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

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

I.

Question 1.
Find the angle between the lines represented by 2x² + xy – 6y² + 7y – 2 = 0.
Solution:
Comparing with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
a = 2, 2g = 0, g = 0
b = – 6, 2f = 7, f = 7/2
c = – 2, 2h = 1, h = 1/2
Angle between the lines is given by Question 2.
Prove that the equation 2x² + 3xy – 2y² + 3x + y + 1 = 0 represents a pair of perpendicular lines.
Solution:
Given a = 2, b = -2
a + b = 0 ⇒ cos α = 0 ⇒ α = π/2
∴ The given lines are perpendicular. II.

Question 1.
Prove that the equation 3x² + 7xy + 2y² + 5x + 5y + 2 = 0 represents a pair of straight lines and find the co-ordinates of the’point of intersection.
Solution:
The given equation is
3x² + 7xy + 2y² + 5x + 5y + 2 = 0
Comparing a = 3 2f = 5 ⇒ f = $$\frac{5}{2}$$
b = 2 2g = 5 ⇒ g = $$\frac{5}{2}$$
c = 2 2h = 7 ⇒ h = $$\frac{7}{2}$$
∆ = abc + 2fgh – af² – bg² – ch²
= 3(2)(2) + 2.$$\frac{5}{2}$$.$$\frac{5}{2}$$.$$\frac{7}{2}$$ – 3.$$\frac{25}{4}$$ – 2.$$\frac{25}{4}$$ – 2.$$\frac{49}{4}$$ = 0
= $$\frac{1}{4}$$(48 + 175 – 75 – 50 – 98)
= $$\frac{1}{2}$$(223 – 223) = 0 ∴ The given equation represents a pair of lines point of intersection is Question 2.
Find the value of k, if the equation 2x² + kxy – 6y² + 3x + y + l = 0 represents a pair of straight lines. Find the point of intersection of the lines and the angle between the straight lines for this value of k.
Solution:
The given equation is
2x² + kxy – by² + 3x + y+ 1 = 0
a = 2 2f = 1 ⇒ f = $$\frac{1}{2}$$
b = -6 2g = 3 ⇒ g = $$\frac{3}{2}$$
c = 1 2h = k ⇒ h = $$\frac{k}{2}$$
The given equation represents a pair of straight lines abc + 2fgh – af² – bg² – ch² = 0
-12 + 2.$$\frac{1}{2}$$.$$\frac{3}{2}$$(+$$\frac{k}{2}$$ -2.$$\frac{1}{4}$$ + 6.$$\frac{9}{4}$$ – $$\frac{k^{2}}{4}$$ = 0
– 48 + 3k – 2 + 54 – k² = 0
-k² + 3k + 4 = 0 ⇒ k² – 3k – 4 = 0
(k – 4) (k + 1) = 0
k = 4 or – 1.

Case (i): k = – 1
Point of’intersection is $$\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)$$ Point of intersection is ($$\frac{-5}{7}$$, $$\frac{1}{7}$$)
Angle between the lines Case (ii): k = 4 Point of intersection is P(-$$\frac{5}{8}$$, –$$\frac{1}{8}$$) Question 3.
Show that the equation x² – y² – x + 3y – 2 = 0 represents a pair of perpendicular lines and find their equations.
Solution:
Comparing a= 1, f = $$\frac{3}{2}$$
b = -1, g = –$$\frac{1}{2}$$
c = -2, h = 0
abc + 2fgh – af² – bg² – ch²
= 1 (-1) (-2) + 0- 1.$$\frac{9}{4}$$ + 1.$$\frac{1}{4}$$ + 0
= + 2 – $$\frac{9}{4}$$ + $$\frac{1}{4}$$ = 0
h² – ab = 0 – 1(-1) = 1 > 0,
f² – be = $$\frac{9}{4}$$ – 2 = $$\frac{1}{4}$$ = 1 > 0
g² – ac = $$\frac{1}{4}$$ + 2 = $$\frac{9}{4}$$ > 0
a + b = 1 – 1 = 0
The given equation represent a pair of per-pendicular lines
Let x² – y² – x + 3y – 2
= (x + y + c1) (x – y + c2)
Equating the co-efficients of x
⇒ c1 + c2 = – 1
Equating the co-efficients of y
⇒ – c1 + c2 = 3
Adding 2c2 = 2 ⇒ c2 = 1
c1 + c2 = – 1 ⇒ c1 + 1 = – 1
c1 = – 2
Equations of the lines are x + y – 2 = 0 and x – y + 1 = 0 Question 4.
Show that the lines x² + 2xy – 35y² – 4x + 44y – 12 = 0 are 5x + 2y – 8 = 0 are concurrent.
Solution:
Equations of the given lines are
x² + 2xy- 35y² -4x + 44y- 12 = 0
a = 1, f = 22
b = – 35, g = – 2
c = – 12, h = 1
Point or intersection is $$\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)$$ Point of intersection of the given lines is P($$\frac{4}{3}$$, $$\frac{2}{3}$$)
5x + 2y – 8 = 5.$$\frac{4}{3}$$ + 2.$$\frac{2}{3}$$ – 8
= $$\frac{20+4-24}{3}$$ = 0
P lies on the third line 5x + 2y – 8 = 0
∴ The given lines are concurrent. Question 5.
Find the distances between the following pairs of parallel straight lines :
i) 9x² – 6xy + y² + 18x – 6y + 8 = 0
Solution:
Distance between parallel lines = $$2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}$$ ii) x² + 2√3xy + 3y² – 3x – 3√3y – 4 = 0
Solution:
Distance between parallel lines = $$2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}$$ Question 5.
Show that the two pairs of lines 3x² + 8xy – 3y² = 0 and 3x² + 8xy – 3y² + 2x – 4y – 1 = 0 form a square. Solution:
Combined equation of CA and CB is
3x² + 8xy – 3y²
(x + 3y) (3x – y) = 0
3x – y = 0, x + 3y = 0
Equation of OA is 3x – y = 0 ………. (1)
Equation of OB is x + 3y = 0 ……….(2)
Combined equation of CA and CB is
3x² + 8xy – 3y² + 2x – 4y + 1 =0
Let 3x² + 8xy – 3y² + 2x – 4y + 1 = (3x – y + c1) (x + 3y + c2)
Equating the co-efficients of
x, we get c1 + 3c2 = 2
y, we have 3c1 + c2 = -4 Equation of BC is 3x – y – 1 = 0 ………. (3)
Equation of AC is x + 3y + 1 = 0 ………. (4)
OA and BC differ by a constant ⇒ OA parallel to BC
OB and CA differ by a constant ⇒ OB parallel to AC
From combined equation of OA and OB
OACB is a rectangle a + b = 3 – 3 = 0
OA = Length of the -L lar from O to AC = $$\frac{|0+0+1|}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}$$
OB = Length of the perpendicular from O to BC = $$\frac{|0+0-1|}{\sqrt{9+1}}=\frac{1}{\sqrt{10}}$$
OA = OB and OACB is a rectangle
OACB is a square.

III.

Question 1.
Find the product of the lengths of the perpendiculars drawn from (2, 1) upon the lines
12x² + 25xy + 12y² + 10x + 11y + 2 = 0 Solution:
Combined equation of AB, AC is
12x² + 25xy+ 12y² + 10x + 11y + 2 = 0
12x² + 25xy +12y²
= 12x² + 16xy + 9xy + 12y² = 0
= 4x (3x + 4y) + 3y (3x + 4y)
= (3x + 4y) (4x + 3y)
Let 12x² + 25xy + 12y² + 10x + 11y + 2
= (3x + 4y + c1) (4x + 3y + c2)
Equating the co-efficients of x, we get
4c1 + 3c2 = 10 ………. (1)
Equating the co-efficients of y, we get
3c1 + 4c2 = 11 ………… (2)
i.e., 4c1 + 3c2 – 10 = 0
3c1 + 4c2 – 11 = 0  Equation of AB is 3x + 4y + 1 = 0
Equation of AC is 4x + 3y + 2 = 0
PQ = Length of the perpendicular from P(2, 1) on
AB = $$\frac{6+4+1}{\sqrt{9+16}}=\frac{11}{5}$$
PQ = Length of the perpendiculars from P(2, 1) on
AC = $$\frac{|8+3+2|}{\sqrt{16+9}}=\frac{13}{5}$$
Product of the length of the perpendiculars
= PQ × PR= $$\frac{11}{5}$$ × $$\frac{13}{5}$$ = $$\frac{143}{25}$$ Question 2.
Show that the straight lines y² – 4y + 3 = 0 and
x² + 4xy + 4y² + 5x + 10y + 4 = 0 form a parallelogram and find the lengths of its sides. Solution:
Equation of the first pair of lines is
y² – 4y + 3 = 0
(y – 1) (y – 3) = 0
y – 1 = 0 or y – 3 = 0
Equation of AB is y – 1 = 0 ……….. (1)
Equation of CD is y – 3 = 0 ………. (2)
Equations of AB and CD differ by a constant.
∴ AB and CD are parallel.
Equation of the second pair of lines is
x² + 4xy + 4y² + 5x + 10y + 4 = 0
(x + 2y)² + 5(x + 2y) + 4 = 0
(x + 2y)² + 4 (x + 2y) + (x + 2y) + 4 = 0
(x + 2y)(x + 2y + 4) + 1(x + 2y + 4) = 0
(x + 2y + 1) (x + 2y + 4) = 0
x + 2y + 1 = 0, x + 2y + 4 = 0
Equation of AD is x + 2y + 1 = 0 ……… (3)
Equation of BC is x + 2y + 4 = 0 ……… (4)
Solving (1), (3) x + 2 + 1 = 0
x = – 3
Co-ordinates of A are (-3, 1)
Solving (2), (3) x + 6 + 1 = 0
x = -7
Co-ordinates of DC are (-7, 3)
Solving (1), (4) x + 2 + 4 = 0
x = – 6
Co-ordinates of B are (-6, 1) Lengths of the sides of the parallelogram are 3, 2√5.

Question 3.
Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0 is $$\frac{|c|}{\sqrt{(a-b)^{2}+4 h^{2}}}$$
Solution:
Let ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents the lines
l1x + m1y + n1 = 0 ……….. (1)
l2x + m2y + n2 = 0 ………. (2)
⇒ ax² + 2hxy + by² + 2gx + 2fy + c ≡ (l1x + m1y + n1) (l2x + m2y + n2)
l1l2 = a, m1m2 = b, l1m2 + l2m1 = 2h,
l1n2 + l2n1 = 2g, m1n2 + m2n1 = 2f, n1n2 = c
Ir distance from origin to (1) is = $$\frac{\left|n_{1}\right|}{\sqrt{l_{1}^{2}+m_{1}^{2}}}$$
Irdistance from origin to (2) is = $$\frac{\left|n_{2}\right|}{\sqrt{I_{2}^{2}+m_{2}^{2}}}$$
Product of perpendiculars  Question 4.
If the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of intersecting lines, then show that the square of the distance of their point of intersection from the origin is $$\frac{c(a+b)-f^{2}-g^{2}}{a b-h^{2}}$$. Also slow that the square of this distance is $$\frac{f^{2}+g^{2}}{h^{2}+b^{2}}$$ if the given lines are perpendicular.
Solution:
Let the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0
represent the lines
l1x + m1y + n1 = 0 …….. (1)
l2x + m2y + n2 = 0 …….. (2)
(l1x + m1y + n1)(l2x + m2y + n2) = ax² + 2hxy + by² + 2gx + 2fy + c
l1l2 = a, m1 m2 = b, n1n2 = c
l1m2 + l2m1 = 2h, l1n2 + l12n1 = 2g, m1n2 + m2n1 = 2f
Solving (1) and (2)
$$\frac{x}{m_{1} n_{2}-m_{2} n_{1}}=\frac{y}{I_{2} n_{1}-l_{1} n_{2}}=\frac{1}{l_{1} m_{2}-I_{2} m_{1}}$$
The point of intersection, P  If the given pair of lines are perpendicular, then a + b = 0
∴ a = -b
$$\mathrm{OP}^{2}=\frac{0-r^{2}-g^{2}}{(-b) b-h^{2}}=\frac{r^{2}+g^{2}}{h^{2}+b^{2}}$$