Given lines are
3x – 2y = 5 … (1)
and 3x + 2y = 5 … (2)
Let (h, k) is any point, whose distances from the lines (1) and (2) are equal. Therefore
\(\frac{|3 h-2 k-5|}{\sqrt{9+4}}=\frac{|3 h+2 k-5|}{\sqrt{9+4}} \text { or }|3 h-2 k-5|=|3 h+2 k-5|\)
which gives 3h – 2k – 5 = 3h + 2k – 5 or – (3h – 2k – 5) = 3h + 2k – 5.
Solving these two relations we get k = 0 or h \(\frac{5}{3}\)Thus, the point (h, k) satisfies the equations y = 0 or x =\(\frac{5}{3}\) which represent straight lines. Hence, path of the point equidistant from the lines (1) and (2) is a straight line.

Given lines are
5x – y + 4 = 0 ... (1)
3x + 4y – 4 = 0 ... (2)
Let the required line intersects the lines (1) and (2) at the points, (\(\alpha\)₁, \(\beta\)₁) and (\(\alpha\)₂, \(\beta\)₂), respectively
Therefore \(5 \alpha_{1}-\beta_{1}+4=0 \text { and } \)\(3 \alpha_{2}+4 \beta_{2}-4=0\)
or \(\beta_{1}=5 \alpha_{1}+4 \text { and } \beta_{2}=\frac{4-3 \alpha_{2}}{4}\)
We are given that the mid point of the segment of the required line between (\(\alpha\)₁, \(\beta\)₁) and (\(\alpha\)₂, \(\beta\)₂)
Therefore \(\frac{\alpha_{1}+\alpha_{2}}{2}=1 \text { and } \frac{\beta_{1}+\beta_{2}}{2}=5\)
or \(\alpha_{1}+\alpha_{2}=2 \text { and } \frac{5 \alpha_{1}+4+\frac{4-3 \alpha_{2}}{4}}{2}=5\)
or \(\alpha_{1}+\alpha_{2}=2 \text { and } 20 \alpha_{1}-3 \alpha_{2}=20\) ..(3)
Solving equations in (3) for \(\alpha\)₁ and \(\alpha\)₂, we get
\(\alpha_{1}=\frac{26}{23} \text { and } \alpha_{2}=\frac{20}{23} \text { and hence, } \beta_{1}=5 \cdot \frac{26}{23}+4=\frac{222}{23} \text { . }\)
Equation of the required line passing through (1, 5) and (\(\alpha\)₁, \(\beta\)₁) is
\(y-5=\frac{\beta_{1}-5}{\alpha_{1}-1}(x-1) \text { or } y-5=\frac{\frac{222}{23}-5}{\frac{26}{23}-1}(x-1)\)
or 107x – 3y – 92 = 0,
which is the equation of required line

Given line is 4x – y = 0 ... (1)
In order to find the distance of the line (1) from the point P (4, 1) along another line, we have to find the point of intersection of both the lines. For this purpose, we will first find the equation of the second line (Fig 9.16). Slope of second line is tan 135° = –1. Equation of the line with slope – 1 through the point P (4, 1) is

y – 1 = – 1 (x – 4) or x + y – 5 = 0 ... (2)
Solving (1) and (2), we get x = 1 and y = 4 so that point of intersection of the two lines is Q (1, 4). Now, distance of line (1) from the point P (4, 1) along the line (2)
= the distance between the points P (4, 1) and Q (1, 4).
\(=\sqrt{(1-4)^{2}+(4-1)^{2}}=3 \sqrt{2} \text { units. }\)

Here A = 3, B = –4, C₁ = 7 and C₂ = 5. Therefore, the required distance is
\(d=\frac{|7-5|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{2}{5}\)

(a) The slope of the line through (3, – 2) and (– 1, 4) is
\(m=\frac{4-(-2)}{-1-3}=\frac{6}{-4}=-\frac{3}{2}\)
(b) The slope of the line through the points (3, – 2) and (7, – 2) is
\(m=\frac{-2-(-2)}{7-3}=\frac{0}{4}=0\)
(c) The slope of the line through the points (3, – 2) and (3, 4) is
\(m=\frac{4-(-2)}{3-3}=\frac{6}{0}\), which is not defined.
(d) Here inclination of the line \(\alpha\) = 60°. Therefore, slope of the line is
m = tan 60° = \(\sqrt{3}\)

Let A and d be the first term and common difference of AP and x, R be the first term and common ratio of GP, respectively.
According to the given condition
A + (p -1) d = a ..(i)
A + (q-1)d = b ..(ii)
A + (r -1) d = c..(iii)
and \(a=x R^{p-1} \)..(iv)
\(b=x R^{q-1} \)..(v)
\(c=x R^{r-1}\)..(vi)
On subtracting Eq. (ii) from Eq. (i),we get
d(p -1 - q + 1)= a - b
\(\Rightarrow a-b=d(p-q)\)..(vii)
On subtracting Eq. (iii) from Eq. (ii),we get
\(d(q-1-r+1) =b-c\)
\(\Rightarrow b-c =d .(q-r)\)
On subtracting Eq. (i) from Eq. (iii), we get
\(d(r-1-p+1)=c-a \Rightarrow c-a=d(r-p)\)
Now, we have to prove \(a^{b-c} b^{c-a} c^{a-b}=1\)
LHS = \(a^{b-c} b^{c-a} c^{a-b}\)
Using Eqs. (iv), (v), (vi) and (vii), (viii), (ix),we get
LHS = \(\left(x R^{p-1}\right)^{d(q-r)}\left(x R^{q-1}\right)^{d(r-p)}\left(x R^{r-1}\right)^{d(p-q)}\)
\(=x^{d(q-r)+d(r-p)+d(p-q)}\)
\(R^{(p-1) d(q-r)+(q-1) d(r-p)+(r-1) d(p-q)}\)
\(\left.=x^{d(q-r+r}-p+p-q\right)\)
\(R^{d(p q-p r-q+r+q r-p q-r+p+r p-r q-p+q)}\)
\(=x^{0} R^{0}=1=\mathrm{RHS}\)
Hence proved.

Let the given series be \(S= \frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots\)
Then, nth term \(\left(T_{n}\right)=\frac{1}{n(n+1)}\)
Now, we will split the denominator of the nth term into two parts or we will write T_n as the difference of two terms.
\(\therefore T_{n} =\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)} \)
\(=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)
On putting n = 1, 2, 3, 4, ... successively, we get
\(T_{1}=\frac{1}{1}-\frac{1}{2}, \quad T_{2}=\frac{1}{2}-\frac{1}{3}, T_{3}=\frac{1}{3}-\frac{1}{4}\)
.............
..............
\(T_{n}=\frac{1}{n}-\frac{1}{n+1}\)
On adding all these terms, we get
\(S =T_{1}+T_{2}+T_{3}+\ldots+T_{n} \)
\(=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\ldots+\left(\frac{1}{n}+\frac{1}{n+1}\right) \)
\(=1-\frac{1}{n+1} \)
\(=\frac{1}{1}-\frac{1}{n+1} \)
\(=\frac{n+1-1}{n+1} \)
\(\Rightarrow S =\frac{n}{n+1}\)

Here a = 2, r = 2 and n = 10
Using the sum formula \(\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}\)
We have S₁₀ = 2(210 – 1) = 2046
Hence, the number of ancestors preceding the person is 2046.

We have \(\sqrt{3}x+y=1\)
\(\Rightarrow y=-\sqrt{3}x+1\)
\(\therefore m_1=-\sqrt{3}\)
Also  \(x+\sqrt{3}y=1\)
\(\Rightarrow \sqrt{3}y=-x+1\)
\(\Rightarrow y=\frac{-1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}\)
\(\therefore m_2=\frac{-1}{\sqrt{3}}\)
Let \(\theta\) be the angle between the lines. Then
\(\tan\theta=|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3}(\frac{-1}{\sqrt{3}})}|\)
\(=|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}|=|\frac{-2}{\sqrt{3}}\times \frac{1}{2}|\)
\(=|\frac{-1}{\sqrt{3}}|=\frac{1}{\sqrt{3}}\)
\(\tan\theta=\tan 30^o\) and \(\tan(180^o-30^o)\)
\(\theta=30^o\) and \(150^o\)

Here,  \(p=\frac { \left| 0.cos\theta -0.(-sin\theta )-k\quad cos\quad 2\theta \right| }{ \sqrt { { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta } }\)
\( \Rightarrow p=\left| k\quad cos\quad 2\theta \right| \)
and \(q=\frac { \left| 0.sec\theta +0.cosec\theta -k \right| }{ \sqrt { { sec }^{ 2 }\theta +{ cosec }^{ 2 }\theta } }\)
\( \therefore q=\frac { \left| k \right| }{ \sqrt { \frac { 1 }{ { cos }^{ 2 }\theta } +\frac { 1 }{ { sin }^{ 2 }\theta } } } =\frac { \left| k \right| }{ \sqrt { \frac { { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta .{ cos }^{ 2 }\theta } } } \)
\(=\frac { \left| k \right| .\left| sin\theta cos\theta \right| }{ 1 } \times \frac { 2 }{ 2 } =\frac { 1 }{ 2 } \left| k\left( 2sin\theta \quad cos\theta \right) \right| \)
\(\Rightarrow q=\frac { 1 }{ 2 } \left| k\quad sin\quad 2\theta \right| \)
\(\Rightarrow 2q=\left| k\quad sin\quad 2\theta \right|\)
\( \therefore { p }^{ 2 }+{ 4q }^{ 2 }={ k }^{ 2 }{ cos }^{ 2 }2\theta +{ k }^{ 2 }{ sin }^{ 2 }2\theta \)

Equation of the line in intercept form is
\(\frac { x }{ a } +\frac { y }{ b } =1\) ....(i)
Given, a + b = 9   ........(ii)
and line (i) passes through the point (2,2)
\(\frac { 2 }{ a } +\frac { 2 }{ b } =1\)  ...(ii)
\(\Longrightarrow \frac { 2 }{ a } +\frac { 2 }{ 9-a } =1 \quad [from\quad Eq.(ii)]\)
a² - 9a + 18 = 0
a = 6 or 3
b = 3 or 6
Hence, Eq.(i) becomes \(\frac { x }{ 6 } +\frac { y }{ 3 } =1or\frac { x }{ 3 } +\frac { y }{ 6 } =1\)
x + 2y = 6 or 2x+ y = 6

Total cost of the tractor = Rs12000
Cash amount paid = Rs 6000
Balance amount = Rs12000 - Rs 6000 = Rs 6000
Interest of 1st instalment =Rs\(\frac { 6000\times 12\times 1 }{ 100 } \)=Rs 720
Amount of 1st instalment =Rs500+Rs720=Rs1220
Interest of second instalment =Rs \(\frac { 5500\times 12\times 1 }{ 100 } \)=Rs 660
Amount of second instalment =Rs 500+Rs 600=Rs1160
Interest of the 3rd instalment \(=\frac { 5000\times 12\times 1 }{ 100 } \)=Rs 600
Amount of third instalment = 500 + 600 = Rs1100
The sequence of instalment is 1220, 1160, 1100 ...
Here a = 1220 d = 1160 - 1220 = - 60 and n = 12
∴ S_n=\(\frac { n }{ 2 } \)[2a+(n-1)d]
∴ S₁₂=\(\frac { 12 }{ 2 } \)[2 x 1220+(12-1) x -60]
=6[2400-660]=Rs 10680
Thus total cost of tractor =10680+6000=Rs 16680

5+55+555+...n terms
\(=\frac { 5 }{ 9 } \)(9+99+999+...n terms)
\(=\frac { 5 }{ 9 } \) [(10-1)+(10²-1)+(10³-1)+...+(10ⁿ-1)]
\(=\frac { 5 }{ 9 } \) [(10+10²+10³+...+10ⁿ)-(1+1+...+1) n terms]
\(=\frac { 5 }{ 9 } \) \(\left[ 10\left( \frac { { 10 }^{ n }-1 }{ 10-1 } \right) -n \right] \)
Ans.\(\frac { 50 }{ 81 } ({ 10 }^{ n }-1)-\frac { 5n }{ 9 } \)

b,c,d are in GP.
\(\Rightarrow \) C² = bd...(ii)
Similarly, \(\frac { 1 }{ c } ,\frac { 1 }{ d } ,\frac { 1 }{ e } \)are in AP.
\(\Rightarrow \)  \(\frac { 2 }{ d } =\frac { 1 }{ c } +\frac { 1 }{ e } \)
\(\Rightarrow \)  \(d=\frac { 2ce }{ c+e } \)
On putting the values of b and d from Eq.(i) and (iii), in Eq.(ii), we get
\({ c }^{ 2 }=\left( \frac { a+c }{ 2 } \right) \times \left( \frac { 2ce }{ c+e } \right) \Rightarrow { c }^{ 2 }=ae\)
Therefore, a,c,e are in GP.

Let the GP be A,AR,AR²,AR³....
Given , first term , 
A = a ....(i)
and nth term , AR^n-1 = b ....(ii)
Now , p = Product of n terms
p = A\(\times\)AR¹\(\times\)AR²\(\times\)AR³\(\times\).....\(\times\)n terms
p = A^{1+1+1+1+.....+n} terms R^{1+2+3+....(n-1)}
p = AⁿR \(\frac{n(n-1)}{2}\)
p2 = Aⁿ Aⁿ R^n(n-1) = Aⁿ (AR^n-1)n
p2 = aⁿbⁿ  [using Eqs.(i)  and (ii)]

 Here, \(\frac{a}{r}\) + a + ar =  \(\frac{39}{10}\) and \((\frac{a}{r})\) \(\times\) ( a ) \(\times\) ( ar ) =1
\(\Rightarrow\) a³ = 1  \(\Rightarrow\)  a =1
On putting the value of a = 1 in Eq.(i) we get 
\(\frac{1}{r}\) + 1 + r = \(\frac{39}{10}\) \(\Rightarrow\)  \(\frac{1+r+r^2}{r}\) = \(\frac{39}{10}\)
\(\Rightarrow\) 10r²- 29r + 10 = 0 \(\Rightarrow\)  ( 5r - 2 )( 2r - 5 ) = 0
\(\therefore\) r  = \(\frac{2}{5}  \) or \(\frac{39}{10}\)
When a = 1 and = \( \frac{2}{5}\) , then the numbers are \(\frac{5}{2}\), 1,  \(\frac{2}{5}\)
When a = 1 and = \(\frac{5}{2}\), then the numbers are \(\frac{2}{5}\), 1,  \(\frac{5}{2}\)

Given, a+ar=4
and T₅=4T₃ \(\Rightarrow \)  ar^5-1 = 4ar^3-1 \(\Rightarrow \) r \(\pm\) 2
If r = 2 , then from eq .(i),
a+a(2)=-4 \(\Rightarrow \) 3a=-4
a=- \(\frac{4}{3}\)
If r =2, then from Eq.(i)
 a + a ( 2 ) = -4 \(\Rightarrow \)-a = -a
\(\therefore\)  a =- \( \frac{4}{3}\)
If r = -2, then from Eq.(i),
a + a ( -2 ) = -4
\(\Rightarrow \) -a = -4
\(\therefore\)  a = 4
Hence , GP is -\(\frac{4}{3}\),  \(\frac{8}{3}\),  \(\frac{16}{3}\),..... or 4,-8,16,..

a_k = kth term from the beginning = a₁r^k-1
a_n-k+1 = kth term from the end = a_n \((\frac{1}{r}k-1)\)
where1 a_ka_n-k+1=(a₁r^k-1)a_n\((\frac{1}{r}k-1)\) = a₁a_n
for all k satisfying 1

Let a and b two numbers and A₁, A₂,.........A_2n be 2n ( even numbers ) AMs.Then a + b = \(\frac{13}{6} \) and 
A₁ + A₂ +......+ A_2n = 2n \((\frac {a+b}{2})\) = \((\frac{13}{6})\)  n
According to the question, 
A₁ + A₂ +......+ A_2n=2n+1 \(\Longrightarrow \)  n=6
Ans. 12

Given, 1 = a₁ = a₂ and a_n = a_{n - 1} + a_{n - 2}, n > 2
On Putting n = 3, 4, 5 , 6 respectively, we get
For n =3, a₃ = a_{3 - 1} + a_{3 -2} = a₂ + a₁ = 1 + 1 = 2
For n =4, a₄ = a_{4 - 1} + a_{4 -2} = a₃ + a₂ = 2 + 1 = 3
For n =5, a₅ = a_{5 - 1} + a_{5 -2} = a₄ + a₃ = 3 + 2 = 5
For n =6, a₆ = a_{6 - 1} + a_{6 -2} = a₅ + a₄ = 5 + 3 = 8
Now, \(\frac { { a }_{ n+1 } }{ { a }_{ n } } \) ,  for n = 1, 2, 3, 4,5.
For n = 1, \(\frac { { a }_{ 2 } }{ { a }_{ 1 } } =\frac { 1 }{ 1 } =1\) :
For n =2,  \(\frac { { a }_{ 3 } }{ { a }_{ 2 } } =\frac { 2 }{ 1 } =2\) :
For n =3,  \(\frac { { a }_{ 4 } }{ { a }_{ 3 } } =\frac { 3 }{ 2 }\) :
For n =4,  \(\frac { { a }_{ 5 } }{ { a }_{ 4 } } =\frac { 5 }{ 3 }\) :
For n =5,  \(\frac { { a }_{ 6 } }{ { a }_{ 5 } } =\frac { 8 }{ 5 }\)
Hence, the required terms are 1, 2, \(\frac { 3 }{ 2 }\), \(\frac { 5 }{ 3 }\) and \(\frac { 8 }{ 5 }\)