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Vertices of a parallelogram ABCD are: A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) Length of side AB 

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

Length of side AB = \(\sqrt{(3 - 1)^2 + (6 + 2)^2}\) = \(\sqrt(4 + 64)\) 

= \(\sqrt68\) units

Length of side BC = \(\sqrt{(5 - 3)^2 + (10 - 6)^2}\) = \(\sqrt(4 + 16)\) 

= \(\sqrt20\) units

Length of side CD = \(\sqrt{(3 - 5)^2 + (2 - 10)^2}\) = √(4 + 64) 

= √68 units

Length of side DA = \(\sqrt{(3 - 1)^2 + (2 + 2)^2}\) = √(4+16) 

= √20 units 

Length of diagonal BD = \(\sqrt{(3 - 3)^2 + (2 - 6)^2}\) 

= √16 = 4 units 

Length of diagonal AC = \(\sqrt{(5 - 1)^2 + (10 + 2)^2}\) = √(16+144)

= √160 units

Opposite sides of the quadrilateral formed by the given four points are equal i.e. (AB = CD) & (DA = BC)Also, the diagonals BD & AC are unequal.Therefore, the given points form a parallelogram.

Correct option is (D) 1

If a line makes an angle of \(\theta\) with positive X-axis then its slope is given by \(tan\,\theta.\)

\(\therefore\) Slope of line which makes angle of \(45^\circ\) with positive X-axis is \(m=tan\,45^\circ=1.\)

Correct option is D) 1

Correct answer is

(C) 1/√3

i. Angle made with the positive direction of

X-axis (θ) = 45°

Slope of the line (m) = tan θ

∴ m = tan 45° = 1

∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis 

(θ) = 60°

Slope of the line (m) = tan θ

∴ m = tan 60° = √3

∴ The slope of the line is √3.

iii. Angle made with the positive direction of

X-axis (θ) = 90°

Slope of the line (m) = tan θ

∴ m = tan 90°

But, the value of tan 90° is not defined.

∴ The slope of the line cannot be determined.

(B) 8

Distance formula: d² = (x₂ – x₁)² + (y₂ – y₁)²

According to the question,

We have,

x₁ = 0, x₂ = 0

y₁ = 6, y₂ = – 2

d² = (0 – 0)² + ( – 2 – 6)²

d= √((0)²+ (-8)²)

d = √64

d = 8 units

Therefore, the distance between A (0, 6) and B (0, 2) is 8

(i) We know that, the point whose abscissa is 0 will lie on Y-axis. So, the required points whose abscissa is 0 are A, L and O.

(ii) We know that, the point whose ordinate is 0 will lie on X-axis. So, the required points, whose ordinate is 0 are G,l and O.

(iii) Here, abscissa ‘-5’ is negative, which shows that point with abscissa -5 will lie in II and III quadrants. So, the required points whose abscissa is -5, are D and H.

Note: We know that, origin O is the intersection point of both axes. So, we can consider it on X-axis as well as on Y-axis.

A (3, 1) 

From Point A draw a perpendicular to x-axis we get 3 and perpendicular to y-axis we get 1. Therefore coordinates of point A is (3, 1).

B(6, 0)

Since Point B lies on x-axis six places away from origin. Therefore co-ordinates of point B is (6, 0).

C(0, 6)

Since Point C lies on y-axis six places away from origin. Therefore co-ordinates of point C is (0, 6).

D(-3, 0)

Since Point D lies on x-axis three places away from origin on left side. Therefore co-ordinates of point D is (-3,0). 

E(-4, 3)

From Point E draw a perpendicular to x-axis we get-4 and perpendicular to y-axis we get 3. Therefore coordinates of point E is (-4, 3).

F(-2, -4)

From Point F draw a perpendicular to x-axis we get-2 and perpendicular to y-axis we get -4. Therefore coordinates of point F is (-2, -4).

G(0, -5)

Since Point G lies on y-axis 5 places away from origin in the downward disrection since value of the coordinate is negative. Therefore co-ordinates of point G is (0, -5).

H(3, -6)

From Point H draw a perpendicular to x-axis we get 3 and perpendicular to y-axis we get -6. Therefore coordinates of point H is (3, -6).

P(7, -3)

From Point P draw a perpendicular to x-axis we get 7 and perpendicular to y-axis we get -3. Therefore coordinates of point P is (7, -3).

i) False Correct statement: In the Cartesian plane the horizontal line is called X – axis. 

ii) True 

iii) True 

iv) False Correct statement: The point (2, -3) lies in the fourth quadrant. 

v) False Correct statement: (- 5, – 8) lies in the third quadrant. 

vi) True

Correct option is (C) (a, 0)

Correct option is (B) 45°

Let \(\theta\) be the angle made by the line y = x with the positive direction of X-axis. Then slope of line is \(m=tan\,\theta.\)

Given line is y = x.

By comparing with slope-intercept form of line (i.e., y = mx), we get m = 1.

But slope of line is \(m=tan\,\theta.\)

\(\therefore\) \(tan\,\theta=1=tan\;45^\circ\)

\(\Rightarrow\) \(\theta=45^\circ\)

Hence, angle made by the line y = x with the positive direction of X-axis is \(45^\circ.\)

Correct option is B) 45°

Correct option is (C) 2

To locate the exact position of a point, we required to have 2 coordinates (or 2 parameters or references).

Correct option is  C) 2

Correct option is (C) 2

To locate the exact position of a point, we need x-coordinate and y-coordinate of that point.

Hence, we need two references.

Hence, to locate the exact position of a point the number of references we need is 2.

Correct option is C) 2

Let A ( 3k − 1, k − 2 ) , B ( k, k − 7 ) and C ( k − 1, −k − 2 ) be the given points.For points to be collinear area of triangle formed by the vertices must be zero. 

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 = 0 

⇒ ( 3k−1 ) [ ( k−7 ) − ( −k−2 ) ] + k [ ( −k−2 ) − ( k−2 ) ] + ( k−1 ) [ ( k−2 ) − ( k−7 ) ] =0

⇒ ( 3k−1 ) [ k−7 + k + 2 ] + k [ −k−2 − k+ 2 ] + ( k−1 ) [ k−2 − k + 7 ] =0 ⇒ ( 3k−1 ) ( 2k−5 ) + k (−2k ) + 5 ( k −1 ) =0

⇒ 6k² - 15k -2k + 5 - 2k² + 5k - 5 = 0 

⇒ 6k²−17k + 5−2k² + 5k−5 = 0 

⇒ 4k²−12k = 0 

⇒ 4k ( k−3 ) = 0 

⇒ k = 0 or k−3 = 0 

⇒ k = 0 or k = 3 

Hence, 

the value of k is 0 or 3.

The given points A(−1, −4), B(b, c) and C(5, −1) are collinear. 

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)| 

Given that area of ∆ABC = 0 

∴ −1[c − (− 1)]+b[− 1 − ( − 4)] + 5( − 4 − c) = 0 

∴ − c − 1 + 3b − 20 − 5c = 0 

3b − 6c = 21 

∴b − 2c = 7 …(1) 

Also it is given that 2b + c = 4 …(2) 

Solving 1 and 2 simultaneously, we get, 

2(7 + 2c) + c = 4 

14 + 4c + c = 4 

5c = − 10 

c = − 2 

∴ b = 3 

Hence, 

value of b and c are 3 and -2 respectively

(c) 4

The distance of a point (x. y) from x axis is |y|

Here, the point is (-3, 4) So, its distance from x-axis is |4| = 4

Coordinates of points are A(3, k), B(k, 5) and P(k-1, 2) 

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

⇒ PA = PB

\(\sqrt{(k - 1-3)^2 + (2 - k)^2}\) = \(\sqrt{(k - 1-k)^2 + (2 - 5)^2}\)

On squaring both sides, we get

(k - 1 - 3)² + (2 - k)² = (k - 1 - k)² + (2 - 5)²

(k - 4)² + (2 - k)² = (-1)² + (-3)²

k² - 8k + 16 +  k² - 4k + 4 = 1 + 9

k² - 6k + 5 = 0

k² - 5k - k + 5 = 0

k(k - 5) - 1(k - 5) = 0

(k - 5)(k - 1) = 0

k = 1, 5

Coordinates of points are A(0, 2), B(3, p) and C(p, 5) 

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

⇒ AB = AC

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

On squaring both sides, we get

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

9 + p² - 4p + = p² + 9

- 4p = - 4

p = 1

AB = \(\sqrt{(3 - 0)^2 + (1 - 2)^2}\) = \(\sqrt{9 + 1}\) = \(\sqrt{10}\) units

Coordinates of points are P(2, 2) A(- 2, k) and B(- 2k, -3) 

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

⇒ PA = PB

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

On squaring both sides, we get

(- 2 - 2)² + (k - 2)² = (- 2k - 2)² + (- 3 - 2)²

16 + k² - 4k + 4 = 4k² + 8k + 4 + 25

k² + 4k + 3 = 0

k² + 3k + k + 3 = 0

k(k + 3) + 1(k + 3) = 0

(k + 1)(k + 3) = 0

k = -1,- 3

ap = \(\sqrt{(-2 - 2)^2 + (-1 - 2)^2}\) = \(\sqrt{16 + 9}\) = 5 units

P on x- axis, since ordinate is zero. 

Q on y - axis, since abscissa is zero. 

R on x - axis, since ordinate is zero. 

 S on y - axis, since abscissa is zero.

(i) Since each side of square is 2a. 

Coordinates of A are (0, 0), since it coincides with origin.

Coordinates of B are (2a, 0), for a point along x-axis ordinate is zero. 

Coordinates of C are (2a, 2a), since this point is equi-distance from x-axis and y-axis. 

Coordinates of D are (0, 2a), since abscissa is zero and ordinate is 2a. 

(ii) Each side of square is a units. 

Coordinates of A are (a, a), since this point lies in Ist coordinate.

 Coordinates of B are (-a, a), since this point lies in IInd coordinate. 

Coordinates of C are (-a, -a), since this point lies in IIIrd coordinate. 

Coordinates of D are (a, -a), since this point lies in IVth coordinate.

(i) Coordinate of the vertices of the square of side 2a are:

A(0,0), B(2a,0), C(2a,2a) and D(0,2a)

(ii) Coordinate of the vertices of the square of side 2a are:

A(a,a), B(-a,a), C(-a,-a) and D(a,-a)

Correct option is (B) \((\frac{-c}m,0)\)

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

\(\therefore\) For intersection point of line y = mx+c with X-axis, put y = 0 in y = mx+c, we get

mx+c = 0

\(\Rightarrow x=\frac{-c}m\)

Hence, intersection point of line y = mx+c & X-axis is \((\frac{-c}m,0).\)

Hence, the line y = mx+c meets X-axis at the point \((\frac{-c}m,0).\)

Correct option is B) (-c/m, 0)

Correct option (a) (-3,5)

Explanation:

Centroid, circumcentre and orthocenter are collinear such that centroid divides the circumcenter and orthocentre in the ratio 1:2. 

Coordinate axes intersect each other at 90° or coordinate axes are perpendicular to eact other.

Since the ordinate of both the given points is 0, therefore both the points lie on x - axis.

As (x , y) is the mid-point

x = (3 + k)/ 2 and y = (4 + 7)/ 2 = 11/2

Also,

Given that the mid-point lies on the line 2x + 2y + 1 = 0

2[(3 + k)/ 2] + 2(11/2) + 1 = 0

3 + k + 11 + 1 = 0

Thus, k = -15

As (a, b) is the mid-point of the line segment A(10, -6) and B(k, 4)

So,

(a, b) = (10 + k / 2, -6 + 4/ 2)

a = (10 + k)/ 2 and b = -1

2a = 10 + k

k = 2a – 10

Given, a – 2b = 18

Using b = -1 in the above relation we get,

a – 2(-1) = 18

a = 18 – 2 = 16

So,

k = 2(16) – 10 = 32 – 10 = 22

Thus,

AB = √[(22 – 10)² + (4 + 6)²] 

= √[(12)² + (10)²] 

= √[144 + 100] 

= 2√61 units

Correct option is (D) √3/2

Slope of line joining points \((-\alpha, \alpha)\) and \((\alpha, \alpha+\alpha \sqrt3)\) is

\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{\alpha+\alpha\sqrt3-\alpha}{\alpha-(-\alpha)}\)

\(=\frac{\alpha\sqrt3}{2\alpha}=\frac{\sqrt3}{2}\)

Correct option is D) √3/2

Let A ≡ (x, y), B ≡ (2, 1), C ≡ (3, –2) 

Area of ΔABC = \(\frac{1}{2}\) |{x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)}|

= \(\frac{1}{2}\) | \(x\)(1 + 2) + 2(–2 – y) + 3(y – 1) | = \(\frac{1}{2}\) | 3\(x\) – 4 – 2y + 3y – 3 | = \(\frac{1}{2}\) | 3\(x\) + y – 7 |

Given \(\frac{1}{2}\) | 3\(x\) + y – 7 | = 5

⇒ | 3\(x\) + y – 7 | = 10 ⇒ 3\(x\) + y – 7 = 10   or  –(3\(x\) + y – 7) = 10 ⇒ 3\(x\) + y = 17   or  3\(x\) + y = –3

Case I. 3\(x\) + y = 17. Also given y = \(x\) + 3 

∴ 3x + \(x\) + 3 = 17 ⇒ 4\(x\) = 14 ⇒ \(x\) = \(\frac{7}{2}\) ⇒ y = \(\frac{7}{2}\) + 3 = \(\frac{13}{2}\)

Case II. 3\(x\) + y = –3, y = \(x\) + 3

∴ 3\(x\) + \(x\) + 3 = –3 ⇒ 4\(x\) = – 6 ⇒ \(x\) = \(\frac{-3}{2}\) ⇒ y = \(\frac{-3}{2}\) + 3 = \(\frac{3}{2}\)

∴ Co-ordinates of A are \(\bigg(\frac{7}{2},\frac{13}{2}\bigg)\) or \(\bigg(\frac{-3}{2},\frac{3}{2}\bigg)\)

For three points to be collinear, the area of the triangle formed by the three points should be zero. 

∴ Area of D formed by the given three points

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

= \(\frac{1}{2}\) [a(c – b) + b(a – c) + c(b – a)] = \(\frac{1}{2}\) [ac – ab + ba – bc + cb – ca] = 0.

Hence (a, b + c), (b, c + a) and (c, a + b) are collinear.

Let A(x₁, y₁) = (3, 4), B(x₂, y₂) ≡ (0, 5), C(x₃, y₃) ≡ (2, –1)and D(x₄, y₄) ≡ (3, –2) be the vertices of quadrilateral ABCD.

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

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

= \(\frac{1}{2}\) |{(15 – 0) + (0 – 10) + (– 4 + 3) + (12 + 6)}|

= \(\frac{1}{2}\) |{15 – 11 + 0 + 18}| = \(\frac{1}{2}\)x 22 = 11 sq. units.

Let A(–36, 7), B(20, 7) and C(0, –8) be the vertices of the given triangle.

Then, a = BC = \(\sqrt{(0-20)^2+(-8-7)^2}\) = \(\sqrt{400+225}\) = \(\sqrt{625}\) = 25

b = AC =\(\sqrt{(0-36)^2+(-8-7)^2}\) = \(\sqrt{1296+225}\) = \(\sqrt{1521}\) = 39

c = AB = \(\sqrt{(20+36)^2+(7-7)^2}\) = \(\sqrt{56^2}\) = 56.

The co-ordinates of the incentre of the ΔABC are

\(\bigg[\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\bigg]\) = \(\bigg(\frac{25(-36)+39(20)+56\times0}{25+39+56},\frac{25(7)+39(7)+56(-8)}{25+39+56}\bigg)\)

= \(\bigg(\frac{-900+780}{120},\frac{175+273-448}{120}\bigg)\) = \(\bigg(\frac{-120}{120},\frac{0}{120}\bigg)\) = (–1, 0)

(b) \(\frac{\sqrt{17}}{2}\)

Equation of the line through the points (5, 0) and (0, 3) 

y – 0 = \(\frac{3-0}{0-5}\) (x - 5)

⇒ y = \(\frac{-3}{5}\)(x - 5)

⇒ 5y + 3x – 15 = 0 

∴ Distance of perpendicular from point (4, 4) on the line

 5y + 3x – 15 = 0 is \(\bigg|\frac{5\times4+3\times4-15}{\sqrt{5^2+3^2}}\bigg|\)

= \(\frac{|20+12-15|}{\sqrt{25+9}{}}\) = \(\frac{17}{\sqrt{34}}\) units. = \(\frac{\sqrt{17}}{2}\) units.

(a) 60º

The two lines are: y = \((2-\sqrt3)\) \(x\) + 5            ...(i) 

y = \((2-\sqrt3)\) \(x\) – 7               ...(ii) 

Slope of line (i), m₁ = \(2-\sqrt3\)

Slope of line (ii), m₂ = \(2+\sqrt3\)

If θ is the angle between the two lines, then 

tan θ = \(\big|\frac{m_1-m_2}{1+m_1m_2}\big|\) = \(\bigg|\frac{(2-\sqrt3)-(2+\sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}\bigg|\)

= \(\bigg|\frac{-2\sqrt3}{1+1}\bigg|=\sqrt3\)

∴ θ = –1 tan ( 3) = 60º.

The coordinates of the points P, Q, R, S, T and O are as follows:

P = (1, 1)

Q = (-3, 0)

R = (-2, -3)

S = (2, 1)

T = (4, -2)

O = (0, 0)

(i) Point (3, 0) lies in the first quadrant.

False

Justification:

The ordinate of the point (3, 0) is zero.

Hence, the point lies on the x-axis

(ii) Points (1, –1) and (–1, 1) lie in the same quadrant.

False

Justification:

(1, -1) lies in IV quadrant

(-1, 1) lies in II quadrant.

(iii) The coordinates of a point whose ordinate is – ½ and abscissa is 1 are – ½ , 1.

False

Justification:

The coordinates of a point whose ordinate is – ½ and abscissa is 1 is (1, -1/2).

(iv) A point lies on y-axis at a distance of 2 units from the x-axis. Its coordinates are (2, 0).

False

Justification:

A point lies on y-axis at a distance of 2 units from the x-axis. Its coordinates are (0, 2).

(v) (–1, 7) is a point in the II quadrant.

True

Justification:

(–1, 7) is a point in the II quadrant.

(b) (-2,4)

Here, given point P lies in II quadrant, so its abscissa will be negative and ordinate wilt be positive. Also, its perpendicular distance from X-axis is 4, so y-coordinate of P is 4 and its perpendicular distance from Y-axis is 2, so x-coordinate is -2. Hence, coordinates of P are (-2, 4).

(i) B = (–5, 2), C(–2, –3), E = (3, –1) 

(ii) F

(iii) 1 

(iv) 0

(i) True, because a point on the y-axis is of the form (0, y). 

(ii) False, because the perpendicular distance of a point from the x-axis is its ordinate. Hence it is 3, not 4.

(c) L

In point (-5, 3), x-coordinate is negative and y-coordinate is positive, so it will lie in II quadrant. Now, we see that perpendicular distance of L from V-axis is 5 and from X-axis is 3. So, the required point is L.

(b) (0,4)

Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

(c) P, R and T

We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

(a) 3

We know that, abscissa or the x-coordinate of a point is its perpendicular distance from the Y-axis. So, perpendicular distance of the point P(3, 4)from Y-axis = Abscissa = 3

(c) (0,-5)

Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.

Hence, the required point is (0, – 5).

D. any number

Explanation:

Abscissa of all the points on the x-axis can be any number.

Hence, (D) is the correct option.

A. on the negative direction of the x-axis

Explanation:

Point (– 10, 0) lies on the negative direction of x-axis.

Hence, (A) is the correct option.

C. on the y-axis

Explanation:

Since the abscissa is 0, Point (0, –7) lies on y-axis.

Hence, (C) is the correct option.

B. Second Quadrant

Explanation:

(-3,5) is of form (-x,y).

In the point (-3, 5) abscissa is negative and ordinate is positive. So, it lies in the second quadrant.

Hence, (B) is the correct option.

Correct answer is (b) 4

The diagonals of a parallelogram bisect each other. The vertices of the 11gm ABCD are

A(1,3), B(-1,2) and C(2,5) and D(x,4)

Here, AC and BD are the diagonals. So

\(\frac{1\,+\,2}{2}=\frac{-1\,+\,x}{2}\)

\(\Rightarrow\) x - 1 = 3

\(\Rightarrow\) x = 1 + 3 = 4