Using the distance formula,

(8 – 3)²+ (y - (-2)² = 89

25 + (y + 2)² = 89

(y + 2)² = 89 - 25

(y + 2)² =64

Taking square roots,

y + 2 = ± 8

Therefore, y = 6 or y = -10

Hence, the possible values of yare 6 and -10.

\(\begin{aligned} & \mathrm{AB}=\sqrt{(16+0)}=4 ; \\ & \mathrm{BC}=\sqrt{(4+9)}=\sqrt{13} ; \\ & \mathrm{CD}=\sqrt{(16+0)}=4 ; \\ & \mathrm{DA}=\sqrt{(4+9)}=\sqrt{13} . \end{aligned}\)

AB = CD (4) and BC = DA \(\sqrt{13}\),

So, opposite sides are equal but the diagonals AC = \(\sqrt{(36+9)}=\sqrt{45} \text { and } \mathrm{BD}=\sqrt{(4+9)}=\sqrt{13}\) are unequal So, ABCD is a parallelogram (NOT a rectangle, since the diagonals are unequal).

(A) Centre = midpoint of PR

\(=\left(\frac{(-2+6)}{2}, \frac{(1+7)}{2}\right)\)

= (2, 4).

(B) \(\begin{aligned} P R & =\sqrt{(64+36)} \\ & =\sqrt{100}=10 . \end{aligned}\)

(C) Side of square = \(\frac{\text { diagonal }}{\sqrt{2}}\)

\(=\frac{10}{\sqrt{2}}=5 \sqrt{2}\)

= 7.07 units

(A) \(A B=\sqrt{(64+0)}=8 ;\)

\(\begin{aligned} B C & =\sqrt{(16+36)} \\ & =\sqrt{52}=2 \sqrt{13} \\ C A & =\sqrt{(16+36)}=2 \sqrt{13} . \end{aligned}\)

(B) Perimeter = \(8+2 \sqrt{13}+2 \sqrt{13}\)

\(=8+4 \sqrt{13}\)

= 22.42 units.

(C) BC = CA, so the triangle is isosceles.

\(\begin{aligned} P Q & =\sqrt{(25+16)}=\sqrt{41} ; \\ Q R & =\sqrt{(25+16)}=\sqrt{41} ; \\ P R & =\sqrt{(100+64)} \\ & =\sqrt{164}=2 \sqrt{41} . \end{aligned}\)

Since \(P Q+Q R=2 \sqrt{41}=P R\) the points are colinear with Q lying between P and R.

\(\begin{aligned} & A B=\sqrt{(9+16)}=5 ; \\ & B C=\sqrt{(9+16)}=5 ; \\ & A C=\sqrt{(36+64)}=10 . \end{aligned}\)

Check: AB + BC= 10 = AC.

The points are collinear, and B is the midpoint of the segment AC.

P(-4, 3): The x-coordinate is negative and the y-coordinate is positive (x < 0,y> 0). Therefore, P lies in Quadrant Il. It is 4 units to the left of
the y-axis and 3 units above the x-axis.
Q(2, -7): The X-coordinate is positive and the y-coordinate is negative (x > 0,y< 0). Therefore, Q lies in Quadrant IV. It is 2 units to the right of the y-axis and 7 units below the x-axis.

R(0, 5): The x-coordinate is 0 and the y-coordinate is positive.

Therefore, R lies on the positive y-axis. It is 5 units above the origin and does not belong to any quadrant. S(-6, -1): Both coordinates are negative (x< 0, y<0). Therefore, S lies in Quadrant lIl. It is 6 units to the left of the y-axis and 1 unit below the x-axis.

(A) Both coordinates are positive (x > 0, y> 0). Therefore, the point lies in Quadrant I.

(B) The X-coordinate is negative and the y-coordinate is positive (x<0.y> 0). Therefore, the point lies in Quadrant II

(C) Both coordinates are negative (x <0,y< 0). Therefore, the point lies in Quadrant III.

(D) The x-coordinate is positive and the y-coordinate is negative (x>0.y< 0). Therefore, the point lies in Quadrant IV.

(E) The x-coordinate is 0 and the y-coordinate is positive. Therefore, the point lies on the positive y-axis.
(F) The y-coordinate is 0 and the x-coordinate is negative. Therefore, the point lies on the negative x-axis.

The streets are spaced 10 metres apart.

Distance of the shop from the Great Bath towards the east = 3 x 10 = 30 m

Distance of the shop from the Great Bath towards the north = 2 x 10 = 20 m
Therefore, the coordinate of the merchant's shop is (30, 20).

(A) Yes, two perpendicular axes that cross at a fixed origin (the city centre), distances measured along each give a unique (x, y) for any location

(B) No, two railway tracks are parallel, not perpendicular. Two parallel reference lines cannot uniquely specify a point. Coordinate axes must be perpendicular.

(C) Yes, it can form a 2-D coordinate system. The north-south and east-west grid lines provide two perpendicular reference directions. The position of a place can be determined uniquely by specifying its location along these two directions.

Distance of the bench from the fountain:

\(\begin{aligned} \mathrm{FB} & =\sqrt{(6-0)^2+(8-0)^2} \\ & =\sqrt{36+64}=\sqrt{100}=10 \end{aligned}\)

Distance of the swing from the fountain

\(\begin{aligned} \mathrm{FS} & =\sqrt{(-8-0)^2+(6-0)^2} \\ & =\sqrt{64+36}=\sqrt{100}=10 \end{aligned}\)

Since FB = FS = 10 units, the bench and the swing are equidistant from the fountain.

Let B = (x, y).

For x-coordinate \(2=\frac{(-4+x)}{2}\)

⇒ x = 8;

For y-coordinate \(5=\frac{(7+y)}{2 y}=3 .\)

So, B = (8,3).

\(\mathrm{AB}=\sqrt{\left[(0-a)^2+(b-0)^2\right]}\)

\(\sqrt{\left(a^2+b^2\right)}\)

(2, 3) and (3, 2) represent different points because the first number denotes the x-coordinate and the second number denotes the y-coordinate.

S The shortest distance to reach point P will be OP.

From the figure,

AP = 8 units

OA = PB 6 units

By Baudhayana Pythagoras theorem, in △OPA

OP² = AP²+ OA²

⇒ OP² = (8)²+ (6)²

⇒ OP² = 64+ 36

⇒ OP² = 100

⇒ \(\mathrm{OP}=\sqrt{100}\)

⇒ OP = 10 units

Using the midpoint the formula

(i) \(\begin{aligned} M & =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\ M & =\left(\frac{2+6}{2}, \frac{4+10}{2}\right) \\ & =\left(\frac{8}{2}, \frac{14}{2}\right)=(4,7) \end{aligned}\)

(ii) \(\begin{aligned} M & =\left(\frac{-3+3}{2}, \frac{5+(-5)}{2}\right) \\ & =\left(\frac{0}{2}, \frac{0}{2}\right)=(0,0) \end{aligned}\)

(iii) \(\begin{aligned} M & =\left(\frac{-6+(-2)}{2}, \frac{2+8}{2}\right) \\ & =\left(\frac{-8}{2}, \frac{10}{2}\right)=(-4,5) \end{aligned}\)

(iv) \(\begin{aligned} M & =\left(\frac{1+5}{2}, \frac{3+9}{2}\right) \\ & =\left(\frac{6}{2}, \frac{12}{2}\right)=(3,6) \end{aligned}\)

Let S and L be the endpoints of a line segment, and let M(7, 7) be its midpoint.

\(\mathrm{SL}=\sqrt{64+36}=\sqrt{100}=10\)

Since M is the midpoint of SL,

\(S M=M L=\frac{S L}{2}=\frac{10}{2}=5\)

Thus, the distance from S to the midpoint M is 5 units, which is half the length of the segment SL

The distance formula is derived from the Baudhayana-Pythagoras Theorem.

(-4, -9) has both coordinates negative, so it lies in Quadrant (lII), not in (i).

\(\begin{aligned} \text { Midpoint } & =\left(\frac{(4+(-2)}{2}\right) \cdot\left(\frac{6+2}{2}\right) \\ & =(1,4) \\ \text { Distance } & =\sqrt{\left[(-2-4)^2+(2-6)^2\right]} \\ & =\sqrt{(36+16)}=\sqrt{52}=2 \sqrt{13} . \end{aligned}\)

Since,

\(\begin{aligned} P Q & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ & =\sqrt{(12-0)^2+[0-(-5)]^2} \\ & =\sqrt{(144+25)}=\sqrt{169}=13 . \end{aligned}\)

A point whose y-coordinate is 0 always lies on the x-axis.

So, (-4, O) lies on the x-axis (specifically on the negative side of the x-axis because x= -4).

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation:

The midpoint of the segment joining (a, b) and (-a, -b) is the origin.
Using the midpoint formula,

\(\begin{aligned} & =\left(\frac{a+(-a)}{2}, \frac{b+(-b)}{2}\right) \\ & =\left(\frac{0}{2}, \frac{0}{2}\right)=(0,0) \end{aligned}\)

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation:

An equilateral triangle is a triangle in which all three sides are equal. Since

PQ = QR= PR,

all three sides of APOR are equal. Therefore, △POR is an equilateral triangle.

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation: Let the points be A(3, 0), B(0,4) and C(3, 4)

Using the distance formula:

\(\begin{aligned} A B & =\sqrt{(3-3)^2+(4-0)^2}=4 \\ B C & =\sqrt{(3-0)^2+(4-4)^2}=3 \\ A B & =\sqrt{(3-0)^2+(0-4)^2} \\ & =\sqrt{9+16}=5 \end{aligned}\)

Thus, the three side lengths are 3, 4, and 5.

Since

3²+4²=5²

by the converse of the Baudhayana-Pythagoras Theorem, the triangle is right-angled. Equivalently. AC is vertical and BC is horizontal, so they are perpendicular and angle \(\angle \mathrm{C}=90^{\circ} .\)

Hence, the Assertion is true. The Reason is also true and correctly explains why the triangle is right-angled.

(d) Assertion (A) is false but reason (R) is true.

Explanation:

\(\begin{aligned} & A B=\sqrt{(2-1)^2+(2-1)^2}=\sqrt{2} \\ & B C=\sqrt{(3-2)^2+(3-2)^2}=\sqrt{2} \end{aligned}\)

\(A C=\sqrt{(3-1)^2+(3-1)^2}=\sqrt{8}=2 \sqrt{2}\)

Since, \(A B+B C=\sqrt{2}+\sqrt{2}=2 \sqrt{2}=A C,\)

the three points are collinear.

Therefore, the Assertion "the points are non-collinear" is false, while the Reason is true.

(c) Assertion (A) is true but reason (R) is false.

Explanation :

The distance between A(3, 4) and B(0, 0) is

\(\begin{aligned} A B & =\sqrt{(3-0)^2+(4-0)^2} \\ & =\sqrt{9+16}=\sqrt{25}=5 \end{aligned}\)

Hence, assertion (A) is true.

However, the distance of a point P (x, y) from the origin is

\(\begin{aligned} &\sqrt{x^2+y^2}, \text { not } \sqrt{x^2-y^2} \text {. }\\ &\text { Therefore, reason ( } R \text { ) is false. } \end{aligned}\)

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation:

The point (0, -5) has x-coordinate 0. By definition, every point on the y-axis has x-coordinate 0. Therefore, (0, -5) lies on the y-axis.

(2, 2)

Explanation :

\(\text { Midpoint }=\left(\frac{(-3+7)}{2}, \frac{(5+(-1))}{2}\right)=(2,2) .\)

\((0,0): \sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\right]}\)

True

Explanation:

The distance from (x, y) to the origin (0, 0) is \(\sqrt{\left[(x-0)^2+(y-0)^2\right.}=\sqrt{\left(x^2+y^2\right)} .\)

False

Explanation:

(0, 7) lies on the positive y-axis, not in Quadrant I, because points in Quadrant I must satisfy x>0 and y> 0.

(A) \(\begin{aligned} P R & =\sqrt{(8-2)^2+(3-5)^2} \\ & =2 \sqrt{10} \end{aligned}\)

(B) Coordinates of Q (4, 4).

The mid-point of PR is (5,4)

∴ Q is not the mid-point of PR

(C) Let the point be (x, 0)

\(\text { So, } \sqrt{(2-x)^2+25}=\sqrt{(4-x)^2+16}\)

Hence, x = \(\frac{3}{4} \text {. Therefore, the point is }\left(\frac{3}{4}, 0\right)\)

(A) Area of triangle = \(\frac{1}{2} \times \text { base × height }\)

Base = OR= (4-1) = 3 units

Height = PQ = (4- 1) = 3 units

∴ \(\text { Area }=\frac{1}{2} \times 3 \times 3\)

= 4.5 sq. units

(B) The shape of park constructed for adults is square.

AB = BC = CD = DA =3 units

Hence, side of square =3 units

Diagonal, AC = \(\sqrt{2} \times \text { side }\)

\(=\sqrt{2} \times 3 \text { units }\)

\(=3 \sqrt{2} \text { units }\)

According to question,

1 square box = 10 units

So, AB = BC= CD = DA

=3x 10 units

= 30 units

Area of square = side²

= (30)²

= 900 sq. units

(C) QR =3 units, PQ=3 units

Apply, Pythagoras theorem

PR² = PQ²+OR²

⇒ (3)² + (3)²

⇒ = 9+9 = 18

\(P R=3 \sqrt{2} \text { units }\)

(A) In IV quadrant, abscissa (x-coordinate) is always positive and ordinate (y-coordinate) is negative. Since, Abscissa > Ordinate

Hence, Sumit is in the IV quadrant.

(B) In | quadrant, abscissa (x-coordinate) is always negative and ordinate y-coordinate) is always positive.

Since, Abscissa < Ordinate

Hence, Raju is in the ll quadrant.

(C) In I quadrant Abscissa (K-coordinate) is always negative and ordinate y-coordinate) is negative.

Since, Abscissa < 0

Also, ordinate < 0

Hence, Deepti is located in lIl quadrant.

In I quadrant, abscissa (x-coordinate) is always positive and ordinate (y-coordinate) is positive

Since, Abscissa > 0

Also, Ordinate > 0

Hence, Ramya is located in I quadrant

(A) (d) Cheetah and Ostrich

Explanation: From the graph cheetah and ostrich are parallel toy-axis.

(B) (c) Cheetah

Explanation: Ostrich moves 11 units in North direction, ie., along y-axis. So, ostrich will meet cheetah and the new coordinates of ostrich are (2, 6).

(C) The coordinates of giraffe are (-5, -4), Now moving 6 units towards East direction along X-axis so, giraffe's new coordinates are (-5 + 6, -4) = (1, -4). Hence, giraffe will be in IV quadrant.

(A) (5, -7): x> 0,y<0, so the point lies in Quadrant IV.
(B) (-9,-4): both coordinates are negative, so the point lies in Quadrant III.

(C) (-2.6): x <0.y> 0, so the point lies in Quadrant II.
(D) (8, 1): both coordinates are positive, so the point lies in Quadrant I.

\(\begin{aligned} & A B=\sqrt{(9+16)}=5 ; \\ & B C=\sqrt{(16+0)}=4 ; \end{aligned}\)

\(\mathrm{AC}=\sqrt{(49+16)}=\sqrt{16}\)

Check: AB² + BC² = 25 + 16 = 41

≠ AC².

The equality is not satisfied. Therefore, the triangle does not satisfy Pythagoras theorem and hence is not a right-angled triangle.

Since the point lies on the x-axis, its y-coordinate is 0.

Distance from the origin is \(|x|=13\)

Therefore, X = ±13

Hence, the possible coordinates are (13, 0) and (-13, 0).

Let \(B=(x, y) \cdot 0=\frac{(-6+x)}{2}\)

⇒ \(x=6.4=\frac{(7+y)}{2}\)

⇒ y=1. So B= (6, 1).

Distance from origin = \(\sqrt{x^2+y^2}\)

\(\begin{aligned} & =\sqrt{\left[(-7)^2+24^2\right]} \\ & =\sqrt{(49+576)} \\ & =\sqrt{625}=25 \text { units. } \end{aligned}\)

\(\begin{aligned} \text { Midpoint } & =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\ & =\left(\frac{(8+(-4))}{2}, \frac{(-3+7)}{2}\right) \end{aligned}\)

= (2, 2).

\(\begin{aligned} \text { Distance } & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ & =\sqrt{(25+144)}=\sqrt{169} \end{aligned}\)

= 13 units.

P (-3, 0): on the negative x-axis.

Q (0, 4): on the positive y-axis.

R (0, -5): on the negative y-axis.

On plotting the points in the cartesian plane:

(A) A horizontal line and a vertical line are perpendicular. So, AM⊥MP.

(B) AM has slope 0, so AM is parallel to the x-axis. Also, MP is parallel to the y-axis.

(C) M(-5, -2) and P(-5, 2) has same x-coordinate and opposite y-coordinates. They are mirror images in the x-axis.

Specific case: P(2,4), Q(8, 10).

Midpoint M = (5, 7).

\(\begin{aligned} & \mathrm{PM}=\sqrt{(9+9)}=\sqrt{18} \\ & \mathrm{QM}=\sqrt{(9+9)}=\sqrt{18} . \text { Equal. } \end{aligned}\)

(B) General proof.

\(\begin{aligned} \mathrm{M} & =\left(\frac{\left(x_1+x_2\right)}{2}, \frac{\left(y_1+y_2\right)}{2}\right) \\ \mathrm{PM} & =\sqrt{\left[\left(\frac{\left(x_1+x_2\right)}{2}-x_1\right)^2+\left(\frac{\left(y_1^*+y_2\right)}{2}-y_1\right)^2\right]} \\ & =\sqrt{\left[\left(\frac{\left(x_2-x_1\right)}{2}\right)^2+\left(\frac{\left(y_2-y_1\right)}{2}\right)^2\right]} \\ & =\left(\frac{1}{2}\right) \sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\right.}=\frac{\mathrm{PQ}}{2} . \end{aligned}\)

By symmetry. QM\(=\frac{P Q}{2}\) Hence PM=QM.

(C) In computer graphics, smooth lines are drawn by repeatedly finding midpoints. First, the midpoint between two pixels is found and plotted. Then the midpoints of the smaller line segments are found again. By repeating this process many times, the line appears smooth between the two endpoints.

(A) \(P R=\sqrt{\left(80^2+60^2\right)}=\sqrt{(6400+3600)}\)

\(=\sqrt{10000}=100 ;\)

\(Q S=\sqrt{\left(80^2+60^2\right)}=100 .\)

(B) Centre = midpoint of PR = (40, 30).

(C) Midpoint of PQ = (40, 0); midpoint of RS = (40, 60).

(D) Distance between flagpoles = \(\sqrt{(0+3600)}=60\) units (= the width ofthe field, as expected).

Compute all four sides:

\(\begin{aligned} & \mathrm{AB}=\sqrt{(9+25)}=\sqrt{34} ; \\ & \mathrm{BC}=\sqrt{(25+9)}=\sqrt{34} ; \\ & \mathrm{CD}=\sqrt{(9+25)}=\sqrt{34} ; \\ & \mathrm{DA}=\sqrt{(25+9)}=\sqrt{34} ; \end{aligned}\)

Since all four sides are equal, the figure is at least a rhombus.

Now check the diagonals:

\(\begin{aligned} & A C=\sqrt{(4+64)}=\sqrt{68} \\ & B D=\sqrt{(64+4)}=\sqrt{68} \end{aligned}\)

Both diagonals are equal. A rhombus with equal diagonals is a square. Therefore, ABCD is a square.