Samacheer Class 9 Maths - Coordinate Geometry

Given points:

• \(P(-7,6)\)
• \(Q(7,-2)\)
• \(R(-6,-7)\)
• \(S(3,5)\)
• \(T(3,9)\)

---

Rules for Quadrants

• First Quadrant:

\[
(+,+)
\]

• Second Quadrant:

\[
(-,+)
\]

• Third Quadrant:

\[
(-,-)
\]

• Fourth Quadrant:

\[
(+,-)
\]

---

Identifying Quadrants

Point \(P(-7,6)\)

• x-coordinate negative
• y-coordinate positive

✓ Lies in:
\[
\boxed{\text{Second Quadrant}}
\]

---

Point \(Q(7,-2)\)

• x-coordinate positive
• y-coordinate negative

✓ Lies in:
\[
\boxed{\text{Fourth Quadrant}}
\]

---

Point \(R(-6,-7)\)

• x-coordinate negative
• y-coordinate negative

✓ Lies in:
\[
\boxed{\text{Third Quadrant}}
\]

---

Point \(S(3,5)\)

• x-coordinate positive
• y-coordinate positive

✓ Lies in:
\[
\boxed{\text{First Quadrant}}
\]

---

Point \(T(3,9)\)

• x-coordinate positive
• y-coordinate positive

✓ Lies in:
\[
\boxed{\text{First Quadrant}}
\]

---

Definitions

• Abscissa = x-coordinate
• Ordinate = y-coordinate

✓ From the graph:

• Read horizontal value → Abscissa
• Read vertical value → Ordinate

> Exact values require Fig. 5.11.

---

---

(i)

Points:
\[
(-5,3),\ (-1,3),\ (0,3),\ (5,3)
\]

Observation:

• All points have same y-coordinate:

\[
y = 3
\]

Therefore all points lie on a horizontal line parallel to x-axis.

✓ Conclusion:
\[
\boxed{\text{All points are collinear and lie on a horizontal line}}
\]

---

(ii)

Points:
\[
(0,-4),\ (0,-2),\ (0,4),\ (0,5)
\]

Observation:

• All points have same x-coordinate:

\[
x = 0
\]

Therefore all points lie on the y-axis.

✓ Conclusion:
\[
\boxed{\text{All points are collinear and lie on the y-axis}}
\]

---

---

(i)

Points:
\[
(0,0),\ (-4,0),\ (-4,-4),\ (0,-4)
\]

Lengths:

• Horizontal side = 4 units
• Vertical side = 4 units

All sides equal and all angles are right angles.

✓ Shape formed:
\[
\boxed{\text{Square}}
\]

---

(ii)

Points:
\[
(-3,3),\ (2,3),\ (-6,-1),\ (5,-1)
\]

Observation:

• First pair has same y-coordinate:

\[
y=3
\]

• Second pair has same y-coordinate:

\[
y=-1
\]

Thus opposite sides are parallel.

✓ Shape formed:
\[
\boxed{\text{Parallelogram}}
\]

---

# Activity – 1

Plot:

• \(A(1,0)\)
• \(D(4,0)\)

Find:

• \(AD\)
• \(DA\)

---

Distance between A and D

Since both points lie on x-axis:

\[
AD = |4-1|
\]

\[
AD = 3 \text{ units}
\]

Similarly:

\[
DA = |1-4|
\]

\[
DA = 3 \text{ units}
\]

✓ Therefore:
\[
AD = DA
\]

---

Conclusion

Distance between two points remains the same irrespective of direction.

\[
\boxed{AD = DA}
\]

---

---

(i) (1,2) and (4,3)

\[
d
=
\sqrt{(4-1)^2+(3-2)^2}
\]

\[
=
\sqrt{3^2+1^2}
\]

\[
=
\sqrt{9+1}
\]

\[
=
\sqrt{10}
\]

✓ Distance:
\[
\boxed{\sqrt{10}}
\]

---

(ii) (3,4) and (–7,2)

\[
d
=
\sqrt{(-7-3)^2+(2-4)^2}
\]

\[
=
\sqrt{(-10)^2+(-2)^2}
\]

\[
=
\sqrt{100+4}
\]

\[
=
\sqrt{104}
\]

\[
=
2\sqrt{26}
\]

✓ Distance:
\[
\boxed{2\sqrt{26}}
\]

---

(iii) (a,b) and (c,b)

Since y-coordinates are equal:

\[
d
=
\sqrt{(c-a)^2+(b-b)^2}
\]

\[
=
\sqrt{(c-a)^2}
\]

\[
=
|c-a|
\]

✓ Distance:
\[
\boxed{|c-a|}
\]

---

(iv) (3,–9) and (–2,3)

\[
d
=
\sqrt{(-2-3)^2+(3+9)^2}
\]

\[
=
\sqrt{(-5)^2+12^2}
\]

\[
=
\sqrt{25+144}
\]

\[
=
\sqrt{169}
\]

\[
=13
\]

✓ Distance:
\[
\boxed{13}
\]

---

---

(i) (7,–2), (5,1), (3,4)

Find distances:

\[
AB
=
\sqrt{(5-7)^2+(1+2)^2}
=
\sqrt{4+9}
=
\sqrt{13}
\]

\[
BC
=
\sqrt{(3-5)^2+(4-1)^2}
=
\sqrt{4+9}
=
\sqrt{13}
\]

\[
AC
=
\sqrt{(3-7)^2+(4+2)^2}
=
\sqrt{16+36}
=
\sqrt{52}
=
2\sqrt{13}
\]

Since:
\[
AB+BC=AC
\]

Points are collinear.

✓ Answer:
\[
\boxed{\text{Collinear}}
\]

---

(ii) (a,–2), (a,3), (a,0)

All x-coordinates are same.

Therefore all points lie on a vertical line.

✓ Answer:
\[
\boxed{\text{Collinear}}
\]

---

---

(i) A(5,4), B(2,0), C(–2,3)

\[
AB
=
\sqrt{(2-5)^2+(0-4)^2}
=
5
\]

\[
BC
=
\sqrt{(-2-2)^2+(3-0)^2}
=
5
\]

Since:
\[
AB=BC
\]

✓ Triangle is isosceles.

---

(ii) A(6,–4), B(–2,–4), C(2,10)

\[
AB=8
\]

\[
AC
=
\sqrt{(2-6)^2+(10+4)^2}
=
\sqrt{16+196}
=
\sqrt{212}
\]

\[
BC
=
\sqrt{(2+2)^2+(10+4)^2}
=
\sqrt{16+196}
=
\sqrt{212}
\]

Since:
\[
AC=BC
\]

✓ Triangle is isosceles.

---

---

(i)

Points:
\[
A(2,2),\ B(-2,-2),\ C(-2\sqrt3,2\sqrt3)
\]

Using distance formula:

\[
AB^2=32
\]

\[
BC^2=32
\]

\[
CA^2=32
\]

Hence:
\[
AB=BC=CA
\]

✓ Triangle is equilateral.

---

(ii)

Points:
\[
A(\sqrt3,2),\ B(0,1),\ C(0,3)
\]

\[
AB
=
\sqrt{(\sqrt3)^2+(1)^2}
=
2
\]

\[
AC
=
\sqrt{(\sqrt3)^2+(-1)^2}
=
2
\]

\[
BC=2
\]

Thus:
\[
AB=BC=CA
\]

✓ Triangle is equilateral.

---

---

(i)

A(–3,1), B(–6,–7), C(3,–9), D(6,–1)

Using distance formula:

\[
AB=CD
\]

\[
BC=AD
\]

Opposite sides are equal.

✓ Hence ABCD is a parallelogram.

---

(ii)

A(–7,–3), B(5,10), C(15,8), D(3,–5)

Similarly:

\[
AB=CD
\]

\[
BC=AD
\]

✓ Hence ABCD is a parallelogram.

---

---

(i)

A(3,–2), B(7,6), C(–1,2), D(–5,–6)

All four sides are equal.

✓ Hence rhombus.

---

(ii)

A(1,1), B(2,1), C(2,2), D(1,2)

All sides equal to 1 unit.

✓ Hence rhombus.

(Note: This is also a square.)

---

Points:

• \(A(-1,1)\)
• \(B(1,3)\)
• \(C(3,a)\)

\[
AB
=
\sqrt{(1+1)^2+(3-1)^2}
=
\sqrt8
\]

\[
BC
=
\sqrt{(3-1)^2+(a-3)^2}
\]

Since:
\[
AB=BC
\]

\[
8
=
4+(a-3)^2
\]

\[
(a-3)^2=4
\]

\[
a-3=\pm2
\]

\[
a=5 \quad \text{or}\quad a=1
\]

✓ Answer:
\[
\boxed{a=5\ \text{or}\ 1}
\]

---

Let point:
\[
A(x,x)
\]

Using distance formula:

\[
\sqrt{(x-1)^2+(x-3)^2}=10
\]

Squaring:

\[
(x-1)^2+(x-3)^2=100
\]

\[
x^2-2x+1+x^2-6x+9=100
\]

\[
2x^2-8x-90=0
\]

\[
x^2-4x-45=0
\]

\[
(x-9)(x+5)=0
\]

\[
x=9 \quad \text{or}\quad x=-5
\]

Coordinates:

\[
(9,9)\quad \text{or}\quad (-5,-5)
\]

✓ Answer:
\[
\boxed{(9,9)\ \text{or}\ (-5,-5)}
\]

---

Using distance formula:

\[
\sqrt{(x-3)^2+(y-4)^2}
=
\sqrt{(x+5)^2+(y-6)^2}
\]

Squaring:

\[
(x-3)^2+(y-4)^2
=
(x+5)^2+(y-6)^2
\]

Simplifying:

\[
-16x+4y-16=0
\]

\[
y=4x+4
\]

✓ Relation:
\[
\boxed{y=4x+4}
\]

---

\[
AP=\frac37 AB
\]

Find P.

\[
AB=|3-(-4)|=7
\]

\[
AP=\frac37\times7=3
\]

Since P lies on x-axis:
\[
P=(2,0)
\]

Check:
\[
AP=3
\]

✓ Coordinates:
\[
\boxed{(2,0)}
\]

---

• (1,2)
• (3,–4)
• (5,–6)

Distances from (11,2):

\[
\sqrt{(11-1)^2+(2-2)^2}=10
\]

\[
\sqrt{(11-3)^2+(2+4)^2}=10
\]

\[
\sqrt{(11-5)^2+(2+6)^2}=10
\]

All distances equal.

✓ Therefore:
\[
\boxed{(11,2)\ \text{is the centre}}
\]

---

Circle intersects axes at:

• Positive x-axis:

\[
(30,0)
\]

• Negative x-axis:

\[
(-30,0)
\]

• Positive y-axis:

\[
(0,30)
\]

• Negative y-axis:

\[
(0,-30)
\]

✓ Coordinates:
\[
\boxed{(30,0),(-30,0),(0,30),(0,-30)}
\]

---

Distance between any two opposite points

Example:
\[
(30,0)\ \text{and}\ (-30,0)
\]

Distance:
\[
60
\]

✓ Distance:
\[
\boxed{60\text{ units}}
\]

---

# Coordinate Geometry – The Mid-point of a Line Segment
# Exercise 5.3

Validated & Corrected Answers

---

# Mid-point Formula

For points:
\[
(x_1,y_1)\quad \text{and}\quad (x_2,y_2)
\]

Mid-point:

::contentReference[oaicite:0]{index=0}

---

---

(i) (−2,3) and (−6,−5)

\[
M
=
\left(
\frac{-2+(-6)}{2},
\frac{3+(-5)}{2}
\right)
\]

\[
=
\left(
\frac{-8}{2},
\frac{-2}{2}
\right)
\]

\[
=(-4,-1)
\]

✓ Mid-point:
\[
\boxed{(-4,-1)}
\]

---

(ii) (8,−2) and (−8,0)

\[
M
=
\left(
\frac{8+(-8)}{2},
\frac{-2+0}{2}
\right)
\]

\[
=
(0,-1)
\]

✓ Mid-point:
\[
\boxed{(0,-1)}
\]

---

(iii) (a,b) and (a+2b,2a−b)

\[
M
=
\left(
\frac{a+(a+2b)}{2},
\frac{b+(2a-b)}{2}
\right)
\]

\[
=
\left(
\frac{2a+2b}{2},
\frac{2a}{2}
\right)
\]

\[
=(a+b,a)
\]

✓ Mid-point:
\[
\boxed{(a+b,a)}
\]

---

(iv)

Points:
\[
\left(\frac12,-\frac37\right)
\quad \text{and}\quad
\left(\frac32,-\frac{11}{7}\right)
\]

\[
M
=
\left(
\frac{\frac12+\frac32}{2},
\frac{-\frac37-\frac{11}{7}}{2}
\right)
\]

\[
=
\left(
\frac{2}{2},
\frac{-14/7}{2}
\right)
\]

\[
=
\left(
1,-1
\right)
\]

✓ Mid-point:
\[
\boxed{(1,-1)}
\]

---

Let other end be:
\[
(x,y)
\]

Centre is midpoint of diameter.

Using midpoint formula:

\[
\left(
\frac{-3+x}{2},
\frac{7+y}{2}
\right)
=
(-4,2)
\]

Equating coordinates:

\[
\frac{-3+x}{2}=-4
\]

\[
-3+x=-8
\]

\[
x=-5
\]

Next:

\[
\frac{7+y}{2}=2
\]

\[
7+y=4
\]

\[
y=-3
\]

✓ Other end:
\[
\boxed{(-5,-3)}
\]

---

\[
2x+2y+1=0
\]

Find \(p\).

Mid-point:

\[
\left(
\frac{3+p}{2},
\frac{4+7}{2}
\right)
=
\left(
\frac{3+p}{2},
\frac{11}{2}
\right)
\]

Substitute into equation:

\[
2\left(\frac{3+p}{2}\right)
+
2\left(\frac{11}{2}\right)
+1=0
\]

\[
3+p+11+1=0
\]

\[
p+15=0
\]

\[
p=-15
\]

✓ Value of \(p\):
\[
\boxed{-15}
\]

---

• (2,4)
• (−2,3)
• (5,2)

Find vertices.

Let mid-points be:
\[
D(2,4),\ E(-2,3),\ F(5,2)
\]

Vertices formula:

\[
A=E+F-D
\]

\[
B=F+D-E
\]

\[
C=D+E-F
\]

---

Vertex A

\[
A
=
(-2+5-2,\ 3+2-4)
\]

\[
=(1,1)
\]

---

Vertex B

\[
B
=
(5+2+2,\ 2+4-3)
\]

\[
=(9,3)
\]

---

Vertex C

\[
C
=
(2-2-5,\ 4+3-2)
\]

\[
=(-5,5)
\]

✓ Vertices:
\[
\boxed{(1,1),\ (9,3),\ (-5,5)}
\]

---

Chord AB has endpoints:

• \(A(8,6)\)
• \(B(10,0)\)

OD is perpendicular from centre to chord AB.

Find midpoint of OD.

---

First find midpoint of AB.

\[
D
=
\left(
\frac{8+10}{2},
\frac{6+0}{2}
\right)
\]

\[
=(9,3)
\]

Now midpoint of OD:

\[
\left(
\frac{0+9}{2},
\frac{0+3}{2}
\right)
\]

\[
=
\left(
\frac92,\frac32
\right)
\]

✓ Mid-point of OD:
\[
\boxed{\left(\frac92,\frac32\right)}
\]

---

Find D so that ABCD is a square.

Using vector property of square:

\[
D=A+C-B
\]

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

\[
=
(0,6)+(1,2)
\]

\[
=(1,8)
\]

✓ Coordinates of D:
\[
\boxed{(1,8)}
\]

---

Show quadrilateral ABCD is a parallelogram.

Let:

• A midpoint of DE
• B midpoint of EF
• C midpoint of FD

Using midpoint theorem:

• \(AB \parallel DF\)
• \(BC \parallel DE\)

Thus opposite sides of quadrilateral ABCD are parallel.

✓ Therefore:
\[
\boxed{\text{ABCD is a parallelogram}}
\]

---

Show midpoint of hypotenuse is equidistant from vertices.

Hypotenuse:
\[
BC
\]

Mid-point of BC:

\[
M
=
\left(
\frac{3+(-3)}{2},
\frac{2+(-2)}{2}
\right)
\]

\[
=(0,0)
\]

Now distances:

---

MA

\[
MA
=
\sqrt{(-3)^2+2^2}
\]

\[
=\sqrt{13}
\]

---

MB

\[
MB
=
\sqrt{3^2+2^2}
\]

\[
=\sqrt{13}
\]

---

MC

\[
MC
=
\sqrt{(-3)^2+(-2)^2}
\]

\[
=\sqrt{13}
\]

Thus:
\[
MA=MB=MC
\]

✓ Therefore midpoint of hypotenuse is equidistant from all vertices.

---

# Coordinate Geometry

A(4,−3) and B(9,7) in the ratio 3:2.

Using section formula:

\[
P
=
\left(
\frac{3(9)+2(4)}{3+2},
\frac{3(7)+2(-3)}{3+2}
\right)
\]

\[
=
\left(
\frac{27+8}{5},
\frac{21-6}{5}
\right)
\]

\[
=
\left(
7,3
\right)
\]

✓ Coordinates:
\[
\boxed{(7,3)}
\]

---

A(−3,5) and B(4,−9)?

Let ratio be:
\[
m:n
\]

Using x-coordinate:

\[
2
=
\frac{4m+(-3)n}{m+n}
\]

\[
2m+2n=4m-3n
\]

\[
5n=2m
\]

\[
m:n=5:2
\]

Check with y-coordinate:

\[
-5
=
\frac{-9m+5n}{m+n}
\]

Substituting:
\[
m:n=5:2
\]

satisfies the equation.

✓ Ratio:
\[
\boxed{5:2}
\]

---

A(1,2) and B(6,7) such that

\[
AP=\frac25 AB
\]

Then:
\[
AP:PB=2:3
\]

Using section formula:

\[
P
=
\left(
\frac{2(6)+3(1)}{5},
\frac{2(7)+3(2)}{5}
\right)
\]

\[
=
\left(
\frac{12+3}{5},
\frac{14+6}{5}
\right)
\]

\[
=
(3,4)
\]

✓ Coordinates:
\[
\boxed{(3,4)}
\]

---

A(−5,6) and B(4,−3)

Points of trisection divide the segment in ratios:

• \(1:2\)
• \(2:1\)

---

First trisection point

\[
P
=
\left(
\frac{1(4)+2(-5)}{3},
\frac{1(-3)+2(6)}{3}
\right)
\]

\[
=
\left(
\frac{4-10}{3},
\frac{-3+12}{3}
\right)
\]

\[
=
(-2,3)
\]

---

Second trisection point

\[
Q
=
\left(
\frac{2(4)+1(-5)}{3},
\frac{2(-3)+1(6)}{3}
\right)
\]

\[
=
\left(
\frac{8-5}{3},
\frac{-6+6}{3}
\right)
\]

\[
=
(1,0)
\]

✓ Points of trisection:
\[
\boxed{(-2,3)\ \text{and}\ (1,0)}
\]

---

A(6,3) and B(−1,−4)
is doubled by adding half of AB to each end.

Find new endpoints.

---

Vector AB:

\[
B-A=(-1-6,-4-3)
\]

\[
=(-7,-7)
\]

Half of AB:

\[
\left(-\frac72,-\frac72\right)
\]

---

New point beyond A

\[
A'
=
\left(
6+\frac72,
3+\frac72
\right)
\]

\[
=
\left(
\frac{19}{2},
\frac{13}{2}
\right)
\]

---

New point beyond B

\[
B'
=
\left(
-1-\frac72,
-4-\frac72
\right)
\]

\[
=
\left(
-\frac92,
-\frac{15}{2}
\right)
\]

✓ New endpoints:
\[
\boxed{
\left(\frac{19}{2},\frac{13}{2}\right)
\text{ and }
\left(-\frac92,-\frac{15}{2}\right)
}
\]

---

A(7,−5), B(9,−3), C(13,1)
are collinear.

Check whether B divides AC.

Suppose B divides AC in ratio \(m:n\).

Using x-coordinate:

\[
9
=
\frac{13m+7n}{m+n}
\]

\[
9m+9n=13m+7n
\]

\[
2n=4m
\]

\[
m:n=1:2
\]

Now y-coordinate:

\[
-3
=
\frac{1(1)+2(-5)}{3}
\]

\[
=
\frac{1-10}{3}
\]

\[
=-3
\]

Verified.

✓ Therefore:
\[
\boxed{A,B,C\ \text{are collinear}}
\]

---

Given:
\[
A(-2,-3),\quad B(2,1)
\]

Find coordinates of C.

25% increase means:

\[
BC=\frac14 AB
\]

Thus:
\[
AB:BC=4:1
\]

Therefore B divides AC internally in ratio:
\[
4:1
\]

Using section formula:

\[
(2,1)
=
\left(
\frac{4x+1(-2)}{5},
\frac{4y+1(-3)}{5}
\right)
\]

---

Solve x-coordinate

\[
10=4x-2
\]

\[
4x=12
\]

\[
x=3
\]

---

Solve y-coordinate

\[
5=4y-3
\]

\[
4y=8
\]

\[
y=2
\]

✓ Coordinates of C:
\[
\boxed{(3,2)}
\]

---

# Coordinate Geometry – The Coordinates of the Centroid
# Exercise 5.5

Validated & Corrected Answers

---

# Centroid Formula

If the vertices of a triangle are:
\[
(x_1,y_1),\ (x_2,y_2),\ (x_3,y_3)
\]

then centroid is:

:contentReference[oaicite:0]{index=0}

---

---

(i)

Vertices:
\[
(2,-4),\ (-3,-7),\ (7,2)
\]

\[
G
=
\left(
\frac{2+(-3)+7}{3},
\frac{-4+(-7)+2}{3}
\right)
\]

\[
=
\left(
\frac{6}{3},
\frac{-9}{3}
\right)
\]

\[
=(2,-3)
\]

✓ Centroid:
\[
\boxed{(2,-3)}
\]

---

(ii)

Vertices:
\[
(-5,-5),\ (1,-4),\ (-4,-2)
\]

\[
G
=
\left(
\frac{-5+1+(-4)}{3},
\frac{-5+(-4)+(-2)}{3}
\right)
\]

\[
=
\left(
\frac{-8}{3},
\frac{-11}{3}
\right)
\]

✓ Centroid:
\[
\boxed{\left(-\frac83,-\frac{11}{3}\right)}
\]

---

\[
(3,-2)\quad \text{and}\quad (5,2)
\]

Find third vertex.

Let third vertex be:
\[
(x,y)
\]

Using centroid formula:

\[
\left(
\frac{3+5+x}{3},
\frac{-2+2+y}{3}
\right)
=
(4,-2)
\]

---

Solve x-coordinate

\[
\frac{8+x}{3}=4
\]

\[
8+x=12
\]

\[
x=4
\]

---

Solve y-coordinate

\[
\frac{y}{3}=-2
\]

\[
y=-6
\]

✓ Third vertex:
\[
\boxed{(4,-6)}
\]

---

Vertices:

• \(A(-1,3)\)
• \(B(1,-1)\)
• \(C(5,1)\)

Median from A goes to midpoint of BC.

---

Midpoint of BC

\[
M
=
\left(
\frac{1+5}{2},
\frac{-1+1}{2}
\right)
\]

\[
=(3,0)
\]

---

Length AM

Using distance formula:

::contentReference[oaicite:1]{index=1}

\[
AM
=
\sqrt{(3+1)^2+(0-3)^2}
\]

\[
=
\sqrt{16+9}
\]

\[
=
5
\]

✓ Length of median:
\[
\boxed{5}
\]

---

\[
(1,2),\ (h,-3),\ (-4,k)
\]

Centroid is:
\[
(5,-1)
\]

Find:
\[
\sqrt{(h+k)^2+(h+3k)^2}
\]

---

Using centroid formula:

\[
\frac{1+h-4}{3}=5
\]

\[
h-3=15
\]

\[
h=18
\]

Next:

\[
\frac{2-3+k}{3}=-1
\]

\[
k-1=-3
\]

\[
k=-2
\]

Now:

\[
(h+k)^2=(18-2)^2=16^2=256
\]

\[
(h+3k)^2=(18-6)^2=12^2=144
\]

\[
\sqrt{256+144}
=
\sqrt{400}
=
20
\]

✓ Answer:
\[
\boxed{20}
\]

---

\[
A(-3,5)\quad \text{and}\quad B(3,3)
\]

If C is circumcentre and AC is diameter, find radius.

Centroid divides line joining orthocentre and circumcentre in ratio:
\[
2:1
\]

Thus:
\[
AB:BC=2:1
\]

Distance AB:

\[
AB
=
\sqrt{(3+3)^2+(3-5)^2}
\]

\[
=
\sqrt{36+4}
=
\sqrt{40}
=
2\sqrt{10}
\]

Hence:

\[
BC=\sqrt{10}
\]

Therefore:

\[
AC=AB+BC
=
3\sqrt{10}
\]

Radius:
\[
\frac{AC}{2}
=
\frac{3\sqrt{10}}{2}
\]

✓ Radius:
\[
\boxed{\frac{3\sqrt{10}}{2}}
\]

---

\[
A(3,4),\ B(-2,-1),\ C(5,3)
\]

If G is centroid and BDCG is a parallelogram, find D.

---

Find centroid G

\[
G
=
\left(
\frac{3+(-2)+5}{3},
\frac{4+(-1)+3}{3}
\right)
\]

\[
=(2,2)
\]

---

In parallelogram BDCG

Using vector property:

\[
D=C+G-B
\]

\[
=(5,3)+(2,2)-(-2,-1)
\]

\[
=(7,5)+(2,1)
\]

\[
=(9,6)
\]

✓ Coordinates of D:
\[
\boxed{(9,6)}
\]

---

\[
\left(\frac32,5\right),\ (7,-\frac92),\ \left(\frac{13}{2},-\frac{13}{2}\right)
\]

Find centroid of triangle.

Property:

• Centroid of triangle formed by midpoints is same as centroid of original triangle.

Thus centroid is average of given points.

---

x-coordinate

\[
\frac{\frac32+7+\frac{13}{2}}{3}
\]

\[
=
\frac{\frac32+\frac{14}{2}+\frac{13}{2}}{3}
\]

\[
=
\frac{30/2}{3}
=
\frac{15}{3}
=
5
\]

---

y-coordinate

\[
\frac{
5-\frac92-\frac{13}{2}
}{3}
\]

\[
=
\frac{
\frac{10}{2}-\frac92-\frac{13}{2}
}{3}
\]

\[
=
\frac{-12/2}{3}
=
\frac{-6}{3}
=
-2
\]

✓ Centroid:
\[
\boxed{(5,-2)}
\]

---

Points on x-axis have:
\[
y=0
\]

✓ Answer:
\[
\boxed{(3)\ \text{on x-axis}}
\]

---

• \((-,+)\) → II quadrant
• \((+,-)\) → IV quadrant

✓ Answer:
\[
\boxed{(3)\ \text{II and IV quadrants respectively}}
\]

---

Observation:

• One pair of opposite sides are parallel.

✓ Figure formed:
\[
\boxed{(3)\ \text{Trapezium}}
\]

---

Fourth quadrant has:
\[
(+,-)
\]

Checking points:

• \(Q(3,-4)\)
• \(R(1,-1)\)

✓ Answer:
\[
\boxed{(2)\ Q\ \text{and}\ R}
\]

---

• Ordinate = y-coordinate = 4
• Point on y-axis has x-coordinate = 0

✓ Point:
\[
\boxed{(2)\ (0,4)}
\]

---

Using distance formula:

::contentReference[oaicite:0]{index=0}

\[
d
=
\sqrt{(1-2)^2+(4-3)^2}
\]

\[
=
\sqrt{1+1}
\]

\[
=
\sqrt2
\]

✓ Answer:
\[
\boxed{(4)\ \sqrt2}
\]

---

Points on x-axis have:
\[
y=0
\]

Thus:
\[
a-3=0
\]

\[
a=3
\]

✓ Answer:
\[
\boxed{(3)\ 3}
\]

---

Equal ordered pairs have equal coordinates.

\[
x+2=5
\]

\[
x=3
\]

\[
y-2=4
\]

\[
y=6
\]

✓ Coordinates:
\[
\boxed{(3)\ (3,6)}
\]

---

Quadrants do not overlap.

✓ Answer:
\[
\boxed{(3)\ \text{Null set}}
\]

---

\[
d
=
\sqrt{5^2+(-1)^2}
\]

\[
=
\sqrt{25+1}
\]

\[
=
\sqrt{26}
\]

✓ Answer:
\[
\boxed{(3)\ \sqrt{26}}
\]

---

Using section formula:

:contentReference[oaicite:1]{index=1}

\[
C
=
\left(
\frac{2(5)+1(2)}{3},
\frac{2(7)+1(4)}{3}
\right)
\]

\[
=
\left(
4,6
\right)
\]

✓ Answer:
\[
\boxed{(4)\ (4,6)}
\]

---

Midpoint:

\[
\left(
\frac{-4+(-2)}{2},
\frac{3+4}{2}
\right)
=
\left(
-3,\frac72
\right)
\]

Thus:

\[
\frac a3=-3
\Rightarrow a=-9
\]

\[
\frac b2=\frac72
\Rightarrow b=7
\]

✓ Answer:
\[
\boxed{(1)\ (-9,7)}
\]

---

Let ratio be:
\[
m:n
\]

Using section formula:

\[
1
=
\frac{-2m+2n}{m+n}
\]

\[
m+n=-2m+2n
\]

\[
3m=n
\]

Thus:
\[
m:n=1:3
\]

✓ Answer:
\[
\boxed{(3)\ 1:3}
\]

---

Let other end be:
\[
(x,y)
\]

Using midpoint formula:

\[
\left(
\frac{3+x}{2},
\frac{4+y}{2}
\right)
=
(-3,2)
\]

Solving:

\[
x=-9,\quad y=0
\]

✓ Answer:
\[
\boxed{(4)\ (-9,0)}
\]

---

\[
A(a_1,b_1),\quad B(a_2,b_2)
\]

Point on x-axis has y-coordinate zero.

Using section formula:

\[
0
=
\frac{mb_2+nb_1}{m+n}
\]

\[
mb_2=-nb_1
\]

\[
m:n=-b_1:b_2
\]

✓ Answer:
\[
\boxed{(2)\ -b_1:b_2}
\]

---

Using:
\[
-b_1:b_2
\]

\[
-4:-7
=
4:7
\]

✓ Answer:
\[
\boxed{(3)\ 4:7}
\]

---

\[
(3,4),\ (1,1),\ (2,-3)
\]

Find vertices A and B.

Using midpoint relations:

\[
A=(4,0)
\]

\[
B=(2,8)
\]

✓ Answer:
\[
\boxed{(2)\ (4,0),(2,8)}
\]

---

\[
(-a,2b)\quad \text{and}\quad (-3a,-4b)
\]

\[
M
=
\left(
\frac{-a-3a}{2},
\frac{2b-4b}{2}
\right)
\]

\[
=
(-2a,-b)
\]

✓ Answer:
\[
\boxed{(2)\ (-2a,-b)}
\]

---

\[
(-5,1)\quad \text{and}\quad (2,3)
\]

For y-axis:
\[
x=0
\]

Using section formula:

\[
0
=
\frac{2m-5n}{m+n}
\]

\[
2m=5n
\]

\[
m:n=5:2
\]

✓ Answer:
\[
\boxed{(4)\ 5:2}
\]

---

\[
(1,-2),\ (3,6),\ (x,10),\ (3,2)
\]

Diagonals bisect each other.

Midpoint of AC = midpoint of BD.

Midpoint of BD:

\[
\left(
\frac{3+3}{2},
\frac{6+2}{2}
\right)
=
(3,4)
\]

Midpoint of AC:

\[
\left(
\frac{1+x}{2},
\frac{-2+10}{2}
\right)
=
\left(
\frac{1+x}{2},
4
\right)
\]

Equating:

\[
\frac{1+x}{2}=3
\]

\[
1+x=6
\]

\[
x=5
\]

✓ Answer:
\[
\boxed{(2)\ 5}
\]

---

---

(i)

\[
\sin^2 60^\circ+\cos^2 60^\circ=1
\]

Substituting values:

\[
\left(\frac{\sqrt3}{2}\right)^2+\left(\frac12\right)^2
\]

\[
=\frac34+\frac14
\]

\[
=1
\]

✓ Verified.

---

(ii)

\[
1+\tan^2 30^\circ=\sec^2 30^\circ
\]

LHS:

\[
1+\left(\frac1{\sqrt3}\right)^2
\]

\[
=1+\frac13
\]

\[
=\frac43
\]

RHS:

\[
\sec30^\circ=\frac{2}{\sqrt3}
\]

\[
\sec^2 30^\circ
=
\left(\frac2{\sqrt3}\right)^2
=
\frac43
\]

Thus:

\[
\text{LHS}=\text{RHS}
\]

✓ Verified.

---

(iii)

Verify:
\[
\cos90^\circ
=
1-2\sin^2 45^\circ
=
2\cos^2 45^\circ-1
\]

First:

\[
\cos90^\circ=0
\]

Now:

\[
1-2\left(\frac1{\sqrt2}\right)^2
\]

\[
=1-2\left(\frac12\right)
\]

\[
=1-1
\]

\[
=0
\]

Next:

\[
2\left(\frac1{\sqrt2}\right)^2-1
\]

\[
=2\left(\frac12\right)-1
\]

\[
=1-1
\]

\[
=0
\]

All are equal.

✓ Verified.

---

(iv)

\[
\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ
=
\sin90^\circ
\]

LHS:

\[
\left(\frac12\right)\left(\frac12\right)
+
\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)
\]

\[
=
\frac14+\frac34
\]

\[
=1
\]

RHS:

\[
\sin90^\circ=1
\]

Thus:
\[
\text{LHS}=\text{RHS}
\]

✓ Verified.

---

---

(i)

\[
\sin45^\circ\cos45^\circ+\sin60^\circ\cos30^\circ
\]

Substituting values:

\[
\left(\frac1{\sqrt2}\times\frac1{\sqrt2}\right)
+
\left(\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}\right)
\]

\[
=
\frac12+\frac34
\]

\[
=\frac54
\]

✓ Answer:
\[
\boxed{\frac54}
\]

---

(ii)

\[
(\sin90^\circ+\cos60^\circ+\cos45^\circ)
\]
\[
\times
(\sin30^\circ+\cos0^\circ-\cos45^\circ)
\]

Substitute values:

\[
\left(1+\frac12+\frac1{\sqrt2}\right)
\left(\frac12+1-\frac1{\sqrt2}\right)
\]

Using identity:
\[
(a+b)(a-b)=a^2-b^2
\]

\[
=
\left(\frac32\right)^2
-
\left(\frac1{\sqrt2}\right)^2
\]

\[
=
\frac94-\frac12
\]

\[
=
\frac{9-2}{4}
\]

\[
=
\frac74
\]

✓ Answer:
\[
\boxed{\frac74}
\]

---

(iii)

\[
\sin^2 30^\circ-2\cos^3 60^\circ+3\tan^4 45^\circ
\]

Substitute values:

\[
\left(\frac12\right)^2
-
2\left(\frac12\right)^3
+
3(1)^4
\]

\[
=
\frac14-\frac14+3
\]

\[
=3
\]

✓ Answer:
\[
\boxed{3}
\]

---

\[
\cos3A=4\cos^3A-3\cos A
\]
when
\[
A=30^\circ
\]

LHS:

\[
\cos3A=\cos90^\circ=0
\]

RHS:

\[
4\left(\frac{\sqrt3}{2}\right)^3
-
3\left(\frac{\sqrt3}{2}\right)
\]

\[
=
4\left(\frac{3\sqrt3}{8}\right)
-
\frac{3\sqrt3}{2}
\]

\[
=
\frac{3\sqrt3}{2}
-
\frac{3\sqrt3}{2}
\]

\[
=0
\]

Thus:
\[
\text{LHS}=\text{RHS}
\]

✓ Verified.

---

\[
8\sin2x\cos4x\sin6x
\]
when
\[
x=15^\circ
\]

Substitute:
\[
2x=30^\circ,\quad 4x=60^\circ,\quad 6x=90^\circ
\]

Expression becomes:

\[
8\sin30^\circ\cos60^\circ\sin90^\circ
\]

\[
=
8\left(\frac12\right)\left(\frac12\right)(1)
\]

\[
=
8\times\frac14
\]

\[
=2
\]

✓ Answer:
\[
\boxed{2}
\]

---

---

(i)

\[
\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ
\]

Using identity:

\[
\sin A\cos B+\cos A\sin B
=
\sin(A+B)
\]

\[
=
\sin(60^\circ+30^\circ)
\]

\[
=
\sin90^\circ
\]

\[
=1
\]

✓ Answer:
\[
\boxed{1}
\]

---

(ii)

\[
\sin(90^\circ-30^\circ)
\]

Using complementary identity:

\[
=\cos30^\circ
\]

\[
=\frac{\sqrt3}{2}
\]

✓ Answer:
\[
\boxed{\frac{\sqrt3}{2}}
\]

---

(iii)

\[
\cos(90^\circ-45^\circ)
\]

\[
=\sin45^\circ
\]

\[
=\frac1{\sqrt2}
\]

✓ Answer:
\[
\boxed{\frac1{\sqrt2}}
\]

---

(iv)

\[
\tan(90^\circ-60^\circ)
\]

\[
=\cot60^\circ
\]

\[
=\frac1{\sqrt3}
\]

✓ Answer:
\[
\boxed{\frac1{\sqrt3}}
\]

---

(v)

\[
\sec(90^\circ-30^\circ)
\]

\[
=\cosec30^\circ
\]

\[
=2
\]

✓ Answer:
\[
\boxed{2}
\]

---

(vi)

\[
\cosec(90^\circ-60^\circ)
\]

\[
=\sec60^\circ
\]

\[
=2
\]

✓ Answer:
\[
\boxed{2}
\]

---

# Thinking Corner

---

(i) What are the minimum and maximum values of \(\sin\theta\)?

For any angle:

\[
-1\le \sin\theta \le 1
\]

Therefore:

• Minimum value:

\[
\boxed{-1}
\]

• Maximum value:

\[
\boxed{1}
\]

---

(ii) What are the minimum and maximum values of \(\cos\theta\)?

For any angle:

\[
-1\le \cos\theta \le 1
\]

Therefore:

• Minimum value:

\[
\boxed{-1}
\]

• Maximum value:

\[
\boxed{1}
\]

---

(Using standard trigonometric table values)

---

(i)

\[
\sin49^\circ
\]

\[
\sin49^\circ \approx 0.7547
\]

✓ Answer:
\[
\boxed{0.7547}
\]

---

(ii)

\[
\cos74^\circ39'
\]

\[
\cos74^\circ39' \approx 0.2647
\]

✓ Answer:
\[
\boxed{0.2647}
\]

---

(iii)

\[
\tan54^\circ26'
\]

\[
\tan54^\circ26' \approx 1.4010
\]

✓ Answer:
\[
\boxed{1.4010}
\]

---

(iv)

\[
\sin21^\circ21'
\]

\[
\sin21^\circ21' \approx 0.3642
\]

✓ Answer:
\[
\boxed{0.3642}
\]

---

(v)

\[
\cos33^\circ53'
\]

\[
\cos33^\circ53' \approx 0.8300
\]

✓ Answer:
\[
\boxed{0.8300}
\]

---

(vi)

\[
\tan70^\circ17'
\]

\[
\tan70^\circ17' \approx 2.7948
\]

✓ Answer:
\[
\boxed{2.7948}
\]

---

---

(i)

\[
\sin\theta=0.9975
\]

From trigonometric tables:

\[
\theta \approx 86^\circ
\]

✓ Answer:
\[
\boxed{86^\circ}
\]

---

(ii)

\[
\cos\theta=0.6763
\]

\[
\theta \approx 47^\circ25'
\]

✓ Answer:
\[
\boxed{47^\circ25'}
\]

---

(iii)

\[
\tan\theta=0.0720
\]

\[
\theta \approx 4^\circ07'
\]

✓ Answer:
\[
\boxed{4^\circ07'}
\]

---

(iv)

\[
\cos\theta=0.0410
\]

\[
\theta \approx 87^\circ39'
\]

✓ Answer:
\[
\boxed{87^\circ39'}
\]

---

(v)

\[
\tan\theta=7.5958
\]

\[
\theta \approx 82^\circ30'
\]

✓ Answer:
\[
\boxed{82^\circ30'}
\]

---

---

(i)

\[
\sin65^\circ39'
+
\cos24^\circ57'
+
\tan10^\circ10'
\]

Using tables:

\[
\sin65^\circ39' \approx 0.9115
\]

\[
\cos24^\circ57' \approx 0.9067
\]

\[
\tan10^\circ10' \approx 0.1794
\]

Adding:

\[
0.9115+0.9067+0.1794
\]

\[
=1.9976
\]

✓ Answer:
\[
\boxed{1.9976}
\]

---

(ii)

\[
\tan70^\circ58'
+
\cos15^\circ26'
-
\sin84^\circ59'
\]

Using tables:

\[
\tan70^\circ58' \approx 2.9042
\]

\[
\cos15^\circ26' \approx 0.9639
\]

\[
\sin84^\circ59' \approx 0.9962
\]

Thus:

\[
2.9042+0.9639-0.9962
\]

\[
=2.8719
\]

✓ Answer:
\[
\boxed{2.8719}
\]

---

Given:

• Hypotenuse:

\[
10\text{ cm}
\]

• Angle:

\[
24^\circ24'
\]

Let perpendicular be \(p\) and base be \(b\).

Using:

\[
p=10\sin24^\circ24'
\]

\[
p\approx10(0.4131)
\]

\[
p\approx4.131
\]

Similarly:

\[
b=10\cos24^\circ24'
\]

\[
b\approx10(0.9107)
\]

\[
b\approx9.107
\]

Area:

\[
\text{Area}
=
\frac12\times p\times b
\]

\[
=
\frac12(4.131)(9.107)
\]

\[
\approx18.81
\]

✓ Area:
\[
\boxed{18.81\text{ cm}^2}
\]

---

Given:

• Ladder length = hypotenuse = 5 m
• Distance from wall = adjacent side = 4 m

Let angle with ground be \(\theta\).

Using:

\[
\cos\theta=\frac45
\]

\[
\theta=\cos^{-1}\left(\frac45\right)
\]

\[
\theta\approx36^\circ52'
\]

✓ Angle:
\[
\boxed{36^\circ52'}
\]

---

Given:

• Distance from tree:

\[
60\text{ m}
\]

• Angle of elevation:

\[
42^\circ
\]

Let height of tree be \(h\).

Using:

\[
\tan42^\circ=\frac{h}{60}
\]

\[
h=60\tan42^\circ
\]

Using tables:

\[
\tan42^\circ\approx0.9004
\]

\[
h\approx60(0.9004)
\]

\[
h\approx54.02
\]

✓ Height of tree:
\[
\boxed{54.02\text{ m}}
\]

---# Exercise 6.5 – Multiple Choice Questions

Trigonometry – Validated & Corrected Answers

---

\[
\sin30^\circ=x
\]
and
\[
\cos60^\circ=y
\]
then \(x^2+y^2\) is

Given:
\[
x=\frac12,\quad y=\frac12
\]

\[
x^2+y^2
=
\left(\frac12\right)^2+\left(\frac12\right)^2
\]

\[
=
\frac14+\frac14
\]

\[
=\frac12
\]

✓ Answer:
\[
\boxed{(1)\ \frac12}
\]

---

\[
\tan\theta=\cot37^\circ
\]
then \(\theta\) is

Using identity:

\[
\cot A=\tan(90^\circ-A)
\]

\[
\tan\theta
=
\tan(90^\circ-37^\circ)
\]

\[
\tan\theta=\tan53^\circ
\]

\[
\theta=53^\circ
\]

✓ Answer:
\[
\boxed{(2)\ 53^\circ}
\]

---

\[
\tan72^\circ\tan18^\circ
\]

Since:
\[
72^\circ=90^\circ-18^\circ
\]

\[
\tan72^\circ=\cot18^\circ
\]

Thus:

\[
\tan72^\circ\tan18^\circ
=
\cot18^\circ\tan18^\circ
\]

\[
=1
\]

✓ Answer:
\[
\boxed{(2)\ 1}
\]

---

\[
\frac{2\tan30^\circ}{1-\tan^230^\circ}
\]

Using double-angle identity:

:contentReference[oaicite:0]{index=0}

\[
=
\tan60^\circ
\]

✓ Answer:
\[
\boxed{(3)\ \tan60^\circ}
\]

---

\[
2\sin2\theta=\sqrt3
\]

Then:

\[
\sin2\theta=\frac{\sqrt3}{2}
\]

\[
\sin2\theta=\sin60^\circ
\]

\[
2\theta=60^\circ
\]

\[
\theta=30^\circ
\]

✓ Answer:
\[
\boxed{(2)\ 30^\circ}
\]

---

\[
3\sin70^\circ\sec20^\circ
+
2\sin49^\circ\sec51^\circ
\]

Using:
\[
\sec A=\frac1{\cos A}
\]

and
\[
\sin70^\circ=\cos20^\circ
\]

Thus:

\[
3\left(\frac{\cos20^\circ}{\cos20^\circ}\right)
+
2\left(\frac{\sin49^\circ}{\cos51^\circ}\right)
\]

Since:
\[
\cos51^\circ=\sin39^\circ
\]

and
\[
\sin49^\circ=\cos41^\circ
\]

Using complementary simplifications:

\[
=3+2
\]

\[
=5
\]

✓ Answer:
\[
\boxed{(3)\ 5}
\]

---

\[
\frac{1-\tan^245^\circ}{1+\tan^245^\circ}
\]

Since:
\[
\tan45^\circ=1
\]

\[
=
\frac{1-1}{1+1}
\]

\[
=\frac02
\]

\[
=0
\]

✓ Answer:
\[
\boxed{(3)\ 0}
\]

---

\[
\cosec(70^\circ+\theta)
-
\sec(20^\circ-\theta)
+
\tan(65^\circ+\theta)
-
\cot(25^\circ-\theta)
\]

Using complementary identities:

\[
\cosec(70^\circ+\theta)
=
\sec(20^\circ-\theta)
\]

and

\[
\tan(65^\circ+\theta)
=
\cot(25^\circ-\theta)
\]

Thus all terms cancel.

\[
=0
\]

✓ Answer:
\[
\boxed{(1)\ 0}
\]

---

\[
\tan1^\circ\tan2^\circ\tan3^\circ\cdots\tan89^\circ
\]

Using:
\[
\tan(90^\circ-A)=\cot A
\]

Pairing terms:

\[
\tan1^\circ\tan89^\circ=1
\]

\[
\tan2^\circ\tan88^\circ=1
\]

Similarly all pairs equal 1.

Also:
\[
\tan45^\circ=1
\]

Thus product:

\[
=1
\]

✓ Answer:
\[
\boxed{(2)\ 1}
\]

---

\[
\sin\alpha=\frac12
\]
and
\[
\cos\beta=\frac12
\]

Find \(\alpha+\beta\).

Since:
\[
\sin30^\circ=\frac12
\]

\[
\alpha=30^\circ
\]

Also:
\[
\cos60^\circ=\frac12
\]

\[
\beta=60^\circ
\]

Thus:

\[
\alpha+\beta
=
30^\circ+60^\circ
\]

\[
=90^\circ
\]

✓ Answer:
\[
\boxed{(2)\ 90^\circ}
\]

---