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Correct option is (B) (0, 0)

\(\overline{OY}\) is positive y-axis and any point lies on y-axis if its x-coordinate is zero.

Among all given points only (0, 0) has x-coordinate zero.

\(\therefore\) (0, 0) lies on \(\overline{OY}\) in graph.

Correct option is  B) (0, 0)

Correct option is (D) -b/a

Given line is \(\frac{x}{a}+\frac{y}{b} = 1\)

\(\Rightarrow\) \(\frac{y}{b}=1-\frac{x}{a}\)

\(\Rightarrow y=-\frac bax+b\)

By comparing with \(y=mx+c,\) we get \(m=-\frac{b}{a}\)

\(\therefore\) Slope of the line \(\frac{x}{a}+\frac{y}{b} = 1\) is \(m=-\frac{b}{a}.\)

Correct option is D) - b/a

Correct option is (A) ordinate

The y-coordinate of a point is called ordinate of that point.

Correct option is A) ordinate

Correct option is (C) abscissa

The x-coordinate of a point is called abscissa of that point.

Correct option is C) abscissa

Correct option is (A) origin

The meeting point or the intersection point of the two axes in the coordinate plane is called origin.

Correct option is A) origin

Correct option is (C) (1, – 2)

If x > 0 & y < 0 then point (x, y) lies in \(4^{th}\) quadrant (in \(Q_4)\)

Since, 1 > 0 & -2 < 0

\(\Rightarrow\) (1, –2) lies in \(Q_4.\)

Correct option is  C) (1, – 2)

Correct option is (D) Q₄

A point that has x-coordinate positive and y-coordinate negative, will lie in IV quadrant.

\(\because\) x > 0, y < 0 is given.

\(\therefore\) (x, y) lies in IV quadrant.

\(\therefore(x,y)\in Q_4\)

Correct option is D) Q₄

Correct option is (C) abscissa

The distance of a point (x, y) from Y-axis (x = 0 line) is called abscissa (x-coordinate of the point).

Correct option is  C) abscissa

Correct option is (C) (-3, 6)

In \(Q_2\), x-coordinate is negative and y-coordinate is positive.

\(\because\) Distance of point from Y-axis = 3 units

\(\therefore\) Abscissa = -3  \((\because\) x-coordinate is negative)

\(\because\) Distance of point from X-axis = 6 units

\(\therefore\) Ordinate = 6

\(\therefore\) Required point is (-3, 6).

Correct option is  C) (-3, 6)

Correct option is (A) (7, 0)

The equation of x-axis is y = 0 and the equation of y-axis is x = 0.

Among all given points only (7, 0) has y-coordinate 0.

Therefore, (7, 0) lies on x-axis.

Correct option is  A) (7, 0)

Correct option is (D) (-1, –3)

If point (x, y) lies in \(Q_3\) then x < 0 & y < 0.

Among all given point only point (-1, -3) has both negative coordinate.

Thus, (-1, -3) lies in \(Q_3.\)

D) (- 1, – 3)

Correct option is (C) Q₃

A point that has both x & y-coordinates are negative, will lie in III quadrant.

\(\because\) x < 0, y < 0 is given.

\(\therefore\) (x, y) lies in III quadrant.

\(\therefore(x,y)\in Q_3\)

Correct option is C) Q₃

Correct option is (B) on Y-axis

\(\because\) Equation of Y-axis is x = 0.

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

B) on Y-axis

Correct option is (B) 2

The number of references needed to locate the exact position of a point in a plane is 2.

Correct option is  B) 2

Correct option is (D) Square

AB = \(\sqrt{(2-2)^2+(8-4)^2}\) = 4 units

BC = \(\sqrt{(6-2)^2+(8-8)^2}\) = 4 units

CD = \(\sqrt{(6-6)^2+(4-8)^2}\) = 4 units

DA = \(\sqrt{(2-6)^2+(4-4)^2}\) = 4 units

AC = \(\sqrt{(6-2)^2+(8-4)^2}\) = \(\sqrt{16+16}\) = \(4\sqrt2\) units

BD = \(\sqrt{(6-2)^2+(4-8)^2}\) = \(\sqrt{16+16}\) = \(4\sqrt2\) units

Since, AB = BC = CD = DA and AC = BD

\(\therefore\) ABCD is a square.

A) Rectangle

Correct option is (B) Q₂

A point that has both x-coordinates negative and y-coordinate positive, will lie in II quadrant.

\(\because\) x < 0, y > 0 is given.

\(\therefore\) (x, y) lies in II quadrant.

\(\therefore(x,y)\in Q_2\)

Correct option is B) Q₂

Correct option is  B) -3

Correct option is (A) Q₁

\(\because\) x < 0 & y < 0

\(\Rightarrow\) -x > 0 & -y > 0

\(\therefore\) (- x, – y) lies in \(Q_1\)  (Because both coordinates are positive)

Correct option is A) Q₁ 

Correct option is (C) II

(- 2, 3); the x-coordinates (-2) is negative and y-coordinates (3) is positive.

\(\therefore\) (- 2, 3) lies \(2^{nd}\) quadrant.

Correct option is  C) II

Correct option is (B) (0, 0)

x-coordinate is 0 of any point lying on Y-axis.

\(\therefore\) (0, 0) lies on Y-axis.

Correct option is  B) (0, 0)

Correct option is (C) Q₂

If x < 0 & y > 0 then point (x, y) lies in \(2^{nd}\) quadrant.

\(\because\) x = -2 < 0 & y = 4 > 0

Then (-2, 4) lies in \(Q_2.\)

Correct option is  C) Q₂ 

Correct option is (D) Q₄

If x > 0 & y < 0 then point (x, y) lies in \(4^{th}\) quadrant.

\(\because\) x = 5 > 0 & y = -3 < 0

Then (5, –3) lies in \(Q_4.\)

Correct option is  D) Q₄

Correct option is (A) Q₁

A point that has both x & y-coordinates are positive, will lie in I quadrant.

\(\because x>0,y>0\) is given.

\(\therefore\) (x, y) lies in I quadrant.

\(\therefore(x,y)\in Q_1\)

Correct option is A) Q₁

Correct option is (C) Q₄

x > 0, y < 0 then (x, y) lie in \(4^{th}\) quadrant or in \(Q_4.\)

Correct option is  C) Q₄

Correct option is (B) Q₁

\(\because\) (x, y) = (7, 3)

\(\therefore\) x = 7 & y = 3

Now, x+y = 7+3 = 10 > 0 & x - y = 7 - 3 = 4 > 0

\(\therefore\) (x + y, x – y) = (10, 4) lies in \(Q_1.\)

Correct option is B) Q₁

Correct option is (C) Q₃

If x < 0, y < 0 then point (x, y) lies in \(3^{rd}\) quadrant.

Correct option is  C) Q₃

Correct option is (A) (-2, 3)

A point that has x-coordinate negative and y-coordinate positive will lie in II quadrant.

\(\because-2<0\;\&\;3<0\)

\(\therefore\) (-2, 3) lies in II quadrant.

Correct option is A) (-2, 3)

Given points = A (3, 4) and B (3, 8) 

These two points lie on the line parallel to Y – axis. 

∴ Distance between A (3, 4) and B (3, 8)= |y₂ – y₁| 

= |8 – 4| = 4 units.

Correct option is (A) Origin

The meeting point of the two axes in the coordinate plane is called origin.

Correct option is A) Origin

Correct option is (C) (x, 0)

Any point on X-axis is of the form (x, 0) because y-coordinate of any point on X-axis is 0.

Correct option is  C) (x, 0)

Correct option is (D) Q₄

x > 0, y < 0

Then x > 0, -y > 0

\(\Rightarrow\) x - y > 0

Now, y - x = -(x - y)

\(\Rightarrow\) y - x < 0   \((\because\) x - y > 0)

\(\because\) x - y > 0 & y - x < 0

Therefore, (x - y, y - x) lies in \(Q_4\)  \((4^{th}\) quadrant)

Correct option is  D) Q₄

Correct option is (D) IV

Given that (k, 2) lies in II quadrant and we know that any point in II quadrant has x-coordinate negative and y-coordinate positive.

\(\therefore k<0\)

\(\Rightarrow-k>0\)

\(\therefore(-k,-2)\) lies in IV quadrant.

\((\because\) Any point in IV quadrant has x-coordinate positive and y-coordinate negative and here \(-k>0\;\&\;-2<0)\)

Correct option is D) IV

Given points = A (-5, -8) and B (-5, -12) 

These two points lie on the line parallel to Y – axis. 

∴ Distance between A (-5, -8) and B (-5, -12) = |y₂ – y₁| 

= |-12 – (-8)|

= |-12 + 8| 

= |-4| = 4 units.

Correct option is (B) (-5, -4)

A point that has both x & y-coordinates are negative, lies in III quadrant.

\(\because-5<0\;\&\;-4<0\)

\(\therefore\) (-5, -4) lies in III quadrant.

Correct option is B) (-5, -4)

Correct option is (C) y = 3x

\(y=3x\) is line passing through origin.

(Because (0, 0) satisfies the equation y = 3x)

Correct option is  C) y = 3x

Correct option is (B) y = k

The equation of line parallel to X-axis is y = k.

Correct option is  B) y = k

(d) \(5\sqrt{10}\)

. Let P ≡ (x, y). Then, PA = PB and PB = PC. 

∴ PA² = PB² ⇒ (x – 1)² + (y – 3)² = (x + 3)² + (y – 5)² 

⇒ x² – 2x + 1 + y² – 6y + 9 = x² + 6x + 9 + y² – 10y + 25 

⇒ –8x + 4y – 24 = 0 ⇒ 2x – y + 6 = 0            ...(i) 

PB² = PC² ⇒ (x + 3)² + (y – 5)² = (x – 5)² + (y + 1)² 

⇒ x² + 6x + 9 + y² – 10y + 25 = x² – 10x + 25 + y² + 2y + 1 

⇒ 16x – 12y + 8 = 0 ⇒ 4x – 3y + 2 = 0           ...(ii) 

From (i), y = 2x + 6. Putting in (ii), we have 

4x – 3(2x + 6) + 2 = 0 

⇒ 4x – 6x – 18 + 2 = 0 ⇒ –2x –16 = 0 ⇒ x = –8 

∴ y = 2 × (– 8) + 6 = –10. 

∴ PB = \(\sqrt{(-8+3)^2+(-10-5)^2}\) = \(\sqrt{(-5)^2+(-15)^2}\)

= \(\sqrt{250}\) = \(5\sqrt{10}\).

(b) Equilateral, \(2\sqrt3a^2\) sq. units

Let A(a, a), B(–a, –a) and C \((-\sqrt3a,\sqrt3a)\) be the vertices of ΔABC. Then,

AB = \(\sqrt{(a+a)^2+(a+a)^2}\) = \(\sqrt{4a^2+4a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)

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

= \(\sqrt{a^2-2\sqrt3a+3a^2+a^2+2\sqrt3a+3a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)

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

= \(\sqrt{a^2+2\sqrt3a+3a^2+a^2-2\sqrt3a+3a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)

∵ AB = BC = AC, ΔABC is equilateral.

Area = \(\frac{\sqrt3}{4}\) (side)² = \(\frac{\sqrt3}{4}\) x (\(2\sqrt2a\))²

⁼ \(\frac{\sqrt3}{4}\) x 8a² = \(2\sqrt3a^2\).

The point on y - axis that is nearest to the point (- 2, 5) is (0, 5).

(c) (–2, –1)

Calculating the distance of each point from the origin, we have

(a) \(\sqrt{(0-2)^2+(0+3)^2} = \sqrt{4+9} = \sqrt{13}\)

(b) \(\sqrt{(0-6)^2+0} = \sqrt{36} = 6\)

(c) \(\sqrt{(0+2)^2+(0+1)^2} = \sqrt{4+1} = \sqrt{5}\)

(d) \(\sqrt{(0-3)^2+(0-5)^2} = \sqrt{9+25} = \sqrt{34}\)

Clearly, (–2, –1) is the nearest point from the origin.

(d) x = y

Given, AP = AQ 

⇒ AP² = AQ² 

⇒ (x + 3)² + (y – 2)² = (x – 2)² + (y + 3)² 

⇒ x² + 6x + 9 + y² = x² – 4x + 4 + y² + 6y + 9 

⇒ 10x = 10y ⇒ x = y

(c) 11

Given, \(\sqrt{(-6+6)^2+(y+1)^2}\) = 12

⇒ \(\sqrt{y^2+2y+1}\) = 12

⇒ y² + 2y + 1 = 144 ⇒ y² + 2y – 143 = 0 

⇒ y² + 13y – 11y – 143 = 0 

⇒ y (y + 13) – 11 (y + 13) = 0 

⇒ (y – 11) (y + 13) = 0 ⇒ y = 11 or –13 

The required positive integer is 11.

(b) \(\sqrt2\)

Reqd. dist. = \(\sqrt{(sin\,\theta-cos\,\theta)^2+(-cos\,\theta-sin\,\theta)^2}\)

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

= \(\sqrt{2(sin^2\,\theta+cos^2\,\theta)}\) = \(\sqrt2\) (∵ cos² θ + sin² θ = 1)

Let the co-ordinates of the third vertex be (x, y). Then, 

Co-ordinates of the centroid are \(\bigg(\frac{1+2+x}{3},\frac{1-3+y}{3}\bigg)\) = \(\bigg(\frac{3+x}{3},\frac{-2+y}{3}\bigg)\)

Given, \(\bigg(\frac{3+x}{3},\frac{-2+y}{3}\bigg)\) = (2.1)

⇒ \(\frac{3+x}{3} = 2\) and \(\frac{-2+y}{3} = 1\)

⇒ 3 + x = 6 and – 2 + y = 3 

⇒ x = 3 and y = 5 

∴ Co-ordinates of the third vertex are (3, 5).

Let D(x, y) be the midpoints of A(6, 8) and B(2, 4). Let our third given point be C(1, 2). 

By midpoint formula.

x = \(\frac{x_1+x_2}2\), y = \(\frac{y_1+y_2}2\)

For midpoint D of AB,

 x = \(\frac{6+2}2\) and y = \(\frac{8+4}2\)

∴ x =4 and y = 6 

∴D(x, y) ≡ (4, 6) 

Now to find distance between C and D, 

By distance formula,

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

For CD, 

CD = \(\sqrt{(4-1)^2+(6-2)^2}\)

= \(\sqrt{9 + 16}\)

= 5 units

Let P divide the line joining A and B and let it divide the segment in the ratio k: 1

Now, using the section formula for the y – coordinate we have

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

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

2k + 2 = -3k + 5

5k = 3

k = 3/5

Thus, P divides the line segment AB in the ratio of 3: 5

Using value of k, we get the x – coordinate as

x = 12 + 60/ 8 = 72/8 = 9

Therefore, the coordinates of point P is (9, 2)

Let P divide A(-3, 10) and B(6, -8) in the ratio of k: 1

Given coordinates of P as (-1, y)

Now, using the section formula for x – coordinate we have

-1 = 6k – 3/ k + 1

-(k + 1) = 6k – 3

7k = 2

k = 2/7

Thus, the point P divides AB in the ratio of 2: 7

Using value of k, to find the y-coordinate we have

y = (-8k + 10)/ (k + 1)

y = (-8(2/7) + 10)/ (2/7 + 1)

y = -16 + 70/ 2 + 7 = 54/9

y = 6

Therefore, the y-coordinate of P is 6

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

So, PA = PB

PA² = PB²

(3 -0)² + (k -2)² = (k – 0)² + (5 – 2)²

9 + k² + 4 – 4k – k² – 9 = 0

4 – 4k = 0

-4k = -4

Therefore, the value of k = 1

The position of Ayush’s house is (2, 4) and the position of the bank is (5, 8).

So, the distance between the house and the bank,

d₁ = √[(5 – 2)² + (8 – 4)²] 

= √[(3)² + (4)²] 

= √[9 + 16] 

= √25 

= 5 km

The position of the bank is (5, 8) and the position of the school is (13, 14).

So, the distance between the bank and the school,

d₂ = √[(13 – 5)² + (14 – 8)²] 

= √[(8)² + (6)²] 

= √[64 + 36] 

= √100 

= 10 km

The position of the school is (13, 14) and the position of the office is (13, 26).

So, the distance between the school and the office,

d₃ = √[(13 – 13)² + (26 – 14)²] 

= √[(0)² + (12)²] 

= √144 

= 12 km

Let d be the total distance covered by Ayush

d = d₁ + d₂ + d₃ 

= 5 + 10 + 12 

= 27 km

Let the D be the shortest distance from Ayush’s house to the office,

D = √[(13 – 2)² + (26 – 4)²] 

= √[(11)² + (22)²] 

= √[121+ 484] 

= √605 

= 24.6 km

Thus, the extra distance covered by Ayush = d – D = 27 – 24.6 = 2.4 km