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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