Validated & Corrected Answers
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:
\[
y=-1
\]
Only the horizontal pair of opposite sides is parallel. Their lengths are different:
\[
5\text{ units and }11\text{ units}
\]
So the figure is not a parallelogram.
✓ Shape formed:
\[
\boxed{\text{Trapezium}}
\]
# 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.
\[
\boxed{AD = DA}
\]
Validated & Corrected Answers
# Distance Formula
For two points:
\[
(x_1,y_1)\quad \text{and}\quad (x_2,y_2)
\]
Distance between them:
::contentReference[oaicite:0]{index=0}
(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:
\[
-16x+4y-36=0
\]
\[
4y=16x+36
\]
\[
y=4x+9
\]
✓ Relation:
\[
\boxed{y=4x+9}
\]
\[
AP=\frac37 AB
\]
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:
\[
\boxed{(11,2)\ \text{is the centre}}
\]
Intersections with axes: (30,0), (−30,0), (0,30), (0,−30).
Distance between opposite points, e.g. (30,0) and (−30,0) = 60 units.
(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 \(p\):
\[
\boxed{-15}
\]
Let D(2,4), E(−2,3), F(5,2). Vertices: A = E+F−D = (−2+5−2, 3+2−4) = (1,1). B = F+D−E = (5+2+2, 2+4−3) = (9,3). C = D+E−F = (2−2−5, 4+3−2) = (−5,5).
Answer: (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
Validated & Corrected Answers
# Section Formula
If a point \(P(x,y)\) divides the line joining
\[
A(x_1,y_1)\quad \text{and}\quad B(x_2,y_2)
\]
in the ratio \(m:n\),
then:
:contentReference[oaicite:0]{index=0}
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:
\[
\boxed{A,B,C\ \text{are collinear}}
\]
AB vector = B−A = (4,4). BC = (1/4)AB = (1,1). So C = B + BC = (2+1, 1+1) = (3,2).
Answer: (3,2).
(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:
\[
\boxed{\left(-\frac83,-\frac{11}{3}\right)}
\]
Let third vertex be (x,y). ( (3+5+x)/3, (−2+2+y)/3 ) = (4,−2).
From x: (8+x)/3 = 4 ⇒ x = 4. From y: y/3 = −2 ⇒ y = −6. Third vertex = (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:
\[
\boxed{5}
\]
From centroid formulas: (1+h−4)/3 = 5 ⇒ h = 18. And (2−3+k)/3 = −1 ⇒ k = −2. Then (h+k) = 16 and (h+3k) = 12. So √(16^2+12^2) = √(256+144) = √400 = 20.
Answer: 20
Solution.
On the Euler line the centroid G divides the line segment joining the orthocentre H and the circumcentre O in the ratio HG:GO = 2:1. Thus, if A is the orthocentre, B is the centroid and C is the circumcentre, then AB:BC = 2:1.
Distance AB = √[(3 − (−3))² + (3 − 5)²] = √[6² + (−2)²] = √(36+4) = √40 = 2√10.
So BC = AB/2 = (2√10)/2 = √10.
Then AC = AB + BC = 2√10 + √10 = 3√10. Since AC is a diameter, radius = AC/2 = (3√10)/2.
Answer: radius = 3√10 / 2.
Centroid G = ((3−2+5)/3, (4−1+3)/3) = (2,2). For parallelogram BDCG, B+C = D+G ⇒ D = B+C−G = (−2,−1)+(5,3)−(2,2) = (1,0).
Answer: (1,0).
Centroid = average of the three mid-points: x = (3/2 + 7 + 13/2)/3 = 5, y = (5 − 9/2 − 13/2)/3 = −2.
Answer: (5, −2).
Validated & Corrected Answers
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 points have coordinates (+, −). Both Q(3,−4) and R(1,−1) have positive x and negative y, so both lie in the fourth quadrant.
Answer: Q 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:
\[
\boxed{(3)\ 4:7}
\]
If mid-points are M1(3,4), M2(1,1), M3(2,−3), vertices are obtained by sums: (M1+M2−M3)=(3+1−2,4+1−(−3))=(2,8); (M1+M3−M2)=(3+2−1,4−3−1)=(4,0); (M2+M3−M1)=(1+2−3,1−3−4)=(0,−6). Thus vertices: (0,−6), (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:
\[
\boxed{(4)\ 5:2}
\]
Diagonals bisect each other. Midpoint of BD = ((3+3)/2, (6+2)/2) = (3,4). Midpoint of AC = ((1+x)/2, (−2+10)/2) = ((1+x)/2, 4). Equate x: (1+x)/2 = 3 ⇒ 1+x = 6 ⇒ x = 5.
Answer: 5.
Validated & Corrected Answers
# Special Trigonometric Values
| Angle | sin | cos | tan |
|---|---|---|---|
| \(0^\circ\) | 0 | 1 | 0 |
| \(30^\circ\) | \(\frac12\) | \(\frac{\sqrt3}{2}\) | \(\frac1{\sqrt3}\) |
| \(45^\circ\) | \(\frac1{\sqrt2}\) | \(\frac1{\sqrt2}\) | 1 |
| \(60^\circ\) | \(\frac{\sqrt3}{2}\) | \(\frac12\) | \(\sqrt3\) |
| \(90^\circ\) | 1 | 0 | Not defined |
(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·cos45 + sin60·cos30 = (1/2)+(3/4) = 5/4.
(ii) (1+1/2+1/√2)(1/2+1−1/√2) = (3/2)^2 − (1/√2)^2 = 9/4 − 1/2 = 7/4.
(iii) (1/4) − 2(1/8) + 3(1) = 1/4 − 1/4 + 3 = 3.
For A = 30°, LHS = cos 90° = 0. RHS = 4(√3/2)^3 − 3(√3/2) = (3√3/2) − (3√3/2) = 0. Hence LHS = 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:
\[
\boxed{2}
\]
Validated & Corrected Answers
# Complementary Angle Identities
For complementary angles:
\[
\sin(90^\circ-\theta)=\cos\theta
\]
\[
\cos(90^\circ-\theta)=\sin\theta
\]
\[
\tan(90^\circ-\theta)=\cot\theta
\]
\[
\cot(90^\circ-\theta)=\tan\theta
\]
\[
\sec(90^\circ-\theta)=\cosec\theta
\]
\[
\cosec(90^\circ-\theta)=\sec\theta
\]
(i) = sin(60+30) = sin90 = 1.
(ii) = cos30 = √3/2.
(iii) = sin45 = 1/√2.
(iv) = cot60 = 1/√3.
(v) = cosec30 = 2.
(vi) = sec60 = 2.
Validated & Corrected Answers
(i) sin49° ≈ 0.7547.
(ii) cos74°39' ≈ 0.2647.
(iii) tan54°26' ≈ 1.4010.
(iv) sin21°21' ≈ 0.3642.
(v) cos33°53' ≈ 0.8300.
(vi) tan70°17' ≈ 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:
\[
\boxed{82^\circ30'}
\]
Using trigonometric tables (values rounded to 4 decimal places):
(i) sin 65°39' ≈ 0.9115, cos 24°57' ≈ 0.9067, tan 10°10' ≈ 0.1794
Sum = 0.9115 + 0.9067 + 0.1794 = 1.9976
Answer: 1.9976
(ii) tan 70°58' ≈ 2.9042, cos 15°26' ≈ 0.9639, sin 84°59' ≈ 0.9962
Value = 2.9042 + 0.9639 − 0.9962 = 2.8719
Answer: 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
sin 30° = 1/2, cos 60° = 1/2
x^2 + y^2 = (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2
Answer: 1/2
Use cot A = tan(90° − A).
tan θ = cot 37° = tan(90° − 37°) = tan 53° ⇒ θ = 53°
Answer: 53°
Note 72° = 90° − 18°, so tan 72° = cot 18°.
tan 72° · tan 18° = cot 18° · tan 18° = 1
Answer: 1
Use the double-angle identity: (2 tan A)/(1 − tan^2 A) = tan 2A.
Here A = 30°, so value = tan 60°.
Answer: tan 60°
sin 2θ = √3 / 2 = sin 60°
So 2θ = 60° ⇒ θ = 30°
Answer: 30°
\[
3\sin70^\circ\sec20^\circ
+
2\sin49^\circ\sec51^\circ
\]
Using:
\[
\sec A=\frac1{\cos A}
\]
and
\[
\sin70^\circ=\cos20^\circ
\]
Thus:
\[
3\left(\frac{\cos20^\circ}{\cos20^\circ}\right)
+
2\left(\frac{\sin49^\circ}{\cos51^\circ}\right)
\]
The first term becomes:
\[
3\sin70^\circ\sec20^\circ=3
\]
But \(\sin49^\circ\) and \(\cos51^\circ\) are not complementary equal values, so the second term does not simplify to \(2\).
\[
2\sin49^\circ\sec51^\circ
=
2\left(\frac{\sin49^\circ}{\cos51^\circ}\right)
\]
\[
\approx 2\left(\frac{0.7547}{0.6293}\right)
=
2.398
\]
Therefore:
\[
3\sin70^\circ\sec20^\circ+2\sin49^\circ\sec51^\circ
\approx
5.398
\]
✓ Answer:
\[
\boxed{\approx 5.398}
\]
tan 45° = 1
Expression = (1 − 1)/(1 + 1) = 0/2 = 0
Answer: 0
Use complementary identities: cosec(70° + θ) = cosec[90° − (20° − θ)] = sec(20° − θ), and tan(65° + θ) = cot(25° − θ).
Thus terms cancel pairwise, giving 0.
Answer: 0
Pair tan A with tan(90° − A): tan A · tan(90° − A) = tan A · cot A = 1. All pairs give 1, and tan 45° = 1 as the middle term.
Product = 1
Answer: 1
sin α = 1/2 ⇒ α = 30° (principal value). cos β = 1/2 ⇒ β = 60° (principal value).
α + β = 30° + 60° = 90°
Answer: 90°
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