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(c) 3x – 4y + 15 = 0 

Let (m > 0) be the gradient (slope) of the required line. Then, 

Equation of any line through (–5, 0) having slope = m is 

y – 0 = m(x – (–5)) or mx – y + 5m = 0           ...(i) 

Its perpendicular distance from origin is 3

⇒ \(\frac{\pm|\,m.\,0-0+5m\,|}{\sqrt{m^2+(-1)^2}}\) = 3 ⇒ | 5m | = 3\(\sqrt{m^2+1}\)

\(\bigg(\because\text{Distance of}\,(x_1,y_1)\,\text{from line}\,as+by+c=\frac{|as_1+by_1+c|}{\sqrt{a^2+b^2}}\bigg)\)

⇒ 25m² = 9(m² + 1) ⇒ 16m² = 9 ⇒ m = \(\frac{3}{4}\) (∵ m is +ve)

∴ Required equation: y = \(\frac{3}{4}\) (x + 5)

⇒ 3x – 4y + 15 = 0.

Co-ordinate geometry is the system of geometry where the position of points on the plane is described using an ordered pair of numbers.

Let A(x₁, y₁) = (-4, 8), B(x₂, y₂) = (-3, -4), C(x₃, y₃) = (0, –5) and D(x₄, y₄) = (5, 6) be the vertices of quadrilateral ABCD.

Area of quad. ABCD = \(\frac{1}{2}\) |{(x₁ y₂ – x₂ y₁) + (x₂y₃ – x₃y₂) + (x₃y₄ – x₄y₃) + (x₄y₁ – x₁y₄)}|

= \(\frac{1}{2}\) |{(-4 × (-4) – (-3) × 8) + (-3 × (–5) – 0 × (-4)) + (0 × 6 – 5 × (–5)) + (5 × 8 – (-4) × 6)}|

= \(\frac{1}{2}\) |{(16 + 24) + (15 + 0) + (0 + 25) + (40 + 24)}|

= \(\frac{1}{2}\) |{40 + 15 + 25 + 64}| 

= \(\frac{1}{2}\)x 144 

= 72 sq. units.

(d) Both (a) and (c)

Since the line passes through A(a, 0) and B(0, b), it makes intercepts a and b on x-axis and y-axis respectively. 

Let the equation of this line in the intercept from be \(\frac{x}{a}\) + \(\frac{y}{a}\) = 1

By the given condition, AB = \(\sqrt{a^2+b^2}\) = 13 and ab = 60

a² + b² = 169 ⇒ a² + b² + 2ab = 169 + 120 

⇒ (a + b)² = 289 ⇒ a + b = ±17                 ...(i) 

Also, a² + b² – 2ab = 169 – 120 ⇒ (a – b)² = 49 

⇒ a – b = ±7.                         ...(ii) 

∴ From (i) and (ii) a = 12, b = 5 and a = –12, b = –5 

∴ The required equations of the straight line are:

\(\frac{x}{12}\) + \(\frac{y}{5}\) = 1 and \(\frac{x}{-12}\) + \(\frac{y}{-5}\) = 1

⇒ 5x + 12y = 60 and 5x + 12y + 60 = 0.

(b)10 + \(5\sqrt2\)

Perimeter of ΔABC = AB + BC + CA

= \(\sqrt{(0+4)^2+(-1-2)^2}\) + \(\sqrt{(3-0)^2+(3+1)^2}\) + \(\sqrt{(3-4)^2+(3-2)^2}\)

= \(\sqrt{16+9}\) + \(\sqrt{9+16}\) +\(\sqrt{49+1}\)

= \(\sqrt{25}\) + \(\sqrt{25}\) + \(\sqrt{50}\) = 5 + 5 + \(5\sqrt2\) = 10 + \(5\sqrt2\)

The two given lines are ax + by + c₁ = 0 and ax + by + c₂ = 0. 

Any line parallel to these two lines and midway between them is 

ax + by + c = 0              ...(i) 

Putting x = 0, y = \(-\frac{c}{b}\) is a point on line (i)

It is equidistant from the given lines but in opposite directions, so

\(\frac{\big|a\times0+b\times-\frac{c}{b}+c_1\big|}{\sqrt{a^2+b^2}}\) = \(-\frac{\big|a\times0+b\times-\frac{c}{b}+c_2\big|}{\sqrt{a^2+b^2}}\) ⇒ – c + c₁ = c – c₂ ⇒ c = \(\frac{c_1+c_2}{2}\)

∴ Required equation is ax + by + \(\frac{c_1+c_2}{2}\) = 0.

Correct option is (A) \(\sqrt{x_1^2+y_1^2}\)

Distance between points \(A(0,0)\;\&\;B(x_1,y_1)\) is

\(AB=\sqrt{(x_1-0)^2+(y_1-0)^2}\)

\(=\sqrt{x_1^2+y_1^2}\) units

Correct option is A) \(\sqrt{x_1^2+y_1^2}\)

Vertices of the rhombus are: A(- 3, 2), B(- 5, - 5), C(2, -3) and D(4, 4)

We know that diagonals of a rhombus bisect each other, therefore point of intersection of diagonals is: 

Abscissa of Mid point of AC = \(\frac{2 - 3}2\) = \(\frac{ - 1}2\)

Ordinate of Mid point of AC = \(\frac{-3 + 3}2\) = \(\frac{ - 1}2\)

Abscissa of Mid point of BD = \(\frac{4 - 5}2\) = \(\frac{ - 1}2\)

Ordinate of Mid point of BD = \(\frac{4 - 5}2\) = \(\frac{ - 1}2\)

Since the diagonals AC and BD bisect each other at O, therefore it is a rhombus. 

Length of diagonal AC 

= \(\sqrt{(2 + 3)^2 + (-3 - 2)^2}\) = \(\sqrt{25 + 25}\) = \(\sqrt{50}\) = \(5\sqrt{2}\) units

Length of diagonal BD 

= \(\sqrt{(4 + 5)^2 + (4 + 5)^2}\) = \(\sqrt{81 + 81}\) = \(\sqrt{162}\) = \(9\sqrt{2}\) units

Area of rhombus = \(\frac{1}2\times{d1}\times{d2}\) = \(\frac{1}2\times{5\sqrt2}\times{9\sqrt2}\) = 45 sq units

Area of rhombus is 45 sq units

Coordinates of points on a circle are A (3, 0), B(-1, - 6) and C(4,-1) 

Let the coordinates of the centre of the circle be O(x, y) 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Since the distance of the points A, B and C will be equal from the center, therefore 

⇒ OA = OC

\(\sqrt{(x - 3)^2 + (y - 0)^2}\) = \(\sqrt{(x + 1)^2 + (y + 6)^2}\)

On squaring both sides, we get

(x - 3)² + (y - 0)² = (x + 1)² + (y + 6)² 

⇒ x² + 9 - 6x + y² = x² + 2x +1 + y² + 36 + 12y 

⇒ - 8x - 12y = 28 

⇒ 2x + 3y = -7------------- (1) 

Similarly, OC = OB

\(\sqrt{(x - 4)^2 + (y + 1)^2}\) = \(\sqrt{(x + 1)^2 + (y + 6)^2}\)

On squaring both sides, we get 

(x - 4)² + (y + 1)² = (x + 1)² + (y + 6)² 

⇒ x² + 16 - 8x + y² + 1 + 2y = x² + 1 +2x + y² + 36 + 12y 

⇒ - 10x - 10y = 20 

⇒ x + y = - 2 ------------- (2) 

Solving eqn (1) and (2), we get 

x = 1; y = - 3 

Coordinates of circum center are (1, - 3) 

Circum radius of the circle = OA = \(\sqrt{(1 - 3)^2 + (3)^2}\)

= \(\sqrt{4 +13}\)

= \(\sqrt{13}\) units

points A(7, 6) and B(-3, 4) are equidistance from point P. 

Let the coordinates of point are P(x, 0) 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

⇒ PA = PB

\(\sqrt{(x - 7)^2 + (0 - 6)^2}\) = \(\sqrt{(x + 3)^2 + (0 - 4)^2}\)

On squaring both sides, we get 

(x - 7)² + (0 - 6)² = (x + 3)² + (0 - 4)²

x² - 14x + 49 + 36 = x² + 6x + 9 + 16

5x = 15

x = 3

Therefore coordinates are (3, 0)

Points A(-2, 5) and B(2, -3) are equidistant from point P. 

Let the coordinates of point are P(x, 0) 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

⇒ PA = PB

\(\sqrt{(x + 2)^2 + (0- 5)^2}\) = \(\sqrt{(x - 2)^2 + (0+ 3)^2}\)

On squaring both sides, we get 

(x + 2)² + (0 - 5)² = (x - 2)² + (0 + 3)²

x² + 4x + 4 + 25 = x² - 4x + 4 + 9

8x = - 16x = -2 

Hence, 

coordinates are (- 2, 0).

Coordinates are P(x, y), A(5, 1) and B (1, 5) 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

⇒ PA = PB

\(\sqrt{(x - 5)^2 + (y - 1)^2}\) = \(\sqrt{(x - 1)^2 + (y - 5)^2}\)

On squaring both sides, we get

(x -5)² + (y - 1)² = (x - 1)² + (y - 5)²

x² - 10x + 25 + y² - 2y + 1 = x² - 2x + 1 + y² - 10y + 25

- 8x + 8y = 0

x = y proved

Coordinates are P(8, -3) and Q(7,6) 

The point A (3, y) is equidistant. 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

⇒ PA = QA

\(\sqrt{(3 - 8)^2 + (y + 3)^2}\) = \(\sqrt{(3 - 7)^2 + (y - 6)^2}\)

On squaring both sides, we get

(3 - 8)² + (y + 3)² = (3 - 7)2 + (y - 6)²

25 + y² + 6y + 9 = 16 + y² - 12y + 36

y = 1

AQ = \(\sqrt{(3 - 7)^2 + (1 - 6)^2}\) = \(\sqrt{16 + 25}\) = \(\sqrt{41}\) units

Coordinates are A (7,-1) and B (6, 8) 

The point P (x, 3) is equidistant. 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

⇒ PA = PB

\(\sqrt{(x - 7)^2 + (3 + 1)^2}\) = \(\sqrt{(x - 6)^2 + (3 - 8)^2}\)

On squaring both sides, we get

(x - 7)² + (3 + 1)² = (x - 6)² + (3 - 8)²

x² - 14x + 49 + 16 = x² - 12x + 36 + 25

x = 2

AP = \(\sqrt{(x - 7)^2 + (3 + 1)^2}\) = \(\sqrt{25 + 16}\) = \(\sqrt{41}\) units

Slope (m) = \(\frac{(y_2-y_1)}{(x_2-x_1)}\) = \(\frac{6-2}{5-1}\) = \(\frac{4}{4}\) = 1

Also slope (m) = tan θ, where θ is the inclination of the line to the positive direction of the x-axis in the anticlockwise direction. 

tan θ = 1 ⇒ θ = tan ^–11 = 45º.

θ = 45°

Coordinate of foot of hill = (19, 3) let equation of line be y = mx + c

m = tan θ = tan 45° = 1

⇒ y = x + c

Substituting y = 3 & x = 19, 3 = 19 + c ⇒ c = -16

∴ Equation of line:

y = x - 16

= x - y - 16 = 0 

(2) Two parallel and two non-parallel sides.

Let A (-2, 5), B(0, 1) and C (2, -3) be the given points

So, we have

AB = √[(0 – (-2))² + (1 – 5)²] 

= √[(2)² + (-4)²] 

= √[4 + 16] 

= √20 

= 2√5 units

BC = √[(2 – 0)² + (-3 – 1)²] 

= √[(2)² + (-4)²] 

= √[4 + 16] 

= √20 

= 2√5 units

AC = √[(2 – (-2))² + (-3 – 5)²] 

= √[(4)² + (-8)²] 

= √[16 + 64] 

= √80 

= 4√5 units

Now, it’s seen that

AB + BC = AC

2√5 + 2√5 = 4√5

4√5 = 4√5

Therefore, we can conclude that the given points (-2, 5), (0, 1) and (2, -3) are collinear.

Let A (5, 9) and B (-4, 6) be the given points

Let the point on x – axis equidistant from the above points be C(x, 0)

Now, we have

AC = √[(x – 5)² + (0 – 9)²] 

= √[x² – 10x + 25 + 81] 

= √[x² – 10x + 106]

And,

BC = √[(x – (-4))² + (0 – 6)²] 

= √[x² + 8x + 16 + 36] 

= √[x² + 8x + 52]

As AC = BC (given condition)

So, AC² = BC²

x² – 10x + 106 = x² + 8x + 52

18x = 54

x = 3

Therefore, the point on the x-axis is (3, 0)

Since the point is on x-axis, therefore coordinate of y-axis is zero.

 Therefore the point is P(k, 0) which is equidistance from A(5, 9) and B(- 4, 6) 

PA = PB

\(\sqrt{(5 - k)^2 + 9^2}\) = \(\sqrt{(-4 - k)^2 + 6^2}\)

 On squaring both sides

(5 - k)² + 9² = (- 4 - k)² + 6²

25 - 10k + k² + 81 = 16 - 8k + k² + 36

25 - 2k + 81 = 16 + 36 - 25 - 81

k = 27

Therefore coordinate is (27, 0)

Let the third vertex be C (x, y)

And, given A (2, 0) & B (2, 5)

We have,

Length of AB = √[(2 – 2)² + (5 – 0)²] 

= √[(0)² + (5)²] 

= √[0 + 25] 

= 5 units

Length of BC = √[(x – 2)² + (y – 5)²] 

= √[x² – 4x + 4 + y² – 10y + 25]

= √[ x² – 4x + y² – 10y + 29] units

Length of AC = √[(x – 2)² + (y – 0)²] 

= √[x² – 4x + 4 + y²] units

Given that,

AC = BC = 3

So, AC² = BC² = 9

x² – 4x + 4 + y² = x² – 4x + y² – 10y + 29

10y = 25

y = 25/10 = 2.5

And,

AC² = 9

x² – 4x + 4 + y² = 9

x² – 4x + 4 + (2.5)² = 9

x² – 4x + 4 + 6.25 = 9

x² – 4x + 1.25 = 0

D = (-4)² – 4 x 1 x 1.25 

= 16 – 5 

= 11

So, the roots are

x = -(-4) + √11/ 2 

= (4 + 3.31)/ 2 

= 3.65

And,

x = -(-4) – √11/ 2 

= (4 – 3.31)/ 2 

= 0.35

Therefore, the third vertex can be C (3.65, 2.5) or (0.35, 2.5).

Correct option is (A) X-axis

Equation y = 0 represents X-axis.

Correct option is A) X – axis

Correct option is (B) parallel to Y–axis

Equation of Y-axis is x = 0.

\(\therefore\) Equation x = 0 represents Y-axis.

\(\therefore\) Equation x = k represents line parallel to Y-axis.

Correct option is B) parallel to Y – axis

Correct option is (A) parallel to X–axis

\(\because\) Equation y = 0 represents X-axis.

\(\therefore\) Equation y = k represents line parallel to X-axis.

Correct option is A) parallel to X – axis

Here both coordinates are positive therefore point A lies in Ist quadrant. 

(a) The distance d between any two points say P(x₁, y₁) and Q(x₂, y₂) is given by:

d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

⇒ d² = (x₂ – x₁)² + (y₂ – y₁)² ⇒ d = ±\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

(b) The distance of a point P(x₁, y₁) form the origin

= \(\sqrt{(x_2-0)^2+(y_2-0)^2}\) = \(\sqrt{x^2_1+y^2_1}\)

Point lies on \(\overline{OY}\) be (x = 0, y ≥ 0)

example : (0, 2) (0, 3).

Point lies on \(\overline{OX}\) be (x = 0, y ≤ 0)

example : (2, 0) (3, 0).

In the above figure, length of the line segment above the X-axis is 4 – 2 = 2 

And length of line segment below X-axis is 4 – 1 = 3

Let the three points be A(-2, 5), B(3, - 4) and C(7, 10).

Then AB² = (3 + 2)² + (-4 - 5)² = 106

BC² = (7 - 3)² + (10 + 4)² = 212

AC² = (7 + 2)² + (10 - 5)² = 106 

We see that BC² = AB²1 + AC²

212 = 106 + 106 

212 = 212

∠A = 90⁰ 

Thus, ABC is a right triangle, right angled at A.

Correct option is (B) (6, 8)

\(\because\) Points is at a distance of 6 units from Y–axis.

\(\therefore\) x-coordinate of that point is 6.

Among all given points only (6, 8) has x-coordinate 6.

Correct option is B) (6, 8)

Given,

Points A(1/2, 3/2) and B(2, -5)

Let the point P(3/4, 5/12) divide the line segment AB in the ratio k: 1

Then, we know that

P(3/4, 5/12) = (2k + 1/2)/ (k +1) , (2k + 3/2)/ (k + 1)

Now, equating the abscissa we get

3/4 = (2k + 1/2)/ (k +1)

3(k + 1) = 4(2k + 1/2)

3k + 3 = 8k + 2

5k = 1

k = 1/5

Therefore, the ratio in which the point P(3/4, 5/12) divides is 1: 5

Correct option is (C) 1

The y–coordinate of point (-4, -3) is -3 and x–coordinate of point (-4, -3) is -4.

\(\because\) -3 - (-4) = -3 + 4 = 1

\(\therefore\) y–coordinate of point A (-4, -3) exceeds its x-coordinate by 1.

Correct option is  C) 1

Correct option is (D) Q₃

x < 0, y < 0 then (x, y) will lie in \(3^{rd}\) quadrant or \(Q_3.\)

Correct option is  D) Q₃

Point E lies in III quadrant. 

(b) (3,0)

Let a point on the x-axis be (a, 0) . Then, 

Dist. bet. (a, 0) and (7, 6)  = Dist. bet. (a, 0) and (–3, 4) 

(7 – a)² + (6 –0)² = (–3 – a)² + (4 – 0)² 

Now solve.

(b) (2,0)

The co-ordinates of point R which satisfy the equation PQ = QR = PR is the reqd. point.

Since the point is on y-axis, therefore coordinate of x-axis is zero.

Therefore the point is P(0, k) which is equidistance from A(2, 3) and B(- 4, 1) 

PA = PB

\(\sqrt{(2 - 0)^2 + (3 - k)^2}\) = \(\sqrt{(- 4)^2 + (1 - k)^2}\)

\(\sqrt{4 + 9 + k^2 - 6k}\) = \(\sqrt{16 + 1 + k^2 - 2k}\)

On squaring both sides, we get 

- 4k = 4

k = - 1

Therefore coordinate is (0, -1)

Consider A(3, 4), B (3, 8) and C(9, 8). 

Let the coordinates of fourth vertex are D (x, y) 

In a parallelogram diagonals bisect each other 

Coordinate of mid point of AC = X = \(\frac{3 + 9}2\) = \(\frac{12}2\) = 6

Y = \(\frac{4 + 8}2\) = \(\frac{12}2\) = 6

Therefore coordinates of mid point of AC are (6, 6) Coordinate of mid point of BD = X = \(\frac{3 + x}2\)

Y = \(\frac{y + 8}2\)

Coordinates of point D are

\(\frac{3 + x}2\) = 6

x = 12 - 3 = 9

\(\frac{y + 8}2\) = 6

y = 12 - 8 = 4

Therefore coordinates of fourth vertex D are (9, 4)

Let the point is P(0, 2) which is equidistance from A(3, k) and B(k, 5) 

PA = PB

\(\sqrt{(3 - 0)^2 + (k - 2)^2}\) = \(\sqrt{(k - 0)^2 + (5 - 2)^2}\)

On squaring both sides, we get

9 + k² + 4 - 4k = k² + 9

- 4k = - 4

k = 1

Vertices of triangle are A(-2, -3), B(- 1, 0), C(7, - 6) 

Let the coordinates of P are (x, y) 

PA = PB = PC 

PA = PB

\(\sqrt{(x + 2)^2 + (y + 3)^2}\) = \(\sqrt{(x + 1)^2 + (y + 0)^2}\)

On squaring both sides, we get

x² + 4x + 4 +y² +6y + 9

= x² + 2x + 1 + y²

2x + 6y = - 12

x + 3y = - 6.......(1)

PA = PC

\(\sqrt{(x + 2)^2 + (y + 3)^2}\) = \(\sqrt{(x - 7)^2 + (y + 6)^2}\)

On squaring both sides, we get

x² + 4x + 4 + y² + 6y + 9

= x² - 14x + 49 + y² + 12y + 36

18x - 6y = 72

3x - y = 12.......(2)

On solving equations (1) & (2), We get 

x = 3 and y = -3 

Therefore coordinates are (3, -3)

Coordinates of points on a circle are A(2,1), B(5,-8) and C(2,-9). 

Let the coordinates of the centre of the circle be O(x, y) 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Since the distance of the points A, B and C will be equal from the center, therefore 

⇒ OA = OB

\(\sqrt{(x - 2)^2 + (y - 1)^2}\) = \(\sqrt{(x - 5)^2 + (y + 8)^2}\)

On squaring both sides, we get 

⇒ x² + 4 - 4x + y² + 1 - 2y = x² + 25 - 10x + y² + 64 + 16y 

⇒ 6x - 18y - 84 = 0 

⇒ x - 3y - 14 = 0 ------------- (1) 

Similarly, OC = OB

\(\sqrt{(x - 2)^2 + (y + 9)^2}\) = \(\sqrt{(x - 5)^2 + (y + 8)^2}\)

⇒ x² + 4 - 4x + y² + 81 + 18y 

= x² + 25 - 10x + y² + 64 + 16y 

⇒ 6x - 2y - 4 = 0 

⇒ 3x - y - 2 = 0 ------------- (2) 

By solving equations (1) and (2), we get x = -1, y = -5 

So, 

the coordinates of the centre of the circle is (-1, -5). 

Radius of the circle = OA = \(\sqrt{(-1 - 2)^2 + (-5 - 1)^2}\)

= \(\sqrt{9+36}\)

= \(\sqrt{45}\)

= \(3\sqrt5\) units

Since the abscissa of first coordinate is zero, therefore this point lies on y - axis. Ordinate of second point is zero, therefore this point lies on y-axis. We know that both the axes are perpendicular to each other, therefore the angle between these points is 90°.

Let the point P(0, 100) and Q(10, 0) be the given points.

Therefore, The angle subtended by the line segment PQ at the origin O is 90°.

Putting \(x\) = 0 in equation of one of the lines say 9\(x\) + 40y –20 = 0, we get y = \(\frac{1}{2}\)

∴ A point on 9\(x\) + 40y – 20 = 0 is \(\big(0,\frac{1}{2}\big)\)

∴ Distance of \(\big(0,\frac{1}{2}\big)\) from 9\(x\) + 40y + 21 = 0 is \(\frac{\big|9\times0+40\times\frac{1}{2}+21\big|}{\sqrt{9^2+40^2}}\) = \(\frac{|41|}{\sqrt{1681}}\) = \(\frac{41}{41}\) = 1.

∵ Length of perpendicular from point (x₁, y₁) to line a\(x\) + by + c = 0 = \(\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\)

∴ Length of perpendicular from (0, 0) to \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 ⇒ \(\frac{\big|\frac{1}{a}\times0+\frac{1}{b}\times0-1\big|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\) = p

⇒ \(\frac{|-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\) = p

⇒ \(\frac{1}{p}\) = \(\sqrt{\frac{1}{b^2}+\frac{1}{b^2}}\) ⇒ \(\frac{1}{p^2}\) = \(\frac{1}{a^2}\) + \(\frac{1}{b^2}\)

(c) a, b, c are in H.P. only for all m 

As the points A(a, ma), B[b, (m + 1)b] and C[c, (m + 2)c] are collinear. 

Area of Δ ABC should be equal to zero.

⇒ \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0

⇒ \(\frac{1}{2}\)[a {(m + 1) b – (m + 2) c} + b {(m + 2) c – ma} + c {ma – (m + 1)b}] = 0 

⇒ mab + ab – mac – 2ac + mbc + 2bc – mab + mac – mbc – bc = 0 

⇒ ab – 2ac + 2bc – bc = 0 ⇒ ab + bc = 2ac 

⇒ b = \(\frac{2ac}{a+c}\)

∴ a, b, c are harmonic progression (H.P.) for all m.

Given line: x – 2y = 3 ⇒ y = \(\frac{x}{2}\) - \(\frac{3}{2}\)            ....(i)

∴ Its slope = m₁ = \(\frac{1}{2}\)

Let m₂ be the slope of line through (3, 2). Since this line is inclined at 45º to line (i),

tan 45º = \(\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\) = \(\bigg|\frac{\frac{1}{2}-m_2}{1+\frac{1}{2}m_2}\bigg|\)

⇒\(\frac{\frac{1}{2}-m_2}{1+\frac{1}{2}m_2}\) = ± 1 ⇒ \(\frac{1-2m_2}{2+m_2}=±\,1\) 

⇒ 1 – 2m₂ = 2 + m₂ or 1 – 2m₂ = –2 – m₂

⇒ 3m₂ = –1 or –m₂ = –3 

⇒ m₂ = \(-\frac{1}{3}\) or m₂ = 3

Since the line passes through (3, 2), the equation of the required line is

(y – 2) = \(-\frac{1}{3}\)(x – 3) or (y – 2) = 3 (x –3)

⇒ 3y – 6 = – x + 3 or y – 2 = 3x – 9 

⇒ 3y + x – 9 = 0 or y – 3x + 7 = 0.

(d) \(\lambda\) = \(\mu\) 

Let the equation of the line in the intercept from be

\(\frac{x}{\lambda}\)+ \(\frac{y}{\mu}\) = 1

Since it passes through (4, 3) and (2, 5)

\(\frac{4}{\lambda}\) + \(\frac{3}{\mu}\) = 1                    ....(i)

and  \(\frac{2}{\lambda}\) + \(\frac{5}{\mu}\) = 1            ....(ii)

Multiplying (ii) by 2 and subtracting from (i), we get

\(\bigg(\)\(\frac{4}{\lambda}\) + \(\frac{3}{\mu}\)\(\bigg)\) - \(\bigg(\)\(\frac{4}{\lambda}\) + \(\frac{10}{\mu}\)\(\bigg)\)= 1 - 2

⇒ \(\frac{-7}{\mu}\) = -1⇒ \(\mu\) = 7

Putting \(\mu\) = 7 in (i), we get \(\frac{4}{\lambda}\) + \(\frac{3}{7}\) = 1

⇒ \(\frac{4}{\lambda}\) = 1 - \(\frac{3}{7}\) = \(\frac{4}{7}\) = \(\lambda\) = 7

∴ \(\lambda\) = \(\mu\) = 7.

Let the x-intercept = a. Then y-intercept = –1 – a 

The equation of the required line is \(\frac{x}{a}\) + \(\frac{y}{-1-a}\) = 1

Given, it passes through (4, 3), so,

\(\frac{4}{a}\) + \(\frac{3}{-1-a}\) = 1

⇒ – 4 – 4a + 3a = – a – a² 

⇒ a² – 4 = 0 

⇒ (a + 2) (a – 2) = 0 

⇒ a = –2, or 2.

When a = 2, required line is \(\frac{x}{2}\) - \(\frac{y}{3}\) = 1

When a = –2, required line is \(\frac{x}{-2}\) + \(\frac{y}{3}\) = 1

Given : 

R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a). 

To prove: 

x + y = a + b 

Proof:

It is said that the point R(x, y) lies on the line segment joining the points P(a, b) and Q(b, a). Thus, these three points are collinear. 

So 

the area enclosed by them should be 0. 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) is:

Area (△) = \(\frac{1}2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\)

Given that area of ∆PQR = 0 

∴ \(\frac{1}2\) |x(b – a) + a(a – y) + b(y – b)| = 0 

∴ bx – ax + a² - ay + by - b² = 0 

∴ ax + ay –bx – by - a² - b² = 0 

∴ ax + ay –bx – by = a² + b² 

(a – b)(x + y) = (a – b )(a + b) 

∴ x +y = a + b 

Hence proved.