = a

AD = length of the perpendicular from A (2, 3) to the line x + y – 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

(1) \(\left(\frac{3}{2}, \frac{5}{2}\right)\)

(2) \(\left(\frac{2}{5}, \frac{2}{5}\right)\)

(3) \(\left(\frac{3}{5}, \frac{3}{5}\right)\)

(4) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)

Answer:

(4) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)

Explaination:

The equation of the given line is

(p + 2q)x + (p – 3q)y = p – q ……….. (1)

Question 15.

The point on the line 2x – 3y = 5 is equal distance from (1, 2), and (3, 4) is

(1) (7, 3)

(2) (4, 1)

(3) (1, – 1)

(4)(-2, 3)

Answer:

(2) (4, 1)

Explaination:

Let P (h, k) be a point on the line

2x – 3y – 5 = 0

Then 2h – 3k – 5 = 0 ……….. (1)

Let A (1, 2) and B ( 3, 4) be the given points such that PA = PB

Squaring on both sides

(h – 1)^{2} + (k – 2)^{2} = (h – 3)^{2} + (k – 4)^{2}

h^{2} – 2h + 1 + k^{2} – 4k + 4 = h^{2} – 6h + 9 + k^{2} – 8k + 16

-2h + 1 – 4k + 4 = – 6h + 9 – 8k + 16

-2h – 4k + 5 = – 6h – 8k + 25

6h – 2h + 8k – 4k + 5 – 25 = 0

4h + 4k – 20 = 0

h + k – 5 = 0 …………. (2)

To find P (h, k), Solve (1) and (2)

(2) ⇒ h + 1 – 5 = 0

h – 4 = 0 ⇒ h = 4

∴ The required point p(h, k) is (4, 1)

Question 16.

The image of the point (2, 3) in the line y = -x is

(1) (-3, -2)

(2) (-3, 2)

(3) (-2, -3)

(4) (3, 2)

Answer:

(1) (- 3, – 2)

Explaination:

The equation of the given line is y = -x

x + y = 0 ………… (1)

The coordinates of the image of the point (x_{1}, y_{1}) with respect to the line ax + by + c = 0 are given by

∴ The coordinates of the image of the point (2, 3) with respect to the line x + y = 0 are given by

x – 2 = y – 3 = – 5

x – 2 = -5 ⇒ x = -5 + 2 = -3

y – 3 = -5 ⇒ y = -5 + 3 = -2

∴ The required point is (-3, -2)

Question 17.

The length of perpendicular from the origin to the line \(\frac{x}{3}-\frac{y}{4}\) = is

(1) \(\frac{11}{5}\)

(2) \(\frac{5}{12}\)

(3) \(\frac{12}{5}\)

(4) –\(\frac{5}{12}\)

Answer:

(3) \(\frac{12}{5}\)

Explaination:

The equation of the given line is \(\frac{x}{3}-\frac{y}{4}\) = 1

\(\frac{4 x-3 y}{12}\) = 14x – 3y = 12

4x – 3y – 12 = 0

The length of the perpendicular from (0, 0) to the line 4x – 3y – 12 = 0 is

Question 18.

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

(1) \(\frac{3}{2}\)

(2) \(\frac{9}{2}\)

(3) \(\frac{2}{3}\)

(4) \(\frac{2}{9}\)

Answer:

(2) \(\frac{9}{2}\)

Explaination:

The equation of the given line is

2x – 3y + 1 = 0 …………. (1)

The equation of any line perpendicular to (1) is

-3x – 2y + k = 0

3x + 2y – k = 0 ………… (2)

This line passes through the point (1, 3)

∴ (2) ⇒ 3 × 1 + 2 × 3 – k = 0

3 + 6 – k = 0 ⇒ k = 9

∴ (2) ⇒ 3x + 2y – 9 = 0 ………. (3)

2y = – 3x + 9

y = \(-\frac{3}{2} x+\frac{9}{2}\)

∴ The required y – intercept is \(\frac{9}{2}\)

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

(1) k = 3

(2) k = \(\frac{1}{3}\)

(3) k = \(\frac{2}{3}\)

(4) k = \(\frac{3}{2}\)

Answer:

(1) k = 3

Explaination:

The equation of the given lines are

x + (2k – 7)y + 3 = 0 ……….. (1)

3kx + 9y – 5 = 0 …………. (2)

Slope of line (1), m_{1} = \(-\frac{1}{2 k-7}\)

Slope of line (2), m_{2} = \(-\frac{3 \mathbf{k}}{9}\)

Given that lines (1) and (2) are perpendicular

∴ m_{1} m_{2} = -1

k = -3(2k – 7)

k = -6k + 21

6k + k = 21

⇒ 7k = 21

k = \(\frac{21}{7}\) = 3

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

(1) 20 sq. units

(2) 16 sq. units

(3) 25 sq. units

(4) 4 sq. units

Answer:

(2) 16 sq. units

Explanation:

OABC is a square in which the vertex O lies on the origin and the side AB along the line 4x + 3y – 20 = 0

OA = length of the perpendicular from (0, 0) to the line 4x + 3y – 20 = 0

∴ The length of the side of the square is 4 units.

∴ Area of the square = 4 × 4 = 16 sq. units

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 =

(1) –\(\frac{6}{7}\)

(2) \(\frac{6}{7}\)

(3) –\(\frac{7}{6}\)

(4) \(\frac{7}{6}\)

Answer:

(1) –\(\frac{6}{7}\)

Explaination:

The equation of the given pair of lines is

6x^{2} + 41xy – 7y^{2} = 0 ……….. (1)

Let m_{1}, m_{2} be slopes of the individual lines.

Given that the individual lines make angles α and β with x – axis.

∴ m_{1} = tan α and m_{2} = tan β

we have m_{1} m_{2} = \(\frac{a}{b}\) = \(\frac{6}{-7}\)

∴ tan α tan β = \(-\frac{6}{7}\)

Question 22.

The area of the triangle formed by the lines x^{2 }– 4y^{2} = 0 and x = a is

(1) 2a^{2}

(2) \(\frac{\sqrt{3}}{2}\)

(3) \(\frac{1}{2}\)

(4) \(\frac{2}{\sqrt{3}}\)

Answer:

(3) \(\frac{1}{2}\)

Explaination:

The equation of the given pair of straight lines is

x^{2} – 4y^{2} = 0

x^{2} – 4y^{2} = (x + 2y) (x – 2y)

∴ The separate equations of the straight lines are

x + 2y = 0 ………… (1)

x – 2y = 0 …………. (2)

The line x = a intersect the line (1) at A.

a + 2y = 0 ⇒ y = \(-\frac{a}{2}\)

∴ A is \(\left(a,-\frac{a}{2}\right)\)

The line x = a intersect the line (2) at B.

Question 23.

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

(1) -3

(2) -1

(3) 3

(4) 1

Answer:

(1) -3

Explaination:

The given pair of straight lines is 6x^{2} – xy – 4cy^{2} = 0 ………. (1)

One of the separate equation is 3x + 4y = 0 ……… (2)

Let the separate equation of the other line be ax + by = 0 ………. (3)

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

3ax^{2} + 4axy + 3bxy + 4by^{2} = 6x^{2} – xy + 4cy^{2}

Equating the coefficient of y^{2} on both sides

4b = 4c ⇒ c = b

Equating the coefficient of x^{2} on both sides

3a = 6

⇒ a = \(\frac{6}{3}\) = 2

Equating the coefficient of xy both sides

4a + 3b = -1

4 × 2 + 3b = -1 ⇒ 3b = -1 – 8

b = \(-\frac{9}{3}\) = -3

∴ c = -3

Question 24.

θ is acute angle between the lines x^{2} – xy – 6y^{2} = 0 then \(\frac{2 \cos \theta+3 \sin \theta}{4 \cos \theta+5 \cos \theta}\) is

(1) 1

(2) \(-\frac{1}{9}\)

(3) \(\frac{5}{9}\)

(4) \(\frac{1}{9}\)

Answer:

(3) \(\frac{5}{9}\)

Explaination:

The equation of the given pair of lines is

x^{2} -xy – 6y^{2} = 0

a = 1, 2h = -1, b = -6

Given θ is the angle between the lines

Question 25.

The equation of one of the lines represented by the equation x^{2} + 2xy cot θ – y^{2} = 0 is

(I) x – y cot θ = 0

(2) x + y tan θ = 0

(3) x cos θ + y (sin θ + 1) = 0

(4) x sin θ + y(cos θ + 2) = 0

Answer:

(4) x sin θ + y(cos θ + 2) = 0

Explaination:

The equation of given pair of straight line is

x^{2} + 2xy cot θ – y^{2} = 0

x^{2} + (2y cot θ)x + (-y^{2}) = 0

This is a quadratic equation in x. Solving for x, we have