| Book Name | : Ganit Prakash |
| Subject | : Mathematics (Maths) |
| Class | : 9 (Madhyamik/WB) |
| Publisher | : Prof. Nabanita Chatterjee |
| Chapter Name | : Internal And External Division Of Straight Line Segment (20th Chapter) |
Let us work out – 20
1. Find the area of triangular region with vertices given below
(a) (2, -2), (4, 2) and (-1, 3)
(b) (8, 9) (2, 6) and (9, 2)
(c) (1, 2) (3, 0) and origin
Solution: (i)
Let (x_1, y_1)=(2,-2) \\
(x_2, y_2)=(4,2) \\
(x_3, y_3)=(-1,3)
Area of triangular region with these vertices
=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \\
=\frac{1}{2}|2(2-3)+4(3+2)+(-1)(-2-2)| \\
=\frac{1}{2}|2 \times(-1)+4 \times 5+(-1) \times(-4)| \\
\frac{1}{2}|-2+20+4| \\
=\frac{1}{2}|24-2| \\
=\frac{1}{2} \times 22 \text { sq.unit } \\
=11 \text { sq unit (Ans) }
(ii) Let (x_1, y_1)=(8,9) \\
(x_2, y_2)=(2,6) \\
(x_3, y_3)=(9,2)
Area of traingular region with these vertices
=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \\
=\frac{1}{2}|8(6-2)+2(2-9)+9(9-6)| \\
=\frac{1}{2}|8 \times 4+2 \times(-7)+9 \times 3| \\
=\frac{1}{2}|32-14+27| \\
=\frac{1}{2}|59-14| \\
=\frac{1}{2}|45| \\
=\frac{45}{2} \text { sq.unit } \\
=22.5 \text { Sq.units (Ans) }
(iii) \text { Let }(x_1, y_1)=(1,2) \\
(x_2, y_2)=(3,0) \\
(x_3, y_3)=(0,0)
Area of triangular region with these vertices
=\frac{1}{2} \mid x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y(-y_2) \mid. \\
=\frac{1}{2}|1(0-0)+3(0-2)+0(2-0)| \\
=\frac{1}{2}|1.0+3 \cdot(-2)+0 \cdot 2| \\
=\frac{1}{2}|0-6+0| \\
=\frac{1}{2}|-6| \\
=\frac{1}{2} \times 6 \\
=3 \text { Sq.units, (Ans) }
2.Prove that the points (3, -2),(-5, 4) and (-1, 1) are collinear.
Solution: If we find the area with these vertices is zero, then these vertices are collinear.
\therefore \text { Let }(x_1, y_1)=(3,-2) \\
(x_2, y_2)=(-5,4) \\
(x_3, y_3)=(-1,1)
Area of traingular region with these vertices
=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \\
=\frac{1}{2}|3(4-1)-5(1+2)-1(-2-4)| \\
=\frac{1}{2}|3 \cdot 3-5 \cdot 3-1 \cdot(-6)| \\
=\frac{1}{2}|9-15+6| \\
=\frac{1}{2}|15-15| \\
= 0
\therefore Area =0
Hence the given vertices are collinear. (proved)
3. Let us write by calculating at the value of k, the points (1, -1),(2, -1) and (k,-12) lie on same straight line.
Solution:
Let A=(1,-1), B=(2,-1) \quad C=(k,-1)
\because A, B & C lie on same straight line
\therefore Area =0
or, \frac{1}{2}|1(-1+1)+2(-1+1)+k(-1+1)|=0
or, \frac{1}{2}|0+0+0|=0
or, 0 = 0
\therefore Hence k does not exist, so k is any real number. The any real value of k, the given vertices lie on the same straight line.
4. Let us prove that the line joining two points (1, 2) and (-2, -4) passes through origin.
Solution:
If the line joining two point (1, 2) and (-2, -4) passes through origin, then area = 0
So first we find the area
Let (x_1, y_1)=(1,2)
(x_2, y_2)=(-2,-4) \\
(x_3, y_3)=(0,0)
Area of traingular region with these vertices
=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \\
=\frac{1}{2}|1(-4-0)-2(0-2)+0(2+4)| \\
=\frac{1}{2}|-4+4+0|=0 \\
\therefore Area = 0
Hence the line joining two po int s(1, 2) and (-2, -4) passes through origin.( proved)
5. Let us prove that the mid-point of the line segment joining two (2,1) and(6,5) lie on the line joining two points(-4, -5) and (9, 8)
Solution:
Midpoint of the line segment joining two point (2, 1) and ( 6, 5)
=(\frac{2+6}{2}, \frac{1+5}{2}) \\
=(\frac{8}{2}, \frac{6}{2}) \\
= (4, 3)
Let (x_1, y_1)=(4,3)
(x_2, y_2)=(-4,-5) \\
(x_3, y_3)=(9,8)
Area of traingular region with these vertices
=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \\
=\frac{1}{2}|4(-5-8)-4(8-3)+9(3+5)| \\
=\frac{1}{2}|4 \times(-13)-4 \times 5+9 \times 8| \\
=\frac{1}{2}|-52-20+72| \\
=\frac{1}{2}|72-72| \\
= 0
\therefore Area = 0
Hence the line joining two points (-4, -5) and (6, 5) passes though(4, 3)
6. Let us find the area of quadrilateral region formed by the line joining, four given points each
(i) (1, 1), (3, 4), (5, -2), (4, -7),  (ii) (1, 4), (-2, 1), (2, -3), (3, 3)
Solution:
Let (x_1, y_1) =(1,1) \\
(x_2, y_2) =(3,4) \\
(x_3, y_3) =(5,-2) \\
(x_4, y_4) =(4,-7)
Area of quadrilateral region with these vetioes
=\frac{1}{2}|(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1)-(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1)| \\
=\frac{1}{2} \mid\{1 \times 4+3 \times(-2)+5 \times(-7)+4 \times 1\}-\{1 \times 3+4 \times 5+(-2) \times 4+(-7) \times 1 \mid \\
=\frac{1}{2}|(4-6-35+4)-(3+20-8-7)| \\
=\frac{1}{2}|(8-41)-(23-15)| \\
=\frac{1}{2}|-33-8| \\
=\frac{1}{2}|-41| \\
=\frac{41}{2} \text { Sq.unis } \\
=20 \frac{1}{2} \text { Sq.unis } \\
\therefore \text { Area }=20-\frac{1}{2} \text { Sq.units (Ans) }
(ii) Let (x_1, y_1) =(1,4) \\
(x_2, y_2) =(-2,1) \\
(x_3, y_3) =(2,-3) \\
(x_4, y_4) =(3,3)
Area of quadilateral region with these vartices
=\frac{1}{2}|(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1)-(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1)| \\
=\frac{1}{2} \mid\{1 \times 1+(-2) \times(-3)+2 \times 3+3 \times 4)-\{(4) \times(-2)+1 \times 2+(-3) \times 3+3 \times 1 \mid \\
=\frac{1}{2}|(1+6+6+12)-(-8+2-9+3)|
=\frac{1}{2}|25-(5-17)| \\
=\frac{1}{2}|25-(-12)| \\
=\frac{1}{2}|25+12| \\
=\frac{1}{2} \times 37=\frac{37}{2}=18 \frac{1}{2} \text { Sq.units } \\
\text { Area }=18 \frac{1}{2} \text { Sq.units (Ans) }
7. The co-ordinate of the three points A, B, C are (3, 4)(-4, 3) and (8, -6) respectively. Let us find the area of triangle and the perpendicular length drawn from the point A on BC.
Solution:
Given:
Let (x_1, y_1)=(3,4) \\
(x_2, y_2)=(-4,3) \\
(x_3, y_3)=(8,-6)
Area of triangular region with these vertices
=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \\
=\frac{1}{2}|3(3+6)-4(-6-4)+8(4-3)| \\
=\frac{1}{2}|3 \times 9-4(-10)+8 \times 1| \\
=\frac{1}{2}|27+40+8| \\
=\frac{1}{2} \times 75 \\
=\frac{75}{2}=37 \frac{1}{2} \text { sq.units }
Let the length of perpendicular be ‘h’
\therefore \text{B C} =\sqrt{(8+4)^2+(-6-3)^2} \\
=\sqrt{(12)^2+(-9)^2}=\sqrt{144+81}=\sqrt{225}=15 \\
\therefore \text { Area } =\frac{75}{2} \text { sq.units } \\
\text { or, } \frac{1}{2} \times B C \times h=\frac{75}{2} \\
\text { or }, \frac{1}{2} \times 15 \times h=\frac{75}{2} \\
\therefore h = 5
\therefore Length of perpendicular = 5 units ( Ans )
8. In triangle ABC, Co-ordinate of A is (2, 5)and the centroid of triangle is (-2, 1), let us find the coordinate of midpoint of BC.
Solution:
let A = (2, 5)
B =(x_1, y_1) \\
C =(x_2, y_2)
The coordinate of the centroid =(\frac{x_1+x_2+2}{3}, \frac{y_1+y_2+5}{3})
\therefore \frac{x_1+x_2+2}{3}=-2 \quad \frac{y_1+y_2+5}{3}=1 \\
\text { or, } x_1+x_2+2=-6 \quad \text { or, } y_1+y_2+5=3 \\
\text { or, } x_1+x_2=-8 \quad \text { or, } y_1+y_3=-2 \\
\therefore C \text {-ordinate of mid point of } B C=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \\
=(-\frac{8}{2},-\frac{2}{2}) \\
=(-4,-1) \text { (Ans) }
\therefore Co-ordinate of mid point of BC =(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})
=(-\frac{8}{2},-\frac{2}{2})
= (-4, -1)( Ans )
9. The co-ordinate of vertices of a triangle are (4, -3),(-5, 2) and (x, y); let us find the value of x and y, if the centroid of the traingle is at the origin
Solution:
The Co-ordinates of the centroid of the triangle having vertices (4, -3) (-5, 2) and (x, y)
=(\frac{4-5+x}{3}, \frac{-3+2+y}{3}) \\
=(\frac{x-1}{3}, \frac{y-1}{3})
\because The centroid of the triangle is at the origin.
\therefore \frac{x-1}{3}=0 \quad \frac{y-1}{3}=0 \\
or, x-1=0 \quad \text { or, } y-1=0 \\
\therefore x=1 \quad y=1 \\
\therefore(x, y)=(1,1) \quad(\text { Ans }).
10. The vertices of \triangle ABC are A(-1, 5), B(3, 1) and C(5, 7), D, E, F are the mid point of BC, CA and AB recectively. Let us find the area of triangular region \triangle DEF and prove that \triangle ABC=4, DCF
Solution:
D is mid point of BC.
\therefore The Co-ordinate of D = (\frac{3+5}{2}, \frac{1+7}{2})
=(\frac{8}{2}, \frac{8}{2}) \\
= (4, 4)
E is mid point of AC
\therefore The co-ordinate E=(\frac{5-1}{2}, \frac{7+5}{2})
=(\frac{4}{2}, \frac{12}{2}) \\
= (2, 6)
F is mid point of AB
\therefore \text { The co-ordinate } F =(\frac{3-1}{2}, \frac{1+5}{2}) \\
=(\frac{2}{2}, \frac{6}{2}) \\
= (1, 3)
\therefore Area of \triangle ABC
=\frac{1}{2}|-1(1-7)+3(7-5)+5(5-1)| \\
=\frac{1}{2}|(-1)(-6)+3 \times 2+5 \times 4| \\
=\frac{1}{2}|6+6+20| \\
=\frac{1}{2} \times 32 \\
=16 \text { sq.units }
Area of \triangle DEF
=\frac{1}{2}|4(6-3)+2(3-4)+1(4-6)| \\
=\frac{1}{2}|4 \times 3+2 \times(-1)+1 \times(-2)| \\
=\frac{1}{2}|12-2-2| \\
=\frac{1}{2}|12-4| \\
=\frac{1}{2} \times 8 \\
=4 \text { Sq.units }
We see that,\triangle A B C=4 \triangle D E F(\text { proved })
11. M.C.Q:
(i) The area of the triangular region by three points (0,4) (0,0) and (-6,0) is
(a) 24sq.units
(b) 12sq.units
(c) 6sq.units
(d) 8sq.units
Solution:
\text { Area }=\frac{1}{2} \times \text { base } \times \text { height } \\
=\frac{1}{2} \times 4 \times 6 \\
=12 \text { sq.units } \\
(b) is correct answer.
(ii) The co-ordinate of centroid of a triangle formed by the three points (7,-5),(-2,5) and (4,6) is
(a) (3,-2)
(b) (2,3)
(c) (3,2)
(d) (2,-3)
Solution:
Co-ordinate of centroid of a triangle
= (\frac{7-2+4}{3}, \frac{-5+5+6}{3}) \\
=(\frac{9}{3}, \frac{6}{3}) \\
= (3, 2)
\therefore (c) is correct answer
(iii) ABC is a right-angled triangle of which \angle \mathrm{ABC}=90^{\circ}, co-ordinate od A and C are (0,4) and (3,0) respectively then the area of triangle ABC is
(a) 12sq.units
(b) 6sq.units
(c) 24sq.units
(d) 8sq.units
Solution:
\text { Area of } \triangle A B C =\frac{1}{2} \times \text { Base } \times \text { Height } \\
=\frac{1}{2} \times 4 \times 3 \text { sq.units } \\
=6 \text { squnits } \quad(\text { b }) \text { is correct answer }(b) is correct answer
(iv) If (0,0), (4,-3) ard (x, y) are collinear then
(a) x = 8, y = 6
(b) x = 8, y = 6
(c) x = 4, y = 6
(d) x = 8, y = 6
Solution:
\because(0,0),(4,-3) \text { and }(x, y) \text { are collinear }
\therefore Area =0 \text { or, } \frac{1}{2}|0(-3-y)+4(y-0)+x(0+3)|=0
or, \frac{1}{2}|4 y+3 x|=0
or, |3 x+4 y|=0
or, 3x + 4y = 0
Since option (a) satisfies this equation
so (a) is correct answer
(v) If in triangle ABC, the coordinate of vertex A is (7,-4) and centroid of triangle is (1,2), then the Co-ordinate of midpoint of BC is
(a) (-2,-5)
(b) (-2,5)
(c) (2,-5)
(d) (5,-2)
Solution:
Let A = (7, -4)
G = (1, 2)
B=(x_{1}, y_{1}) \\
C=(x_{2}, y_{2})
The coordinate of centroid = (\frac{x_{1}+x_{2}+7}{2}, \frac{y_{1}+y_{2}-4}{3})
\because The centroid is (1, 2)
\therefore \frac{x_{1}+x_{2}+7}{3}=1, \quad \frac{y_{1}+y_{2}-4}{3}=2 \\
\text { or, } x_{1}+x_{2}+7=3, \quad \text { or, } y_{1}+y_{2}-4=6 \\
\therefore x_{1}+x_{2}=3-7 , \quad y_{1}+y_{2}=6+4 \\
= -4, \quad = 10
\therefore The Co-ordinate of the midpoint of BC
=(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) \\
=(\frac{-4}{2}, \frac{10}{2}) \\
= (-2, 5)
\therefore(b) is correct answer
12. Short answer type question:
(i) The coordinate of mid point of the sides of a triangle ABC are (0,1),(1,1) and (1,0); let us find the Co-ordinate of its centroid
Solution: Co-ordinate of the centroid
=(\frac{0+1+1}{3}, \frac{1+1+0}{3}) \\
=(\frac{2}{3}, \frac{2}{3})(\text { Ans })
(ii) The coordinate of the centroid of triangle is (6,9) and two vertices are (15,0) and (0,10) ;let us find the co ordinate of the third vertex.
Solution:
Let third vertex be (x, y)
\text { centroid of the triangle } = (\frac{15+0+x}{3}, \frac{0+10+y}{3}) \\
=(\frac{15+x}{3}, \frac{10+y}{3})
\therefore \frac{15+x}{3}=6, \quad \frac{10+y}{3}=9 \\
\text { or, } 15+x=18, \quad \text { or }, 10+y=27 \\
\text { or, } x=18-15, quad \text { or, } y=27-10 \\
\text { or, } x=3, \quad \text { or, } y=17
\therefore Third vertex of the triangle =(3, 17) (Ans)
(iii) If the three points (a,0), (0,b) and (1,1) are collinear then let be us show that \frac{1}{a}+\frac{1}{b}=1
Solution:
\because The three points (a, 0), (0, b) and (1,1) are collinear \therefore Area = 0
or, |a(b-1)+0(1-0)+1(0-b)|=0
or, |a b-a+0-b|=0
or, a b-a-b=0
or, a+b=a b
or, \frac{a}{a b}+\frac{b}{a b}=1
\therefore \quad \frac{1}{a}+\frac{1}{b}=1 ( proved)
(iv) Let us calculate the area of triangular region formed by the three points (1,4), (-1,2) and (-4,1)
Solution:
\text { Area } =\frac{1}{2}|1(2-1)-1(1-4)-4(4-2)| \\
=\frac{1}{2}|1 \times 1-1 \times(-3)-4 \times 2| \\
=\frac{1}{2}|1+3-8| \\
=\frac{1}{2}|4-8| \\
=\frac{1}{2}|-4| \\
=\frac{1}{2} \times 4=2 \text { sq.units (Ans) }
(v) Let us write the Co-ordinate of centroid triangle formed by the three points (x-y, (y-z), (-x,-y) and (y, z).
Solution:
The centroid of the triangle
= (\frac{x-y-x+y}{3}, \frac{y-z-y+z}{3}) \\
= (\frac{0}{3}, \frac{0}{3}) \\
=(0,0) \text { (Ans) }
