Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(c) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(c)

I.

Question 1.

Find the equation of the lines joining the origin to the points of intersection of x² + y² = 1 and x + y = 1.

Solution:

The given curves are

x² + y² = 1 …….. (1)

x + y = 1 ……….. (2)

Homogenising (1) with the help of (2) Combined equation of OA and OB is

x² + y² = (x + y)²

= x² + y² + 2xy

i.e., 2xy = 0 ⇒ xy – 0

Question 2.

Find the angle between the lines joining the origin to the points of intersection of y2 = x and x + y = 1.

Solution:

Equation of the centre is y² = x …………. (1)

Equation of AB is x + y = 1 ………. (2)

Hamogonsing (1) with the help of O

Combined equation of A & OB is

y² = x(x + y) = x² + xy

x² + xy – y2 = 0

a + b = 1 – 1 = 0

OA, OB an perpendicular

∴ ∠AOB =90°

II.

Question 1.

Show that the lines joining the origin to the points of intersection of the curve x² – xy + y² + 3x + 3y – 2 = 0 and the straight line x – y – √2 = 0 are mutually perpendicular.

Solution:

Equation of the curve is

x² – xy + y² + 3x + 3y – 2 = 0 ……….. (1)

Equation of AB is x – y – √2 = 0

x – y = √2

\(\frac{x-y}{\sqrt{2}}\) = 1 ………. (2)

Homogenising, (1) with the help of (2) combined equation of OA, OB is

x² – xy + y² + 3x.1 + 3y.1 – 2.1² = 0

x² – xy + y² + 3(x + y)\(\frac{x-y}{\sqrt{2}}\) – 2\(\frac{(x-y)^{2}}{2}\) = 0

x² – xy + y² + \(\frac{3}{\sqrt{2}}\)(x² – y²) – (x² – 2xy + y²) = 0

x² – xy + y² + \(\frac{3}{\sqrt{2}}\)x² – \(\frac{3}{\sqrt{2}}\)y² – x² + 2xy – y² = 0

\(\frac{3}{\sqrt{2}}\)x² + xy – \(\frac{3}{\sqrt{2}}\)y² = 0

a + b = \(\frac{3}{\sqrt{2}}\) – \(\frac{3}{\sqrt{2}}\) = 0

∴ OA, OB are perpendicular.

Question 2.

Find the values of k, if the lines joining the origin to the points of intersection of the curve 2x² – 2xy + 3y² + 2x – y – 1 = 0 and the line x + 2y = k are muiually perpendicular.

Solution:

Given equation of the curve is

S ≡ 2x² – 2xy + 3y² + 2x – y – 1 = 0 ……….. (1)

Equation of AB is x + 2y = k

\(\frac{x+2y}{k}\) = 1

Homogenising, (1) with the help of (2), combined equation of OA, OB is

2x² – 2xy + 3y² + 2x.1 – y.1 – 1² = 0

2x² – 2xy + 3y² + 2x\(\frac{x+2y}{k}\) – y\(\frac{x+2y}{k}\) – \(\frac{(x+2y)^{2}}{k^{2}}\) = 0

Multiplying with k²

2k²x² – 2k²xy + 3k²y² + 2kx(x + 2y) – ky(x + 2y) – (x + 2y)² = 0

2k²x² – 2k²xy + 3k²y² + 2kx² + 4kxy – kxy – 2ky² – x² – 4xy – 4y² = 0

(2k² + 2k – 1)x² + (-2k² + 3k – 4) xy + (3k² – 2k – 4) y² = 0

Since OA, OB are perpendicular.

Co-efficient x² + Co-efficient of y² = 0

2k² + 2k – 1 + 3k² – 2k – 4 = 0

5k² = 5 ⇒ k² = 1

∴ k = ± 1

Question 3.

Find the angle between the lines joining the origin to the points of intersection of the curve x² + 2xy + y² + 2x + 2y – 5 = 0 and the line 3x – y + 1 = 0

Solution:

Equation of the curve is

x² + 2xy + y² + 2x + 2y- 5 = 0 …….. (1)

Equation of AB is 3x – y + 1 = 0

y – 3x = 1…………. (2)

Homogenising (1) with the help of (2), combined equation of OA, OB is

x² + 2xy + y² + 2x.1 + 2y.1 – 5.1² = 0

x² + 2xy + y² + 2x(y – 3x) + 2y(y – 3x) – 5 (y – 3x)² = 0

x² + 2xy + y² + 2xy – 6x² + 2y² – 6xy – 5(y² + 9x² – 6xy) = 0

– 5x² – 2xy + 3y² – 5y² – 45x² + 30 xy = 0

-50x² + 28xy – 2y² = 0

i.e, 25x² – 14xy + y² = 0

Suppose θ is the angle between OA and OB

III.

Question 1.

Find the condition for the chord lx + my = 1 of the circle x² + y² = a² (whose centre is the origin) to subtend a right angle at the origin.

Solution:

Equation of the circle is x² + y² = a² ………. (1)

Equation of AB is lx + my = 1 ……….. (2)

Homogenising (1) with the help of (2)

Gombined equation of OA, OB is

x² + y² = a².l²

x² + y² = a² (lx + my)²

= a²(l²x² + m²y² + 2lmxy)

= a²l²x² + a²m²y² + 2a²lmxy

i.e., a²l²x² + 2a²lmxy + a²m²y² – x² – y² = 0

(a²l² – 1) x² + 2a² lmxy + (a²m² – 1)y² = 0

Since OA, OB are perpendicular

Co-efficient of x² + co-efficient of y² = 0

a²l² – 1 + a²m² – 1 = 0

a²(l² + m²) = 2

This is the required condition.

Question 2.

Find the condition for the lines joining the origin to the points of intersection of the circle x² + y² = a² and the line lx + my = 1 to coincide.

Solution:

Equation of the circle is

x² + y² = a² ………. (1)

Equation of AB is lx + my = 1 ……….. (2)

Homogenising (1) with the help of (2)

Combined equation of OA, OB is

x² + y² = a².l²

= a²(lx + my)²

= a²(l²x² + m²y² + 2lmxy)

i.e., x² + y² = a²l²x² + a²m²y² + 2a²lmxy

(a²l² – 1)x² + 2a²lmxy + (a²m² – 1)y² = 0

Since OA, OB are coincide

⇒ h² = ab

a^{4} l²m² = (a²l² – 1)(a²m² – 1)

a^{4} l²m² = a^{4} l²m² – a²l² – a² m² + 1

∴ a²l² – a²m² +1=0

a² (l² + m²) = 1

This is the required condition.

Question 3.

Write down the equation of the pair of straight lines joining the origin to the points of intersection of the line 6x – y + 8 = 0 with the pair of straight lines 3x² + 4xy – 4y² – 11x + 2y + 6 = 0. Show that the lines so obtained make equal angles with the coordinate axes.

Solution:

Given pair of lines is

3x² + 4xy – 4y² – 11x + 2y + 6 = 0 ……….. (1)

Given line is

6x – y + 8 = 0 ⇒ \(\frac{6x-y}{-8}\) = 1

⇒ \(\frac{y-6x}{8}\) = 1

Homogenising (1) w. r. t. (2)

3x² + 4xy – 4y² – (11x – 2y) (\(\frac{y-6x}{8}\)) = 6(\(\frac{y-6x}{8}\))² = 0

64[3x² + 4xy – 4y²] – 8[11xy – 66x² – 2y² + 12xy] + 6[y² + 36x² – 12xy] = 0

936x² + 256 xy – 256 xy – 234y² = 0

∴ 468x² – 117y² = 0 …….. (3)

⇒ 4x² – y² = 0

is eq. of pair of lines joining the origin to the point of intersection in.

The eq. pair of angle bisectors to (3) is

h(x² – y²) – (a – b)xy = 0

0(x² – y²) – f4 – l)xy = 0

⇒ xy = 0

x = 0 or y = 0 [Eqs. is of coordinate axes]

∴ The pair of lines are equally inclined to the co-ordinate axes.