Question 16
In what ratio does the x-axis divide the line segment joining the points ( 4, 6) and ( 1, 7)? Find the co-ordinates of the point of division.
Let A( 4, 6) and B ( 1, 7)
Let P be the point on x-axis which divides AB
Since Point P is on x axis, Hence its y coordinate is 0.
So, it is of the form P(x, 0)
Now, we have to find ratio
Let ratio be k : 1
Hence,
m1 = k , m2 = 1
x1 = 4 , y1 = 6
x2 = 1 , y2 = 7
Also,
x = x , y = 0
Using section formula
y = ( _1 _2 + _2 _1)/( _1 + _2 )
0 = ( 7 + 1 ( 6))/( + 1)
0 = (7 6)/( +1)
0(k + 1)= 7k 6
0 = 7k 6
7k 6 = 0
7k = 6
k = 6/7
Hence,
Ratio = k : 1
Ratio = 6/7 : 1 i.e. 6 : 7
Now, we need to find x also
x = ( 1 2 + 2 1)/( 1 + 2)
= (6 ( 1) + 7 ( 4))/(6 + 7)
= ( 6 28)/13
= ( 34)/13
Hence,
the coordinate of point is P = (x, 0) = (( )/ , )
Question 16
The points A(4, 2), B(7, 2), C(0, 9) and D( 3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.
Given ABCD forms a parallelogram
We need to find length of altitude on base AB
Let us draw an altitude CE from point C to AB
Let CE = h
We need to find h
To find h, we find Area of ABC by two formulas
Area of ABC = 1/2 Base Height
Area of ABC = 1/2 AB h
Also, we find Area ABC by and then compare both areas
Finding AB
x1 = 4 , y1 = 2
x2 = 7 , y2 = 2
AB = (( 2 1)2+( 2 1)2)
= (( 7 4)2+(2 ( 2))2)
= ((3)2+(2+2)2)
= (32+(4)2)
= (9+16)
= 25
= 5
Finding Area ABC
Area of triangle ABC = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ]
Here
x1 = 4 , y1 = 2
x2 = 7 , y2 = 2
x3 = 0 , y3 = 9
Putting values
Area of triangle ABC = 1/2 [ 4(2 9) + 7(9 ( 2)) + 0( 2 2) ]
= 1/2 [ 4( 7) + 7(9 + 2) + 0( 4) ]
= 1/2 [ 4( 7) + 7(11) + 0( 4) ]
= 1/2 [ 28 + 77]
= 49/2
Now, From (1)
Area of ABC = 1/2 AB h
Putting values of Area and AB
49/2 = 1/2 5 h
49/2 2/5 = h
49/5 = h
h = 49/5
h = 9.8 units
Question 16
The points A(4, 2), B(7, 2), C(0, 9) and D( 3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.
Given ABCD forms a parallelogram
We need to find length of altitude on base AB
Let us draw an altitude CE from point C to AB
Let CE = h
We need to find h
To find h, we find Area of ABC by two formulas
Area of ABC = 1/2 Base Height
Area of ABC = 1/2 AB h
Also, we find Area ABC by coordinate geometry formula and then compare both areas
Finding AB
x1 = 4 , y1 = 2
x2 = 7 , y2 = 2
AB = (( 2 1)2+( 2 1)2)
= (( 7 4)2+(2 ( 2))2)
= ((3)2+(2+2)2)
= (32+(4)2)
= (9+16)
= 25
= 5
Finding Area ABC
Area of triangle ABC = 1/2 [ x1(y2 y3) + x2(y3 y1) + x3(y1 y2) ]
Here
x1 = 4 , y1 = 2
x2 = 7 , y2 = 2
x3 = 0 , y3 = 9
Putting values
Area of triangle ABC = 1/2 [ 4(2 9) + 7(9 ( 2)) + 0( 2 2) ]
= 1/2 [ 4( 7) + 7(9 + 2) + 0( 4) ]
= 1/2 [ 4( 7) + 7(11) + 0( 4) ]
= 1/2 [ 28 + 77]
= 49/2
Now, From (1)
Area of ABC = 1/2 AB h
Putting values of Area and AB
49/2 = 1/2 5 h
49/2 2/5 = h
49/5 = h
h = 49/5
h = 9.8 units

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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