Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Pair of Straight Lines Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1B Pair of Straight Lines Important Questions

Question 1.

Does the equation x^{2} + xy + y^{2} = 0 represent a pair of lines?

Solution:

a = 1, b = 1, h = \(\frac{1}{2}\), ⇒ h^{2} = \(\frac{1}{4}\), ab = 1

h^{2} = ab < 0 i.e., h^{2} < ab.

∴ The given equation does not represent a pair of lines

Question 2.

Find the nature of the triangle of formed by the lines x^{2} – 3y^{2} = 0 and x = 2.

Solution:

Combined equation of OA and OB is

x^{2} – 3y^{2} = 0

(x + \(\sqrt{3}\) y) (x – \(\sqrt{3}\) y) = 0

x + \(\sqrt{3}\) y = 0 and x – \(\sqrt{3}\) y = 0

i.e., y = \(\frac{1}{\sqrt{3}}\) x, y = –\(\frac{1}{\sqrt{3}}\) x are equally inclined to the X—axis, the inclination being 30°

∴ ∠OAB – ∠OBA = 60°

∴ The triangle is equilateral

Question 3.

Find the centroid of the triangle formed by the lines 12x^{2} – 20xy + 7y^{2} = 0 and 2x – 3y + 4 = 0.

Solution:

Combined equation of OA and OB is

12x^{2} – 20xy + 7y^{2} = 0 ………………….. (1)

Equation of AB is 2x – 3y + 4 = 0

2x = 3y – 4

Substituted in (1)

3(3y – 4)^{2} – 1oy (3y – 4) + 7y^{2} = 0

3(9y^{2} + 16 – 24y) – 30y^{2} + 40y + 7y^{2} = 0

27y^{2} +48 – 72y – 30y^{2} + 40y + 7y^{2} = 0

4y^{2} – 32y + 48 = 0

y^{2} – 8y + 12 = 0

(y – 2) (y – 6) = 0 ⇒ y = 2 or 6

x = \(\frac{3 y-4}{2}\)

y = 2 ⇒ x = \(\frac{6-4}{2}\) = 1

y = 6 ⇒ x = \(\frac{18-4}{2}\) = 7

7 2

∴ Vertices and O (0, 0), A (1, 2), B( 7, 6)

Centroid of OAB is

(\(\frac{0+1+7}{3}\), \(\frac{0+2+6}{3}\)) = (\(\frac{8}{3}\), \(\frac{8}{3}\))

Question 4.

Prove that the lines represented by the equations x^{2} – 4xy + y^{2} = 0 and x + y = 3 form an equilateral triangle.

Solution:

Since the straight line L: x + y = 3 makes 45° with the negative direction of the X-axis, none of the lines which makes 60° with the line L is vertical. If ‘m’ is the slope of one such straight line, then

\(\sqrt{3}\) = tan 60° = \(\left|\frac{m+1}{1-m}\right|\) and so, m satisfies the equation (m + 1)^{2} = 3(m – 1)^{2}

(or) m^{2} – 4m + 1 = 0 …………………….. (1)

But the straight line having slope’m1 and passing through the origin is

y = mx ……………………. (2)

So the equation of the pair of lines passing through the origin and inclined at 60° with the line L is obtained by eliminating’m’ from the equations (1) and (2). Therefore the combined equation of this pair of lines is

(\(\frac{y}{x}\))^{2} – 4(\(\frac{y}{x}\)) + 1 = 0 (i.e.,) x^{2} – 4xy + y^{2} = 0

Which is the same as the given pair of lines. Hence, the given triad of lines form an equilateral triangle.

Question 5.

Show that the product of the perpendicular distances from a point (α, β) to the pair of straight lines ax^{2} + 2hxy + by^{2} = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\). [May 11, 07; Mar. 07, 04]

Solution:

Let ax^{2} + 2hxy + by^{2} = (l_{1}x + m_{1}y) (l_{2}x + m_{2}y)

Then the separate equations of the lines represented by the equation

ax^{2} + 2hxy + by^{2} = 0 are

L_{1} : l_{1}x + m_{1}y = 0 and L_{2} : l_{1}x + m_{1}y = 0

Also, we have l_{1}l_{2} = a; m_{1}m_{2} = b and

l_{1}m_{2} + l_{2}m_{1} = 2h

d_{1} = length of the perpendicular from (α, β) to L_{1} = \(\frac{\left|l_{1} \alpha+\mathrm{m}_{1} \beta\right|}{\sqrt{l_{1}^{2}+\mathrm{m}_{1}^{2}}}\)

d_{2} = length of the perpendicular from (α, β) to L_{2} = \(\frac{\left|l_{2} \alpha+m_{2} \beta\right|}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\)

Then, the product of the lengths of the perpendiculars from (α, β) to the given pair of lines = d_{1}d_{2}

= \(\frac{\left\|\left(l_{1} \alpha+m_{1} \beta\right)\left(l_{2} \alpha+m_{2} \beta\right)\right\|}{\sqrt{\left(l_{1}^{2}+m_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}\right)}}\)

= \(=\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Question 6.

Let ax^{2} + 2hxy + by^{2} = 0 represent a pair of straight lines. Then show that the equation of the pair of straight lines.

i) Passing through (x_{0}, y_{0}) and parallel to the given pair of lines is

a(x – x_{0})^{2} + 2h(x – x_{0}) (y – y_{0}) + b(y – y_{0})^{2} = 0 and

ii) Passing through (x0, y0) and perpendicular to the given pair of lines is

b(x – x_{0})^{2} – 2h(x – x_{0}) (y – y_{0}) + a(y – y_{0})^{2} = 0

Solution:

Let ax^{2} + 2hxy + by^{2} = (l_{1}x + m_{1}y) (l_{2}x + m_{2}y).

Then the lines L_{1} and L_{2} are represented by the given equation are respectively

l_{1}x + m_{1}y = 0 and l_{2}x + m_{2}y = 0.

Also we have l_{1}l_{2} = a, m_{1}m_{2} = b and

l_{1}m_{2} + l_{2}m_{1} = 2h

(i) The straight lines passing through (x_{0}, y_{0}) and parallel to L_{1} and L_{2} are respectively

l_{1}x + m_{1}y = l_{1}x_{0} + m_{1}y_{0} (or)

l_{1}(x – x_{0}) + m_{1}(y – y_{0}) = 0 and

l_{2}(x -x_{0}) + m_{2}(y – y_{0}) = 0.

Therefore, their combined equation is

[l_{1}(x – x_{0}) + m_{1}(y – y_{0})] [l_{2}(x – x_{0})+ m_{2}(y – y_{0})] = 0

(or) a(x – x_{0})^{2} + 2h(x – x_{0}) (y – y_{0}) + b(y – y_{0})^{2} = 0

(ii) The straight lines passing through (x_{0}, y_{0}) and perpendicular to the pair L_{1} and L_{2} are respectively.

m_{1}x – l_{1}y = m_{1}x_{0} – l_{1}y_{0} (or)

m_{1}(x – x_{0}) – l_{1}(y – y_{0}) = 0 and

m_{2}(x – x_{0}) – l_{2}(y – y_{0}) = 0.

Hence their combined equation is

[m_{1}(x – x_{0}) – l_{1}(y – y_{0})] [m_{2}(x – x_{0}) – l_{2}(y – y_{0})] = 0

(i.e.,) b(x – x_{0})^{2} – 2h(x – x_{0}) (y – y_{0}) + a(y – y_{0})^{2} = 0

[Note : The pair of lines passing through the origin and perpendicular to the pair of lines given by ax^{2} + 2hxy + by^{2} = 0 is bx^{2} – 2hxy + ay^{2} = 0].

Question 7.

Show that the area of the triangle formed by the lines ax^{2} + 2hxy + by^{2} = 0 and lx + my + n = 0 is \(\left|\frac{n^{2} \sqrt{h^{2}-a b}}{a m^{2}-2 h / m+b l^{2}}\right|\)

Solution:

Let \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the pair of straight lines represented by the equation

ax^{2} + 2hxy + by^{2} = 0 (see figure) and \(\overleftrightarrow{\mathrm{AB}}\) be the line lx + my + n = 0

Let ax^{2} + 2hxy + by^{2} = (l_{1}x + m_{1}y) (l_{2}x + m_{2}y) and\(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the lines.

l_{1}x + m_{1}y = 0 and l_{2}x + m_{2}y = 0 respectively.

Let A = (x_{1}, y_{1}) and B = (x_{2}, y_{2}).

Then l_{1}x_{1} + m_{1}y_{1} = 0 and l_{1}x_{1} + m_{1}y_{1} + n = 0.

So, by the rule of cross-multiplication, we obtain

Question 8.

Two equal sides of an isosceles triangle are 7x – y + 3 = 0 and x + y – 3 = 0 and its third side passes through the point (1,0). Find the equation of the third side.

Solution:

Let the lines 7x – y + 3 – 0 and x + y – 3 = 0 intersect at A. If we draw lines (not passing through A) perpendicular to each of the bisectors of the angles at A, we get isosceles triangles, equal sides being along the given lines.

(∆ABF ≅ ∆AFC and ∆ADG ≅ ∆AGE)

Of them, we require those triangles whose third sides pass through (1, 0).

The equations of the bisectors of the angles between 7x – y + 3 = 0 and x + y – 3 = 0 are

⇒ 7x – y + 3 = ± 5(x + y – 3)

⇒ x – 3y + 9 = 0 and 3x + y – 3 = 0.

The third sides will be those lines perpendicular to the bisectors and intersecting at (1, 0).

The side perpendicular to x – 3y + 9 = 0 and passing through (1, 0) is 3x + y – 3 =0. The other one is (x – 1) -3(y – 0) = 0 i.e., x – 3y – 1 = 0. Therefore 3x + y- 3 = 0 and x- 3y— 1 = 0 are the required ones. [In the Figure ∆ABC and ∆ADE are isosceles with \(\overline{B C}\) and \(\overline{D E}\) as third sides].

Question 9.

Find the angle between the straight lines represented by 2x^{2} + 5xy + 2y^{2} – 5x – 7y + 3 = 0.

Solution:

a = 2, b = 2, h = \(\frac{5}{2}\)

cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}=\frac{|2+2|}{\sqrt{(2-2)^{2}+4 \cdot \frac{25}{4}}}\) = \(\frac{4}{5}\)

θ = cos^{-1} (\(\frac{4}{5}\))

Question 10.

Find the equation of the pair of lines passing through the origin and parallel to the pair of lines 2x^{2} + 3xy – 2y^{2} – 5x + 5y – 3 = 0

Solution:

Equation of the given pair of lines is

2x^{2} + 3xy – 2y^{2} – 5x + 5y – 3 = 0

Equation of the pair of parallel lines passing through the origin is ax^{2} + 2hxy + by^{2} = 0 is 2x^{2} + 3xy – 2y^{2} = 0

Question 11.

Find the equation of the pair of lines passing through the origin and perpendicular to the pair of lines ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

Solution:

Equation to the pair of lines parallel to the given lines and passing through the origin is ax^{2} + 2hxy + by^{2} = 0

Equation of the pair of lines perpendicular to the given lines and passing through the origin is bx^{2} – 2hxy + ay^{2} = 0

Question 12.

If x^{2} + xy – 2y^{2} + 4x – y + k = 0 represents a pair of straight lines, find k.

Solution:

a = 1, b = -2, c = k; f = –\(\frac{1}{2}\), g = 2, h = \(\frac{1}{2}\)

The condition is

abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

2k + 2(-\(\frac{1}{2}\)) . 2\(\frac{1}{2}\) – 1.\(\frac{1}{4}\) + 2.\(\frac{4}{a}\) – k\(\frac{1}{4}\) = 0

-8k – 4 – 1 + 8 – k = 0

9k = 27 ⇒ k = 3

Question 13.

Prove that the equation 2x^{2} + xy – 6y^{2} + 7y – 2 = 0 represents a pair of straight lines.

Solution:

a = 2

b = -6

c = -2

f = \(\frac{7}{2}\)

g = 0

h = \(\frac{1}{2}\)

abc + 2fgh – af^{2} – bg^{2} – ch^{2}

= 2(-6) (-2) + 2.\(\frac{7}{2}\) 0\(\frac{1}{2}\) – 2(\(\frac{7}{2}\))^{2} + 6.0 + 2.(\(\frac{1}{2}\))^{2}

= 24 – \(\frac{49}{2}\) + \(\frac{1}{2}\) = 0

h^{2} – ab = \(\frac{1}{4}\) + 12 > 0,

g^{2} – ac = 0 + 4 = 4 > 0

f^{2} – bc = \(\frac{49}{4}\) – 12 = \(\frac{1}{4}\) > 0

The given equation represents a pair of line.

Question 14.

Prove that the equation 2x^{2} + 3xy – 2y^{2} – x + 3y – 1 = 0 represents a pair of perpendicular straight lines.

Solution:

a = 2

b = -2

c = -1

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

g = –\(\frac{1}{2}\)

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

abc + 2fgh – af^{2} – bg^{2} – ch^{2}

= 2(-2) (-1) + 2.\(\frac{3}{2}\) (-\(\frac{1}{2}\)).\(\frac{3}{2}\) – 2(\(\frac{9}{4}\)) + 2.\(\frac{1}{4}\) + \(\frac{1.9}{2}\)

= 4.-\(\frac{9}{4}\) – 2.\(\frac{9}{4}\) + \(\frac{1}{2}\) + \(\frac{9}{4}\)

= \(\frac{9}{2}\) – \(\frac{9}{2}\) = 0

h^{2} – ab = \(\frac{9}{4}\) + 4 = \(\frac{25}{4}\) > 0,

g^{2} – ac = \(\frac{1}{4}\) + 2 > 0

f^{2} – bc = \(\frac{9}{4}\) – 2 = \(\frac{1}{4}\) > 0

a + b = 2 – 2 = 0

The given equation represents a pair of perpendicular line.

Question 15.

Show that the equation 2x^{2} – 13xy – 7y^{2} + x + 23y – 6 = 0 represents a pair of straight lines. Also find the angle between them and the co-ordinates of the point of intersection of the lines.

Solution:

Here a = 2

b = -7

c = -6

f = \(\frac{23}{2}\)

g = \(\frac{1}{2}\)

h = –\(\frac{13}{2}\)

abc + 2fgh – af^{2} – bg^{2} – ch^{2}

= 2(-7) (-6) + 2.\(\frac{23}{2}\) . \(\frac{23}{2}\) . (-\(\frac{13}{2}\)) – 2\(\frac{529}{4}\) + 7.\(\frac{1}{4}\) + 6 . \(\frac{169}{4}\)

= \(\frac{1}{4}\) (336 – 299 – 1058 + 7 + 1014)

= \(\frac{1}{4}\) (1357 – 1357) = 0

h^{2} – ab = \(\frac{169}{4}\) + 14 > 0,

g^{2} – ac = \(\frac{1}{4}\) + 12 > 0,

f^{2} – bc = \(\frac{529}{4}\) – 42 > 0

The given equation represents a pair of lines

= \(\frac{-13-92}{-56-169}\) = \(\frac{-105}{-225}\) = \(\frac{7}{15}\)

Point of intersection is P (\(\frac{19}{15}\), \(\frac{7}{15}\))

Question 16.

Find that value of λ for which the equation λx^{2} – 10xy + 12y^{2} + 5x – 16y -3 = 0 represents a pair of straight lines.

Solution:

Here a = λ

b = 12

c = -3

f = -8

g = \(\frac{5}{2}\)

h = -5

The given equation represents a pair of lines

abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

-36λ + 2(-8)\(\frac{5}{2}\) (-5) – λ . 64 – 12 . \(\frac{25}{4}\) + 3 . 25 = 0

-36λ + 200 – 64λ – 75 + 75 = 0

100λ = 200

⇒ λ = 2 ⇒ a = 2

h^{2} – ab = 25 – 24 = 1 > 0

f^{2} – bc = 64 + 36 = 100 > 0

g^{2} – ac = \(\frac{25}{4}\) + 6 = \(\frac{49}{4}\) > 0

∴ The given equation represents a pair of lines for λ = 2.

Question 17.

Show that die pairs of straight lines 6x^{2} – 5xy – 6y^{2} = 0 and 6x^{2} – 5xy – 6y^{2} + x + 5y – 1 = 0 form a square.

Solution:

H ≅ 6x^{2} – 5xy – 6y^{2} = (3x + 2y) (2x – 3y)

and S ≅ 6x^{2} – 5xy – 6y^{2} + x + 5y – 1

= (3x + 2y- 1) (2x – 3y + 1).

∴ H = 0 represents the lines 3x + 2y = 0 and 2x – 3y = 0 which are perpendicular and S = 0 represents the lines 3x + 2y – 1 = 0, 2x – 3y + 1 = 0 which are also perpendicular. These two pairs of perpendicular lines, therefore, determine a rectangle.

Also the distance between the pair of opposite sides 3x + 2y = 0 and 3x + 2y – 1 = 0 is \(\frac{1}{\sqrt{13}}\) and this is the same as the distance between the other pair of opposite sides 2x – 3y = 0 and 2x – 3y + 1 = 0 of the rectangle. Hence the rectangle is a square.

Question 18.

Show that the equation 8x^{2} – 24xy + 18y^{2} – 6x + 9y – 5 = 0 represents a pair of parallel straight lines and find the distance between them.

Solution:

S = 8x^{2} – 24xy + 18y^{2} – 6x + 9y – 5

= 2(2x – 3y)^{2} – 3(2x – 3y) – 5

= [2(2x – 3y) – 5] [(2x – 3y) + 1]

= (4x – 6y – 5) (2x – 3y + 1) = 0

The lines are 4x – 6y – 5 = 0 and 2x – 3y + 1 = 0

\(\frac{a_{1}}{a_{2}}\) = \(\frac{4}{2}\), \(\frac{b_{1}}{b_{2}}\) = \(\frac{-6}{-3}\) = 2

\(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\)

∴ The given equation represents a pair of parallel lines.

Distance between the lines

= 2\(\sqrt{\frac{g^{2}-a c}{a(a+b)}}\) = 2\(\sqrt{\frac{9+40}{8(8+18)}}\) = \(\frac{2.7}{2 \sqrt{52}}\) = \(\frac{7}{\sqrt{52}}\)

Question 19.

If the pair of lines represented by ax^{2} + 2hxy + by^{2} = 0 and ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 form a rhombus, prove that (a – b) fg + h(f^{2} – g^{2}) = 0.

Solution:

Combined equation of OA and OB is ax^{2} + 2hxy + by^{2} = 0

Combined equation of AC, BC is ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

Point of intersection is C(\(\frac{h f-b g}{a b-h^{2}}\), \(\frac{g h-a f}{a b-h^{2}}\))

Equation of the diagonal is y = \(\frac{g h-a f}{h f-b g}\) . x

y(hf – bg) = x(gh – af)

(gh – af) x – (hf- bg) y = 0

A and B are points on both this pair of lines.

Combined equation of AB is

2gx + 2fy + c = 0

OACB is a rhombus

OC and AB are perpendicular

2g(gh – af) – 2f(.hf – bg) = 0

hg^{2} – afg – hf^{2} + bfg = 0

(a – b) fg + h(f^{2} – g^{2}) = 0

Question 20.

If two of the sides of a parallelogram are represented by ax^{2} + 2hxy + by^{2} = 0 and px + qy = 1 is one of its diagonals, prove that the other diagonal is y(bp – hq) = x(aq – hp).

Solution:

Let OACB be the parallelogram two of whose sides \(\overleftrightarrow{O A}\), \(\overleftrightarrow{O B}\) are represented by the equation H ≡ ax^{2} + 2hxy + by^{2} = 0. Since the other pair of sides \(\overleftrightarrow{A C}\) and \(\overleftrightarrow{B C}\) are respectively parallel to \(\overleftrightarrow{O B}\) and \(\overleftrightarrow{O A}\), their combined equation will be of the form S ≡ ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0. Then the equation of the diagonal \(\overleftrightarrow{A B}\) is 2gx + 2fy + c = 0. But this line is given to be px + qy = 1 (or) -pcx – qcy + c = 0 where c ≠ 0

Therefore 2g = – pc, 2f = – qc …………….. (1)

The vertex C of the parallelogram = (\(\frac{h f-b g}{a b-h^{2}}\), \(\frac{g f-a f}{a b-h^{2}}\))

∴ Equation of the diagonal \(\overleftrightarrow{O C}\) is

(gh – af) x = (hf- bg)y

i.e., c(-ph + aq) x = c(-hq + bp)y (By 1)

(or) (aq – hp) x = (bp – hq) y (since c ≠ 0)