In Δ POQ, DE || OQ    [given]
\(\therefore \frac{P E}{E Q}=\frac{P D}{D O}\) ...(i)
[by basic proportionality theorem] 
In ΔPOR,    DF || OR     [given]
\(\therefore \frac{P F}{F R}=\frac{P D}{D O}\) ...(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get 
\(\frac{P E}{E Q}=\frac{P F}{F R}\)
In ΔPQR, we have \(\frac{P E}{E Q}=\frac{P F}{F R}\)
∴  EF || QR
[by Converse of basic proportionally theorem]
Hence proved.

Given point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).
So,   AP = BP
\(\Rightarrow\)   AP² = BP²
\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
       \(\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)
\(\Rightarrow\)   x² + 49 - 14x + y² + 1 - 2y
                = x² + 9 - 6x + y² + 25 - 10y
\(\Rightarrow\)  -14x - 2y + 50 = -6x - 10y + 34
\(\Rightarrow\) -6x - 10y + 14x + 2y = 50 - 34
\(\Rightarrow\)              8x - 8y = 16
\(\Rightarrow\)                 x - y = 2
                                [dividing by 8 on both sides]
Hence, the relation between x and y is x - y = 2.

9x²-6a²x+(a⁴-b⁴) =0
Here A=9 , B=-6a²,C=a⁴-b⁴
D=B²- 4AC
=36a⁴-36a⁴+36b⁴ =36b^{4 
\(\therefore \quad x=\frac { -B+\sqrt { D } }{ 2A } ,\frac { -B-\sqrt { D } }{ 2A } \)}
\(\Rightarrow \quad x=\frac { 6a^{ 2 }+6b^{ 2 } }{ 18 } ,\frac { 6a^{ 2 }-6b^{ 2 } }{ 18 } \)
\(\Rightarrow x=\frac { a^{ 2 }+b^{ 2 } }{ 3 } ,\frac { a^{ 2 }-b^{ 2 } }{ 3 } \)

Let digit at unit's place be x and digit at ten's place by y
\(\therefore\) Number 10y+x
ATQ
10y+x=7(x+y)
\(\Rightarrow\) 10y+x=7x+7y
\(\Rightarrow\) 6x=3y
\(\Rightarrow\) y=2x    ...(i)
Also 10y+x=3xy-12
\(\Rightarrow\) 10 x 2x + x=3x.2x-12
\(\Rightarrow\) 6x²-21x-12=0
\(\Rightarrow\) 2x²-7x-4=0
\(\Rightarrow\) 2x²-8x+x-4=0
\(\Rightarrow\) 2x(x-4)+1(x-4)=0
\(\Rightarrow\) (x-4)(2x+1)=0
\(\Rightarrow\) x=4 or x=\(-\frac{1}{2}\) (rejecting)
When x=4, y=2  x 4=8
\(\therefore\) Number is 10 x 8+4=84

Let AB be a diameter of a given circle and LM and PQ be the tangent lines drawn to the circle at points A and B, respectively.

To prove LM || PQ
Proof We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OA \(\perp\) PQ and OB \(\perp\) LM
\(\Rightarrow\) AB \(\perp\) PQ
and AB \(\perp\) LM
\(\Rightarrow\) \(\angle\)PAB = 90°
and  \(\angle\)ABM = 90°
\(\Rightarrow\) \(\angle\)PAB = \(\angle\)ABM
[each = 90°]
But these are alternate angles.
\(\therefore\) PQ || LM
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Hence proved.

(i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP. Let us denote the number of TV sets manufactured in the nth year by a_n.
Then, a₃ = 600 and a₇ = 700
or,   a + 2d = 600
and   a + 6d = 700
Solving these equations, we get d = 25 and a = 550.
Therefore, production of TV sets in the first year is 550.
(ii) Now   a₁₀ = a + 9d = 550 + 9 x 25 = 775
So, production of TV sets in the 10th year is 775.
(iii) Also  \(S_{7}=\frac{7}{2}[2 \times 550+(7-1) \times 25]\)
\(=\frac{7}{2}[1100+150]=4375\)
Thus, the total production of TV sets in first 7 years is 4375.

Given, in △ABC and △PQR, AD and PM are their medians, respectively.
\(\therefore \quad \frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M}\)  ...(i)
To prove △ABC ~ △PQR
Construction Produce AD to E such that AD = DE and produce PM to N such that PM = MN.
Join BE, CE, QN and RN.

In above figures, quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.
\(\therefore\) BE = AC and QN = PR
\(\begin{array}{ll} \Rightarrow & \frac{B E}{A C}=1 \quad \text { and } \frac{Q N}{P R}=1 \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{B E}{A C}=\frac{Q N}{P R} \text { or } \frac{B E}{Q N}=\frac{A C}{P R} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{B E}{Q N}=\frac{A B}{P Q} \\ \end{array}\)      [from Eq. (i)]
\(\begin{array}{ll} \text { or } & \frac{A B}{P Q}=\frac{B E}{Q N} \end{array}\)          ....(ii)
From Eq. (i), we get
\(\frac{A B}{P Q}=\frac{A D}{P M}=\frac{2 A D}{2 P M}=\frac{A E}{P N}\) [since, diagonals bisect each other]
\(\Rightarrow \quad \frac{A B}{P Q}=\frac{A E}{P N}\)    ...(iii)
From Eqs. (ii) and (iii),
\(\frac{A B}{P Q}=\frac{B E}{Q N}=\frac{A E}{P N}\)
\(\Rightarrow \quad \triangle A B E \sim \triangle P Q N \Rightarrow \angle 1=\angle 2\)  ...(iv)
[since, corresponding angles of two similar triangles are equal]
Similarly, we can prove that
\(\begin{aligned} & \triangle A C E \sim \triangle P R N \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \angle 3=\angle 4 \end{aligned}\)   ....(v)
On adding Egs. (iv) and (v), we get
\(\begin{array}{rlrl} \angle 1+\angle 3 & =\angle 2+\angle 4 \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \angle B A C =\angle Q P R \\ \end{array}\)
and \(\begin{array}{rlrl} \frac{A B}{P Q} & =\frac{A C}{P R} \end{array}\)    [from Eq. (i)]
\(\therefore\)   \(\triangle ABC \sim \triangle PQR\) [by SAS similarity criterion]
Hence proved.

Let X-axis divides the line joining the points A(1, -5) and B(-4, 5)in the ratio k : 1 at point M.

By section formula, we get
Coordinates of M = \(\left(\frac{-4 k+1}{k+1}, \frac{5 k-5}{k+1}\right)\)   ...(i)
Since, point M lies on X-axis.
\(\therefore\) y-coordinate of M = 0
On equating y-coordinate of M to 0, we get
\(\frac{5 k-5}{k+1}=0 \Rightarrow 5 k-5=0 \Rightarrow k=1\)
On putting k = 1 in Eq. (i) we get
\(\begin{aligned} M=\left(\frac{-4 \times 1+1}{1+1}, \frac{5 \times 1-5}{1+1}\right) & =\left(\frac{-4+1}{2}, \frac{5-5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad M & =\left(\frac{-3}{2}, 0\right) \end{aligned}\)
So, the required ratio is 1 : 1 and the coordinates of point of division M is \(\left(\frac{-3}{2}, 0\right)\)

Given XY and X'Y' are two parallel tangents. Another tangent AB touches the circle at C and intersect XY at A and X'Y'at B.
To prove \(\angle\)AOB = 90°
Proof We know that, tangents drawn from an external point to a circle are equal in length.
\(\therefore\)   AP = AC   [\(\because\) A is an external point] ...(i)
Thus, in \(\Delta\)APO and \(\Delta\)ACO, AP = AC   [from Eq. (i)]
AO = AO  [common sides]
OP = OC [radii of circle]
\(\Delta\)APO \(\cong\)\(\Delta\)ACO  [by SSS congruence rule]
Then, \(\angle\)OAP = \(\angle\)OAC    [by CPCT]        ...(ii)
\(\Rightarrow\) \(\angle\)PAC = 2 \(\angle\)CAO                             ....(iii)
Similarly, we can prove that \(\angle\)CBO = \(\angle\)OBQ
\(\Rightarrow\)   \(\angle\)CBQ = 2 \(\angle\)CBO                           ...(iv)
since, XY || X'Y'      [given]
\(\therefore\) \(\angle\)PAC +  \(\angle\)QBC = 180°
[\(\because\) sum of interior angles on the same side of transversal is 180°]
\(\Rightarrow\) 2 \(\angle\)CAO + 2 \(\angle\)CBO = 180°    [from Eqs. (iii) and (iv)]
\(\Rightarrow\)   \(\angle\)CAO + \(\angle\)CBO = 90°                    ...(v)
Now, in \(\Delta\)AOB, \(\angle\)CAO + \(\angle\) CBO + \(\angle\)AOB = 180°
[by angle sum property of triangle]
\(\Rightarrow\) \(\angle\)CAO + \(\angle\)CBO = 180° - \(\angle\)AOB    ...(vi)
From Eqs. (v) and (vi), we get
180° - \(\angle\)AOB = 90° \(\Rightarrow\) \(\angle\)AOB = 90° Hence proved.

Join OT
Let OT intersect PQ at R
\(\because \ T P=T Q\) 
 
In \(\Delta T P Q\) ,
TP = TQ, i.e. two sides are equal
So, \(\Delta T P Q\) is an isosceles triangle
Here, OT is bisector of \(\angle P T Q\) 
So, \(O T \perp P Q\) 
So, PR = RQ
\(P R=\frac{1}{2} P Q=\frac{8}{2}=4 \mathrm{~cm}\) 
In right \(\Delta O R P\) ,
(OP)² = (PR)² + (OR)²
\(\Rightarrow \ (5)^{2}=(4)^{2}+(O R)^{2}\) 
\(\Rightarrow \ (O R)^{2}=25-16=9 \Rightarrow O R=3\) 
Let TP = x
In right \(\Delta P R T,(T P)^{2}=(P R)^{2}+(R T)^{2}\) 
\(\Rightarrow \ x^{2}=16+R T^{2}\)  ...(i)
Now, since TP is a tangent
\(O P \perp T P\) 
In right \(\Delta O P T,(O T)^{2}=(O P)^{2}+(T P)^{2}\) 
\(\Rightarrow \ (O T)^{2}=5^{2}+x^{2}\) 
\(\Rightarrow \ (O R+R T)^{2}=5^{2}+x^{2}\) 
\(\Rightarrow \ (3+R T)^{2}=5^{2}+x^{2}\) 
\(\Rightarrow 9+R T^{2}+6 R T=25+x^{2}\) 
\(\Rightarrow \quad 9+R T^{2}+6 R T=25+16+R T^{2}\) [from Eq. (i)]
\(6 R T=32 \Rightarrow R T=\frac{32}{6}=\frac{16}{3}\) 
From Eq. (i), x²=16+RT²
\(\Rightarrow \ x^{2}=16+\left(\frac{16}{3}\right)\) 
\(\Rightarrow \ x^{2}=\frac{144+256}{9} \Rightarrow x^{2}=\frac{400}{9}\) 
\(\Rightarrow \ x=\frac{20}{3}\)

Given, ΔABC is an isosceles triangle with AB = AC. Also,
we have AD \(\perp\) BC and EF \(\perp\)AC.
To prove ΔABD \(\sim\) ΔECF
Proof Since, in ΔABC, AB = AC
∴ \(\angle\)B = \(\angle\)C     [\(\therefore\) angle opposite to equal sides are equal]
Now consider ΔABD and ΔECF. In this we have
∠ABD = ∠ECF   [\(\because\) \(\angle\)B = \(\angle\)C proved above]
and ∠ADB = ∠EFC       [Each 90°]
∴  ΔABD ∼ ΔECF    [by AA similarly criterion]
Hence proved.

For x + 2y = 3
a₁ = 1, b₁ = 2 ,c₁ = - 3
For (k-1)x + (k +1)y= (k + 2)
a₂ = (k - 1), b₂= (k + 1), c₂ = - (k + 2)
For no solution, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { c_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \quad \frac { 1 }{ k-1 } =\frac { 2 }{ k+1 } \neq \frac { 3 }{ k+2 } \)
           I           II          III
From I and II, \(\frac { 1 }{ k-1 } =\frac { 2 }{ k+1 } \)
\(\Rightarrow\)  k + 1 = 2k - 2
\(\therefore\)  k = 3
From II and III, \(\frac { 2 }{ k+1 } \neq \frac { 3 }{ k+2 } \)
\(\Rightarrow\) 2(k + 2) \(\neq\) 3(k + 1)
\(\Rightarrow\) 2k+4\(\neq\)3k+3
\(\therefore\)  k \(\neq\) 1
From I and III, \(\frac { 1 }{ k-1 } \neq \frac { 3 }{ k+2 } \)
\(\Rightarrow \quad k+2\neq 3k-3\)
\(\therefore \quad k\neq -\frac { 5 }{ 2 } \)
Hence, k = 3 but k \(\neq\)1 and \(\\ \\ k\neq -\frac { 5 }{ 2 } \)

42 km/h

AB =\(\sqrt { { (5-1 })^{ 2 }+(3-0)^{ 2 } } =\sqrt { 16+9 } \) =5
DC = \(\sqrt { { (2+2 })^{ 2 }+(7-4)^{ 2 } } =\sqrt { 16+9 } \) =5
BC = \(\sqrt { { (5-2 })^{ 2 }+(3-7)^{ 2 } } =\sqrt { 9+16 } \) = 5
AD = \(\sqrt { { (1+2 })^{ 2 }+(0-4)^{ 2 } } =\sqrt { 9+16 } \) = 5
Mid - point of AC = \((\frac {1+2}{2} , \frac {0+7}{2})\) = \((\frac {3}{2} , \frac {7}{2})\); Mid-point BD = \((\frac{5-2}{2},\frac{3+4}{2})\) = \(( \frac {3}{2}, \frac {7}{2})\)
Since AB = DC and BC = AD.
Opposite sides are parallel and diagonals bisect each other.
∴ The given points are the vertices of a parallelogram.  

Let first term of the AP = a and common difference = d
Now a₁₂ = -13
\(\Rightarrow \) a+11d = -13
\(\Rightarrow \) a=-13-11d .......... 9i)
Also, a+a+d+a+2d+a+3d=24
\(\Rightarrow \) 4a+6d = 24
\(\Rightarrow \) 4(-13-11d)+6d=24 [using (i)]
\(\Rightarrow \) -52-44d+6d=24 \(\Rightarrow \) -38d=76
\(\Rightarrow \) d=-2
From (i), a = -13+22=9
Now \({ S }_{ 10 }=\frac { 10 }{ 2 } \left( 2a+9d \right) \)
\(=5\left( 2\times 9+9\times -2 \right) =0\)

Here, a = -6, d = - 2 +  6 and a_n = 58
a_n = 58
\(\Rightarrow\)    a + ( n - 1 )d = 58  \(\Rightarrow\) - 6 + ( n - 1 )4 = 58
\(\Rightarrow\)    ( n - 1 )4 = 64  \(\Rightarrow\) n - 1 = 16 \(\Rightarrow\) n = 17 (odd)
\(\therefore\)       Middle term = \(\frac{17+1}{2}=\frac{18}{2}\) = 9th term
\(\therefore\)   9th term is the middle term.
Now,    a₉ = a + 8d = - 6 + 8 X 4 = - 6 + 32 = 26

(i) (a): Let the production during first year be a and let d be the increase in production every year. Then,
a₆ = 16000 \(\Rightarrow\)a + 5d = 16,000     (i)
and a₉ = 22600 \(\Rightarrow\)a + 8d = 22600  (ii)
On substracting (i) from (ii) , we get
3d = 6600 \(\Rightarrow\) d = 2200
Putting d = 2200 in (i) we get,
a + 5 x 2200 = 16000
\(\Rightarrow\) a + 11000 = 16000 \(\Rightarrow\) a = 16000 - 11000 = 5000
Thus , a = 5000 and d = 2200
Production during first year, a = 5000.
(ii) (c): Production during 8th year is given by a₈
= (a + 7d)
= (5000 + 7(2200))
= (5000 + 15400)
= 20400.
(iii) (a): a₂ = (a + d)
= (5000 + 2200) = 7200.
a₃ = (a2 + d)
= 7200 + 2200 = 9400.
Production during first 3 years = 5000 + 7200 + 9400 = 21600
(iv) (c): a_n = 5000 + (n – 1)2200 = 29200
(n – 1)2200 = 29200 – 5000 = 24200
⇒ n – 1  = 11
⇒ n = 12
(v) (d): a₄ = (a + 3d)
= (5000 + 3(2200)) = 5000 + 6600 = 11600.
a₇ = (a6 + d)
= 16000 + 2200 = 18200.
Difference = 18200 – 11600 = 6600

(i) (b)
(ii) (c): In \(\Delta\)OAP and \(\Delta\)OBP,
\(\angle\)OAP = \(\angle\)OBP = 90°
[Since, radius at the point of contact is perpendicular to tangent]
OP = OP (Common)
OA = OB (Radii of circle)
So, \(\angle\)OAP == \(\angle\)OBP (By RHS congruency criterion)
(iii) (b): In quadrilateral OAPB, \(\angle\)AOB = 150°  [Given]
\(\angle\)OAP = \(\angle\)OBP = 90°
\(\therefore\) \(\angle\)APB = 360° - 90° - 90° - 150° = 30°
(iv) (d): We have, \(\angle\)APB = 40°

Now,PA =PB [Since, length of tangents drawn  from an external point are equal]
In \(\Delta\)PAB, \(\angle\)PAB = \(\angle\)PBA = 70° [Angles opposite to equal sides are equal]
Also, \(\angle\)PAB + \(\angle\)BAO = 90°
[Since, radius at the point of contact is perpendicular to tangent]
\(\Rightarrow\) \(\angle\)BAO = 90° - 70° = 20°
(v) (d): We have, \(\angle\)ABO = 45°

\(\because\) AO = OB (Radii of circle)
\(\therefore\) \(\angle\)BAO = \(\angle\)ABO = 45°  [Angles opposite to equal sides are equal]
Now, in \(\Delta\)OAB,
\(\angle\)AOB = 180° - 45° - 45° = 90°
Since, \(\angle\)APB = 360° - 90° - 90° - 90° = 90° i.e., AP\(\perp\) BP
So, OAPB is a square.

(i) (b): The distance between A and C
\(=\sqrt{(8-4)^{2}+(5+3)^{2}}=\sqrt{4^{2}+8^{2}} \)
\(=\sqrt{16+64}=\sqrt{80}=4 \sqrt{5} \text { units }\)
(ii) (a): Let the coordinates of I be (x, y).

Then, by section formula
\(x =\frac{1 \times 8+2 \times 7}{1+2}=\frac{8+14}{3}=\frac{22}{3}\)
\(\text { and } y =\frac{1 \times 5+2 \times 3}{1+2}=\frac{5+6}{3}=\frac{11}{3}\)
Thus, the coordinates of I is \(\left(\frac{22}{3}, \frac{11}{3}\right)\)
(iii) (b): The mid -point of A and C
\(=\left(\frac{8+4}{2}, \frac{5-3}{2}\right)=(6,1)\)
(iv) (b): Let B divides the line segment joining A and C in the ratio k : 1. Then, the coordinates of B will be
\(\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)\)
\(\text { Thus, we have }\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)=(7,3)\)
\(\Rightarrow \frac{8 k+4}{k+1}=7 \text { and } \frac{5 k-3}{k+1}=3 \)
\(\text { Consider, } \frac{8 k+4}{k+1}=7 \Rightarrow 8 k+4=7 k+7 \Rightarrow k=3\)
Hence, the required ratio is 3 : 1
(v) (a):\(\because\) B divides AC in the ratio 3 : 4.
\(\therefore\) A, B, C lie on a straight line.