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Let the coordinates of point A be (x, y)

If AB is the diameter, then the center in the mid-point of the diameter

So,

(2, -3) = (x + 1/ 2, y + 4/ 2)

2 = x + 1/2 and -3 = y + 4/ 2

4 = x + 1 and -6 = y + 4

x = 3 and y = -10

Therefore, the coordinates of A are (3, -10)

Given,

Diameter of the circle = 10√2 units

So, the radius = 5√2 units

Let the center of a circle be 0(2a, a-7) and the circle passes though the point P (11, -9).

Then, OP is the radius of the circle

OP = 5√2

OP² = (5√2) = 50

(11- 2a)² + (-9 – a + 7)² = 50

121 – 44a + 4a² + 4 + a² + 4a = 50

5a² – 40a + 75 = 0

a² – 8a + 15 = 0

(a – 5)(a – 3) = 0 [Factorisation method]

So, a = 5 or a = 3

Suppose the point C(-1, -1) divides the line joining the points A(4, 4) and B(7, 7) in the ratio k : 1 Then, the coordinates of C are (7k+4/k+1,7k+4/k+1) But, we are given that the coordinates of the points C are (-1, -1).

So, 7k+4/k+1=-1=>k=-5/8

Thus, C divides AB externally in the ratio 5 : 8.

Let the required ratio be k : 1. Then the coordinates of the point of division are (5λ+2/k+1,6λ-3/k+1). 

But, it is a point on X-axis on which y-coordinate of every point is zero.

So, 6λ-3/k+1=0

=> k=1/2

Thus, the required ratio is 1/2: 1 or 1 : 2.

1. For the coordinate (-6, 3)

-6 which is the x-coordinate is negative and 3 which is the y-coordinate is positive.

Therefore, (-6, 3) lies in the II quadrant.

2. For the coordinate (-5, -3)

-5 which is the x-coordinate is negative and -3 which is the y-coordinate is negative.

Therefore, (-5, -3) lies in the III quadrant.

3. For the coordinate (11, 6)

11 which is the x-coordinate is positive and 6 which is the y-coordinate is positive.

Therefore, (11, 6) lies in I quadrant.

4. For the coordinate (1, -4)

1 which is the x-coordinate is positive and -4 which is the y-coordinate is negative.

Therefore, (1, -4) lies in the IV quadrant.

5. For the coordinate (-7, -4)

-7 which is the x-coordinate is negative and -4 which is the y-coordinate is negative.

Therefore, (-7, -4) lies in the III quadrant.

6. For the coordinate (4, -1)

4 which is the x-coordinate is positive and -1 which is the y-coordinate is negative.

Therefore, (4, -1) lies in the IV quadrant.

7. For the coordinate (-3, 8)

-3 which is the x-coordinate is negative and 8 which is the y-coordinate is positive.

Therefore, (-3, 8) lies in the II quadrant.

8. For the coordinate (3, -8)

3 which is the x-coordinate is positive and -8 which is the y-coordinate is negative.

Therefore, (3, -8) lies in the IV quadrant.

1. Since the point (2, 0) is of the form (x, 0) it lies on the x-axis.

2. Since the point (0, -5) is of the form (0, y) it lies on the y-axis.

3. Since the point (-4, 0) is of the form (x, 0) it lies on the x-axis.

4. Since the point (0, -1) is of the form (0, y) it lies on the y-axis.

The points which lie on the x-axis are of the form (x, 0)

Therefore, the points which lie on the x-axis are B (4, 0), D (-6, 0) and G (-1, 0).

(i) Abscissa of (0,5) is 0
(ii) Abscissa of (3,7) is 3
(iii) Abscissa of (-2, 4) is -2
(iv) Abscissa of (6, -3) is 6

(i) Ordinate of (4,0) is 0
(ii) Ordinate of (5,2) is 2
(iii) Ordinate of (1,-4) is -4
(iv) Ordinate of (-10,-7) is -7

(a) IV

Explanation:

We know that abscissa is positive and coordinate is negative in IV quadrant.

(c) III

Explanation:

We know that abscissa and coordinates are negative in III quadrant.

(a) I

Explanation:

We know that abscissa and coordinates are positive in I quadrant.

Correct option is (C) Y-axis

Point whose x-coordinate is zero lies on Y-axis.

\(\therefore\) Point (0, 5) lies on Y-axis.

Correct option is C) Y-axis

Correct option is (A) 0

Vertices of triangle are (3, 2), (-6, y) and (3,-2).

\(\therefore\) Centroid of triangle \(=\left(\frac{3+(-6)+3}3,\frac{2+y+(-2)}3\right)\)

\(=(0,\frac y3)\)

But given that centroid of triangle is origin (0, 0).

\(\therefore\) \((0,\frac y3)=(0,0)\)

\(\Rightarrow\frac y3=0\)       (By comparing y-coordinate)

\(\Rightarrow y=0\)

Correct option is A) 0

(a) 45º 

3x + y – 7 = 0 ⇒ y = –3x + 7 ⇒ Slope (m₁) = –3 

x + 2y + 9 = 0 ⇒ y = \(\frac{-x}{2}\) - \(\frac{9}{2}\) ⇒ Slope (m₂) = \(-\frac{1}{2}\)

If θ is the angle between the given lines, then 

tan θ = \(\big|\frac{m_1-m_2}{1+m_1m_2}\big|\) = \(\bigg|\frac{-3-\big(\frac{1}{2}\big)}{1+(-3)\big(-\frac{1}{2}\big)}\bigg|\) = \(\bigg|\frac{-\frac{5}{2}}{1+\frac{3}{2}}\bigg|\)

= \(\bigg|\frac{-\frac{5}{2}}{\frac{5}{2}}\bigg|\) = 1

∴ θ = 45°.

Correct option is (B) (0, 2)

A point can lie on X-axis if its y-coordinate is zero.

Since, y-coordinate of point (0, 2) is \(2\neq0.\)

\(\therefore\) (0, 2) does not lie on X-axis.

Correct option is B) (0, 2)

(c) 3\(x\) + y = 0 

The equations of the two lines whose point of intersection is needed are: 

2\(x\) – y = –5                  ...(i) 

5\(x\) + 3y = 4                 ...(ii) 

3 x Eqn (i) + Eqn (ii) ⇒ (6\(x\) – 3y) + (5\(x\) + 3y) = –15 + 4 

⇒ 11\(x\) = –11 ⇒ \(x\) = –1 

Putting \(x\) = –1 in (i), we get –2 – y = –5 ⇒ y = 3. 

∴ Point of intersection is (–1, 3). 

Slope of line \(x\) – 3y + 21 = 0, i.e., y = \(\frac{x}{3}\) + 7 is \(\frac{1}{3}.\)

⇒ Slope of line perpendicular to line \(x\) – 3y + 21 = 0 is –3 

[∵ m₁ × m₂ = –1] 

∴ Equation of a line through (–1, 3) with slope –3 is 

y – 3 = – 3 (\(x\) + 1) ⇒ y – 3 = –3\(x\) – 3 ⇒ 3\(x\) + y = 0.

i) Water tank 

ii) Mr. J’s house 

iii) In street No. 2, 3rd house on right side. 

iv) In street No. 4, the first house on right side. 

v) In street No. 4, the last house on left side.

Correct option is (D) (-5, 0)

y-coordinate of any point on X-axis is zero.

Put y = 0 in line y = x+5, we obtain

x+5 = 0

\(\Rightarrow\) x = -5

Hence, (-5, 0) is the point of intersection of X-axis and the line y = x+5.

Correct option is D) (-5, 0)

Let A(7, 6) and B(-3, 4) be the given points.

Let P(x, 0) be the point on the x-axis such that PA = PB

So, PA² = PB²

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

x² + 49 – 14x + 36 = x² + 9 + 6x + 16

-20x = -60

x = 3

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

(c) x = 2 

Slope of y-axis = tan 90º                    (∵ y-axis ⊥ x-axis) 

∴ Equation of line passing through (2, –3) parallel to y-axis is (y + 3) = tan 90º (x – 2) 

⇒ (y + 3) = ∞ (x – 2) ⇒ (x – 2) = \(\frac{1}{\infty}\) (y + 3) = 0 ⇒ x = 2.

Correct option is (A) a

Distance between points \(A\,(a\,cos\theta,0)\;\&\;B\,(0,a\,sin\theta)\) is

\(AB=\sqrt{(0-a\,cos\theta)^2+(a\,sin\theta-0)^2}\)

\(=\sqrt{a^2\,cos^2\theta+a^2\,sin^2\theta}\)

\(=\sqrt{a^2\,(cos^2\theta+sin^2\theta)}\)

\(=\sqrt{a^2}\)      \((\because cos^2\theta+sin^2\theta=1)\)

= a

Hence, distance between points \(A\,(a\,cos\theta,0)\;\&\;B\,(0,a\,sin\theta)\) is a.

Correct option is A) a

(c) negative

As the line from the origin is perpendicular to the line 

7x – 3y = 4, so its slope = \(\frac{-1}{\text{slope of }\,7x-3y=4}\)

Slope of 7x – 3y – 4 = \(\frac{7}{3}\)

∴ Slope of line from origin = \(\frac{-1}{+\frac{7}{3}}\) = \(\frac{-3}{7}\)

∴ tan θ = \(\frac{-3}{7}\), where θ is the angle the line makes with the +ve direction of x-axis

⇒ θ = tan^-1 \(\big(\frac{-3}{7}\big)\) = - tan^-1 \(\big(\frac{3}{7}\big)\)

Let A (2, 3) and B (-4, 1) be the given points

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

Now, we have

AC = √[(0 – 2)² + (y – 3)²] 

= √[y² – 6y + 9 + 4] 

= √[y² – 6y + 13]

And,

BC = √[(0 – (-4))² + (y – 1)²] 

= √[y² – 2y + 1 + 16] 

= √[y² – 2y + 17]

As AC = BC (given condition)

So, AC² = BC²

y² – 6y + 13 = y² – 2y + 17

-4y = 4

y = -1

Therefore, the point on the y-axis is (0, -1).

Let P(x, y) be the equidistant point from points A (-5, 4) and B (-1, 6).

So, the mid-point can be the required point

(x, y) = ( (-5 – 1/ 2), (4 + 6)/2 )

(x , y) = (-6/2 , 10/2) = (-3, 5)

Thus, the required point is (-3, 5)

Now,

We also know that, AP = BP

So, AP² = BP²

(x + 5)² + (y – 4)² = (x + 1)² + (y – 6)²

x² + 25 + 10 + y² – 8y + 16 = x² + 2x + 1 + y² – 12y + 36

10x + 41 – 8y = 2x + 37 – 12y

8x + 4y + 4 = 0

2x + y + 1 = 0

Therefore, all the points which lie on the line 2x + y + 1 = 0 are equidistant from A and B.

In the above problem, all points are of the form (K, y) where K is a fixed number and y is a variable. 

∴ All lines in the above problem are parallel to Y – axis. 

Slope of lines parallel to y – axis are not defined.

We observe that the area formed by above all triangles is zero.

(a) (5, 2)

Let A(8, 6) , B(8, –2) and C(2, –2) be the vertices of the given triangle and P(x, y) be the circum-centre of this triangle. 

Then, PA² = PB² = PC² 

Now, PA² = PB² ⇒ (x – 8)² + (y – 6)² = (x – 8) + (y + 2)² 

⇒ x² – 16x + 64 + y² – 12y + 36 = x² – 16x + 64 + y² + 4y + 4 

⇒ 16y = 32 ⇒ y = 2. 

Now, PB² = PC² ⇒ (x – 8)² + (y + 2)² = (x – 2)² + (y + 2)² 

⇒ x² – 16x + 64 + y² + 4y + 4 = x² – 4x + 4 + y² + 4y + 4 

⇒ 12x = 60 ⇒ x = 5. 

∴ Co-ordinates of the circumcentre are (5, 2).

(b) Rhombus

AB = \(\sqrt{(3-1)^2+(5-1)^2}\) = \(\sqrt{4+16}\) = \(\sqrt{20}\) = \(2\sqrt5\)

BC = \(\sqrt{(1-5)^2+(1-3)^2}\) = \(\sqrt{16+4}\) = \(\sqrt{20}\) = \(2\sqrt5\)

CD = \(\sqrt{(5-7)^2+(3-7)^2}\) = \(\sqrt{4+16}\) = \(\sqrt{20}\) = \(2\sqrt5\)

AD = \(\sqrt{(3-7)^2+(5-7)^2}\) = \(\sqrt{16+4}\) = \(\sqrt{20}\) = \(2\sqrt5\)

AC = \(\sqrt{(3-5)^2+(5-3)^2}\) = \(\sqrt{4+4}\) = \(\sqrt{8}\) = \(2\sqrt2\)

BD = \(\sqrt{(1-7)^2+(1-7)^2}\) = \(\sqrt{36+36}\) = \(\sqrt{72}\) = \(6\sqrt2\)

Now, AB = BC = CD = AD ⇒ All sides are equal 

Also, AC ≠ BD ⇒ Diagonals are not equal. 

⇒ ABCD is a rhombus.

Required distance = \(\sqrt{(0-a\,cos 60°)^2+(a\, sin\,60°-0)^2}\)

= \(\sqrt{\big(\frac{-a}{2}\big)^2+\big(\frac{\sqrt3a}{2}\big)^2}\)

= \(\sqrt{\frac{a^2}{4}+\frac{3a^2}{4}}=\sqrt{\frac{4a^2}{4}}\) = a units.

Given, \(\sqrt{(x-2)^2+(-3-(-11))^2} = 10\)

⇒ \(\sqrt{(x-2)^2+64} = 10\)

⇒ \(\sqrt{x^2-4x+4+64}=10\)

⇒ x² – 4x + 68 = 100 

⇒ x² – 4x – 32 = 0 

⇒ (x – 8) (x + 4) = 0 

⇒ x = 8 or – 4

(c) 1, 3

The diagonals of a parallelogram bisect each other, therefore co-ordinates of mid-points of both the diagonals are the same co-ordinates of mid-point of diagonal AC

= \(\bigg(\frac{4-2}{2},\frac{b-1}{2}\bigg) \) = \(\bigg(1,\frac{b-1}{2}\bigg) \)

Co-ordinate of mid-point of diagonal BD

= \(\bigg(\frac{1+a}{2},\frac{2-0}{2}\bigg) \) = \(\bigg(\frac{1+a}{2},1\bigg) \)

⇒ \(\frac{1+a}{2}=1\) and \(\frac{b-1}{2}=1\)

⇒ 1 + a = 2 and b – 1 = 2 

⇒ a = +1 and b = 3

Let P(1, –1), Q(5, 2) and R(9, 5) be the given points. Then

PQ = \(\sqrt{(5-1)^2+(2+1)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5

QR = \(\sqrt{(9-5)^2+(5-2)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5

PR = \(\sqrt{(9-1)^2+(5+1)^2}\) = \(\sqrt{64+36}\) = \(\sqrt{100}\) = 10

⇒ PQ + QR = PR 

⇒ P, Q, R are collinear points.

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

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

For points to be collinear, the Area enclosed by them should be equal to 0 

∴ For given points,

Area = \(\frac{1}2[a(b^2 - c^2) + b(c^2- a^2) + c(a^2-b^2)]\)

Area = \(\frac{1}2\) |(b – c)(a - b)(c – a)| 

Area ≠ 0 

Also it is given that a ≠ b ≠ c,

Hence area of triangle made by these points is never zero. Hence given points are never collinear.

Let three given points be A(a,0), B(0,b) and C(1,1). 

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

= \(\frac{1}2\) |x₁(y₂ - y₃)+x₂(y₃ - y₁)+x₃(y₁ - y₂)| 

Area of ∆ABC 

= \(\frac{1}2\) |a(b – 1) + 1(0 -b)| 

= \(\frac{1}2\) | ab – a –b| 

Here given that \(\frac{1}a+\frac{1}b\) = 1

∴ \(\frac{a+b}{ab}\) = 1

∴ a + b = ab 

Now, 

Area of ∆ABC 

= \(\frac{1}2\) | ab - (a + b)| 

= \(\frac{1}2\) | ab – ab| 

= \(\frac{1}2\) | 0 | 

= 0 sq. units 

We know that if area enclosed by three points is zero, then points are collinear. 

Hence, 

given three points are collinear.

Four points A (6, 3), B (−3, 5) C (4, −2) and D(x, 3x) 

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

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| 

Area of ∆ABC 

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

= \(\frac{1}2\) |42 + 15 -8| 

= \(\frac{49}2\) sq. units 

Area of ∆DBC 

= \(\frac{1}2\) |x(5 – (-2)) + 3(-2 -3x) + 4(3x - 5))| 

= \(\frac{1}2\) |7x +6 + 9x + 12x - 20| 

= \(\frac{1}2\) | 28x -14| 

= ± 7(2x -1 ) 

It is given that   \(\frac{△DBC}{△ABC}\) = \(\frac{1}2\)

∴2 × ∆DBC = ∆ABC 

2 × (± 7(2x -1 )) = \(\frac{49}2\)

∴ ± 4(2x -1 ) = 7 

∴ 4(2x -1 ) = 7 or -4(2x -1 ) = 7 

∴ 8x – 4 =7 or -8x + 4 =7 

∴ 8x =11 or -8x =3 

∴x = \(\frac{11}8\) or x = \(\frac{-3}8\)

Hence, the value of x is \(\frac{11}8\) or \(\frac{-3}8\)

Given: 

the distance between the points P (2, -3) and Q (10, y) is 10 units. 

To find: 

The value of y. 

Solution:

Coordinates are P (2, - 3) and Q (10, y) 

We use distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to find the distance between two points.

 Since PQ = 10 units 

S0,

\(\sqrt{(10 - 2)^2 + (y + 3)^2}\) = 10

On squaring both sides, we get

(10 - 2)² + (y + 3)² = 100

⇒ 82 + (y + 3)² = 100 

⇒ 64 + y² + 6y + 9 = 100 

⇒ 73 + y² + 6y = 100 

⇒ 73 + y² + 6y -100 = 0 

⇒ y² + 6y - 27 = 0

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized. 

⇒ y² + 9y - 3y - 27 = 0 

⇒y(y + 9) - 3(y + 9) = 0 

⇒(y - 3)(y + 9) = 0 

⇒y = 3, - 9

(c) 5 or 4

Radius of the circle = Dist. between the centre and given pt. on the circle

= \(\sqrt{(x-2-4)^2+(x+1-4)^2}\)

= \(\sqrt{(x-6)^2+(x-3)^2}\)

= \(\sqrt{x^2-12x+36+x^2-6x+9}\)

= \(\sqrt{2x^2-18x+45}\)

Given, diameter = 2√5

⇒ radius = \(\frac12\times2\sqrt5 = \sqrt5\)

∴ \(\sqrt{2x^2-18x+45}\) = √5

⇒ 2x² – 18x + 45 = 5 

⇒ 2x² – 18x + 40 = 0 

⇒ x² – 9x + 20 = 0 

⇒ (x – 5) (x – 4) = 0 

⇒ x = 5 or 4

(i) The coordinates of B : (-5, +2) 

(ii) The coordinates of C. : (+5, -5) 

(iii) The point identified by the coordinates (-3, -5) : E. 

(iv) The point identified by the coordinates (2, -4) : G. 

(v) The abscissa of the point D : 6 

(vi) The ordinate of the point H: -3 

(vii) The coordinates of the point L : (0, +5) 

(viii) The coordinates of the point M : (-3, 0).

(i) In a Cartesian plane, to determine the position of any point, horizontal line is called x-axis. Vertical line is y

The y-coordinate is also called the ordinate. 

Horizontal line → xOx’ 

Vertical line → yOy’ 

(ii) The name of each part of the plane formed by these two lines is called Quadrant. 

I Quadrant → xOy 

II Quadrant → yOx’ 

III Quadrant → x’Oy’ 

IV Quadrant → y’Ox 

(iii) The name of the point where these two lines intersect is called origin. 

Coordinates of origin are (0. 0).

(d) 1

For collinearity of points A (–1, 3), B (2, p), C (5, –1) area of ΔABC should be zero, i.e.,

\(\frac12\) | - 1(p +1) + 2(-1 -3) + 5(3-p) | = 0 

⇒ – p – 1 – 8 + 15 – 5 p = 0 

⇒ – 6 p + 6 = 0 ⇒ –6 p = – 6 ⇒ p = 1.

Coordinates of points on a circle are A(6, -6), B(3, -7) and C(3, 3). 

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

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

On squaring both sides, we get

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

x² - 12x + 36 + y² + 12y + 36 

= x² - 6x + 9 + y² - 6y + 9

x - 3y = 9........(1)

Similarly, OA = OB

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

On squaring both sides, we get

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

x² - 12x + 36 + y² + 12y + 36 

= x² - 6x + 9 + y² + 14y + 49

3x + y + 7.......(1)

Solving eqn (1) and (2), we get 

x = 3; y = - 2 

Coordinates of circum center are (3, - 2)

The coordinates are A(-1, 2) and C(3, 2). 

Let the coordinates of the vertex B are (x, y) 

AB = BC 

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

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

On squaring both sides, we get

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

x² + 2x + 1 + y² - 4y + 4 = x² - 6x + y² - 4y + 4

x = 1

In ΔABC 

AB² + BC² = AC² [Using Pythagoras theorem] 

2AB² = AC² [Since AB = BC]

2[(x + 1)² + (y - 2)²] = (3 + 1)² + (2 - 2)²

2[x² + 2x + 1 + y² - 4y + 4] = 16

x² + 2x + 1 + y² - 4y + 4 = 8

x² + 2x + y² - 4y = 3

On substituting x = 1

1 + 2 x 1 + y² - 4y = 3

y² - 4y = 0

y(y - 4) = 0

y = 0, 4

Other coordinates are (1, 0) and (1, 4)

(d) 29 sq. units

Length of a side of the square 

= \(\sqrt{(-1-4)^2+(-5+3)^2}\)

= \(\sqrt{25+4} = \sqrt{29}\) units.

∴ Area = (Side)² = \((\sqrt{29})^2\) sq. units

= 29 sq. units

(x, y) is equidistant from the points (2, 1) and (1, –2) 

⇒ Distance between (x, y) and (2, 1) = Distance between (x, y) and (1, –2)

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

⇒ x² – 4x + 4 + y² – 2y + 1 = x² – 2x + 1 + y² + 4y + 4

⇒ – 4x + 2x – 2y – 4y = 0 ⇒ –2x – 6y = 0 ⇒ x + 3y = 0

Correct option is (B) x – y = 1

Given that slope of required line is m = 1 & it passes through point (3, 2).

Let equation of the required line is

\(y=mx+c\)

\(\Rightarrow\) \(y=x+c\)            \((\because m=1)\)

Put x = 3 & y = 2     \((\because\) Required line passes through (3, 2))

\(\therefore2=3+c\)

\(\Rightarrow c=2-3=-1\)

Put c = -1 in \(y=x+c,\) we get

\(y=x-1\)

\(\Rightarrow x-y=1\)

Hence, the equation of required line is \(x-y=1.\)

Correct option is B) x – y = 1

Let point P(x, 0) is on x-axis and equidistant from A(2, -5) and B (-2, 9)

PA = PB

or PA² = PB²

(2 – x)² + (– 5 – 0)² = (– 2 – x)² + (9 – 0)²

(2 – x)² + (– 5)² = (– 2 – x)² + (9)²

29 + x² – 4x = 85 + x² + 4x

56 = – 8x

or x = – 7

The point on ×-axis is (– 7, 0)

Mid point of the line segment joining the points (3, 2) and (-5, 6)

= [(x₁ + x₂)/2, (y₁ + y₂)/2]

= [(3 - 5)/2, (2 + 6) /2]

= (-1, 4)

Mid point of the line segment joining (3, 2) and (-5, 6) is at one-half of the distance from (3, 2).

Midpoint of the segment joining the points (-1, 4) and ((-5, 6)

= [(-1 + 5)/2, (4 + 6) /2]

= (2, 5)

The point (2, 5) represents three-fourth of the distance from the point (3, 2) to the point (-5, 6).

The correct option is: (B) (7, 5)

Explanation:

Mohit's position will be (7, 5).