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 = (𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 )
0 = (𝑘 × 7 + 1 × (−6))/(𝑘 + 1)
0 = (7𝑘 − 6)/(𝑘 +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 = (𝑚1 𝑥2 + 𝑚2 𝑥1)/(𝑚1 + 𝑚2)
= (6 × (−1) + 7 × (−4))/(6 + 7)
= (−6 − 28)/13
= (−34)/13
Hence,
the coordinate of point is P = (x, 0) = ((−𝟑𝟒)/𝟏𝟑,𝟎)
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 Δ ABC by two formulas
Area of ΔABC = 1/2 × Base × Height
Area of ΔABC = 1/2 × AB × h
Also, we find Area Δ ABC by and then compare both areas
Finding AB
x1 = 4 , y1 = −2
x2 = 7 , y2 = 2
AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
= √(( 7 −4)2+(2 −(−2))2)
= √((3)2+(2+2)2)
= √(32+(4)2)
= √(9+16)
= √25
= 5
Finding Area ΔABC
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 ΔABC = 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 Δ ABC by two formulas
Area of ΔABC = 1/2 × Base × Height
Area of ΔABC = 1/2 × AB × h
Also, we find Area Δ ABC by coordinate geometry formula and then compare both areas
Finding AB
x1 = 4 , y1 = −2
x2 = 7 , y2 = 2
AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
= √(( 7 −4)2+(2 −(−2))2)
= √((3)2+(2+2)2)
= √(32+(4)2)
= √(9+16)
= √25
= 5
Finding Area ΔABC
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 ΔABC = 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

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