Let m₁, m₂ be the slopes of pair of straight lines.
\(m_1+m_2=\frac{-2 h}{b}, \ m_1m_2=\frac{a}{b}\)
given m₂ = 3m₁
\(m_1+3 m_1=\frac{-2 h}{b}\)
\(4 m_1=\frac{-2 h}{b} \Rightarrow m_1=\frac{-h}{2 b}\)
\(m_2=\frac{-3 h}{2 b}\)
\(m_1 m_2=\frac{a}{b}\)
\(\left(\frac{-h}{2 b}\right)\left(\frac{-3 h}{2 b}\right)=\frac{a}{b}\)
\(3 h^2=4 \mathrm{ab}\)
Hence proved

The Condition for concurrent lines is
\(\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=1\)
Consider = \(\left| \begin{matrix} 4 & 3 & -10 \\ 3 & -4 & 5 \\ 5 & 1 & -7 \end{matrix} \right| \)
\(\Rightarrow\) 4(28 - 5) - 3(-21 - 25) - 10(3 + 20)
\(\Rightarrow\)  92 + 138 - 230 = 0
Hence, the given lines are concurrent.

(x + 2) ( x - 5) + (y -2 ) ( y -1) = 0
\(\Rightarrow\) x² -5x + 2x - 10 + y² - y - 2y + 2 = 0
\(\Rightarrow\)  x² + y² - 3x - 3y - 8 = 0
here 2g  = -3    \(\Rightarrow\)  \(g=-\frac { 3 }{ 2 }\)
2f  = -3   \(\Rightarrow\)   \(f=-\frac { 3 }{ 2 } \)
and C = -8
Center of the circle (-g, -f) = \(\left( \frac { 3 }{ 2 } ,\frac { 3 }{ 2 } \right) \)
A radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
Radius of the circle is  \(= \sqrt { \frac { 9 }{ 4 } +\frac { 9 }{ 4 } +8 }\)
\(=\sqrt{\frac{18}{4}+8}=\sqrt{\frac{9}{2}+8}\)
\(r=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}} \text { units }\)

\(n P_r=1680 ; n C_r=70\)
\(\frac{n P_r}{r !}=70\)
\(\frac{1680}{r !}=70\)
\(r !=\frac{1680}{70}=24\)
\(r !=4 ! \Rightarrow r=4\)
\(n P_4=1680\)
\(n(n-1)(n-2)(n-3)=8 \times 7 \times 6 \times 5\)
\(n=8\)

\(\left(3 x+\frac{x^2}{2}\right)^8\)
n = 8
Middle term is \(t_{\frac{n}{2}+1}=t_{\frac{8}{2}+1}=t_5\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
r = 4
\(t_5=8 c_4(3 x)^4\left(\frac{x^2}{2}\right)^4\)
\(=\frac{8 C_4(81)}{16} x^{12}\)

Write the denominator into the product of linear factors.
Here x² - 5x + 6 = (x - 2)(x - 3)
Let \(\frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac{A}{(x-2)}+\frac{B}{(x-3)}\)...(1)
Multiplying both the sides of (1) by (x - 2)(x - 3), we get
7x - 1 = A(x - 3) + B(x - 2)....(2)
Put x = 3 in (2),we get
21 – 1 = B (1)
\(\Rightarrow \) B = 20
Put x = 2 in (2),we get
14 - 1 = A(-1)
\(\Rightarrow \) A = -13
Substituting the values of A and B in (1), we get
\(\frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac{-13}{(x-2)}+\frac{20}{(x-3)}\)

Equation of a circle when end points of the diameter are given is
(x - x₁) (x - x₂) + (y - y₁) (y - y₂) = 0
Here (x₁, y₁) = (4,7) and (x₂, y₂)  = (-2,5) 
(x - 4) (x + 2) + (y - 7) (y - 5 ) = 0
x² + 2x - 4x -8 + y² - 5y - 7y + 35 = 0
x² + y² - 2x - 12y + 27 = 0

Let n denote the number of sides of a regular polygon
Number of diagonals = 44
\(\frac{n(n-3)}{2}=44\)
\(n^2-3 n-88=0\)
\((n-11)(n+8)=0\)
\(n=11\)

3x² - 5xy - 2y² + 17x + y + 10 = 0
Comparing this with ax²+ 2hxy + by²+ 2gx + 2fy + c = 0
We get a=3, 2h = -5, b = -2, 
\(h=\frac { -5 }{ 2 }\)
Let \(\theta\) be the angle between the pair of straight lines then
tan \(\theta\) \(=\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \)
\(\Rightarrow tan\quad \theta =\frac { \pm 2\sqrt { \frac { 25 }{ 4 } +6 } }{ 1 } \)
\(\theta=\tan ^{-1}\left[\not 2\left(\frac{7}{\not 2}\right)\right]=\tan ^{-1}\)

let P (x₁,y₁)  be any point on the locus and A (3, -2) the given point
PA = 4
\(\Rightarrow\) PA² = 16
\(\Rightarrow\)  (x₁ - 3)² + (y₁ +2)²  = 16
\(\Rightarrow\)   \({ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }\) - 6x₁ + 4y₁ + 13 -16 = 0
\(\Rightarrow\)   \({ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }\) - 6x₁ + 4y₁ - 3 = 0
Locus of (x₁,y₁) is x² + y² - 6x + 4y - 3 = 0

4y = x² + 6x + 21
\(4 y=\left(x^2+6 x+9\right)+12\)
\(4 y-12=(x+3)^2\)
(x + 3)² = 4(y - 3)
X = x + 3, Y = y - 3
\(x=X-3 \text { and } y=Y+3\)
\(X^2=4 Y\)
Comparing with \(X^2=4 a Y, 4 a=4\)
a = 1

	Referred to (X, Y)	Referred to (x, y) x = X - 3 y = Y + 3
Axis y = 0	Y = 0	y = 3
Vertex V(0,0)	V(0,0)	V(–3,3)
Focus F(0, a)	Y = –1	y = 2
Length of Latus rectum (4a)	4(1) = 4	4

Equation of the circle be x² + y² + 2gx + 2fy + c = 0  ..(1)
It passes through (0, 1)
0 + 1 + 0 + 2f + c = 0   \(\Rightarrow \) 2f + c = -1 ..(2)
The circle passes through (4, 3)
16 + 9 + 8g + 6f + c = 0  \(\Rightarrow \) 8g + 6f + c = -25 ...(3)
The circle passes through (1,-1)
1 + 1 + 2g - 2f + c = 0  \(\Rightarrow \) 2g - 2f + c = -2 ....(4)
Solving (1), (2) and (3) we get
c = 1   \(\Rightarrow\)  g = -5/2    \(\Rightarrow\)  f = -1
Equation of circle is
x² + y² + 2 \(\left( -\frac { 5 }{ 2 } \right) \)x + 2(-1)y + 1 = 0
x² + y² - 5x - 2y + 1 = 0

\(\left(x+\frac{2^{-}}{x^2}\right)^{17}\)
\(n=17\)
\(T_{r+1}=n C_r \cdot x^{n-r} a^{r h}\)
\(=17 C_r x^{17-r}\left(\frac{2}{x^2}\right)^{\mathrm{r}}\)
\(=17 C_r 2^r x^{17-r-2 r}=17 C_r 2^r x^{17-3 r}\)
To find co-efficient of \(x^{11}\) equate the power of x to 11
17 - 3r = 11
17 - 11 = 3r
6 = 3r \(\Rightarrow\) r = 2
Co-efficient of \(x^{11}=17 C_2 2^2\)
\(=\frac{17 \times 16}{2 \times 1} \times 2^2=544\)

G	L	Number of ways
(6)	(4)
3	2	\(6 C_3 \times 4 C_2\)	120
2	3	\(6 C_2 \times 4 C_3\)	60
1	4	\(6 C_1 \times 4 C_4\)	6
			186

G	L	Number of ways
(6)	(4)
5	0	\(6 C_5 \times 4 C_0\)	6
4	1	\(6 C_4 \times 4 C_1\)	60
3	2	\(6 C_3 \times 4 C_2\)	120
			186

4x² + 12y +9y² - 6x - 9y + 2 = 0
a = 4, 2h = 12
b = 9, h = 6
h² - ab = 36 - 36 = 0
Hence the given equation represents a pair of parallel straight lines.
To find separate equations
\(4 x^2+12 x y+9 y^2=(2 x+3 y)^2\)
\(4 x^2+12 x y+9 y^2-6 x-9 y+2=0\)
\((2 x+3 y)^2-3(2 x+3 y)+2=0\)
Let 2x + 3y = z
\(z^2-3 z+2=0\)
(z - 1) (z - 2) = 0,
Z - 1 = 0,   z - 2 = 0
2x + 3y -1 = 0, 2x + 3y - 2 = 0
Separate equations are 2x + 3y 1 = 0
2x + 3y - 2 = 0

\(\sqrt{4^2+5^2-16}=\sqrt{5^2}=5\)

\(a+b=0 \Rightarrow k-2=0\)

\(t_{r+1} =6 C_r n^{6-r}\left(\frac{2^r}{x^r}\right) \)
\(=6 C_r x^{6-2 r} 2^r \)
\(6-2 r =0 = r =3 \)
\(t_{r+1} =6 C_3 2^3=160 \)

\(\text { M.T } =t_{\frac{n}{2}}+1=t_{5+1}=t_6 \quad r=5 \)
\(t_6 =10 C_5\left(x^5\right)\left(\frac{1}{x^5}\right)=10 C_5 \)

nP₂ = 20
n(n - 1) = 5 x 4
n = 5