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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.