Let P(x, y) be the required point. Using the section formula, we get
x = $\dfrac{(2×4 + 3×(-1))}{(2 + 3)}$ =$\dfrac{ (8 – 3)}{5} $ = 1
y =$\dfrac{ (2×-3 + 3×7)}{(2 + 3)}$ = $\dfrac{(-6 + 21)}{5}$ = 3
Therefore, the point is (1, 3).
Let P $(x_{1}, y_{1}) $ and Q $(x_{2}, y_{2})$ be the points of trisection of the line segment joining the given points,
i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2.
$x_{1}$ =$\dfrac{ (1×(-2) + 2×4)}{3}$ = $\dfrac{(-2 + 8)}{3}$ = $\dfrac{6}{3}$ = 2
$y_{1}$ = $\dfrac{(1×(-3) + 2×(-1))}{(1 + 2)}$ =$\dfrac{ (-3 – 2)}{3}$ = $\dfrac{-5}{3}$
Therefore: P $(x_{1}, y_{1}) $ = P(2, -5/3)
Point Q divides AB internally in the ratio 2:1.
$x_{2}$= $\dfrac{(2×(-2) + 1×4)}{(2 + 1)}$ = $\dfrac{(-4 + 4)}{3}$ = 0
$y_{2}$ = $\dfrac{(2×(-3) + 1×(-1))}{(2 + 1)}$ = $\dfrac{(-6 – 1)}{3}$ =$\dfrac{ -7}{3}$
The coordinates of the point Q is (0, $\dfrac{ -7}{3}$)
From the given instruction, we observed that Niharika posted the green flag at $\dfrac{1}{4}$th of the distance AD,
i.e., ($\dfrac{1}{4}$ ×100) m = 25 m from the starting point of the 2nd line. Therefore, the coordinates of this point are (2, 25).
Similarly, Preet posted a red flag at 1/5 of the distance AD i.e., ($\dfrac{1}{5}$×100) m = 20m from the starting point of the 8th line. Therefore, the coordinates of this point are (8, 20).
Distance between these flags can be calculated by using the distance formula,
Distance between two flags=$\sqrt{(8-2)^{2}+(20-25)^{2}}$=$\sqrt{36+25}$=$\sqrt{61}m$
The point at which Rashmi should post her blue flag is the midpoint of the line joining these points. Let’s say this point is P(x, y).
x = $\dfrac{(2 + 8)}{2}$ = $\dfrac{10}{2}$ = 5 and y = $\dfrac{(20 + 25)}{2}$ = $\dfrac{45}{2}$
Hence, P( x, y) = (5, $\dfrac{45}{2}$)
Therefore, Rashmi should post her blue flag at $\dfrac{45}{2}$ = 22.5m on the 5th line.
Consider the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) to be k :1.
Therefore, -1 = $\dfrac{( 6k-3)}{(k+1)}$
–k – 1 = 6k -3
7k = 2
k = $\dfrac{2}{7}$
Therefore, the required ratio is 2: 7.
Let the ratio in which the line segment joining A (1, – 5) and B ( – 4, 5) is divided by the x-axis be k : 1.
Therefore, the coordinates of the point of division, say P(x, y) is ($\dfrac{(-4k+1)}{(k+1)}$, $\dfrac{(5k-5)}{(k+1)}$).
Or P(x,y)=$\dfrac{-4k+1}{k+1}$,$\dfrac{5k-5}{k+1}$
We know that the y-coordinate of any point on the x-axis is 0.
Therefore, $\dfrac{( 5k – 5)}{(k + 1)}$ = 0
5k = 5
or k = 1
So, the x-axis divides the line segment in the ratio 1:1.
Now, find the coordinates of the point of division:
P (x, y) = ($\dfrac{(-4(1)+1)}{(1+1)}$ , $\dfrac{(5(1)-5)}{(1+1)}$) = (-3/2 , 0)
Let A, B, C and D be the points of a parallelogram : A(1, 2), B(4, y), C(x, 6) and D(3, 5).
Since the diagonals of a parallelogram bisect each other, the midpoint is the same.
To find the value of x and y, solve for midpoint first.
Midpoint of AC = ( $\dfrac{(1+x)}{2}$ , $\dfrac{(2+6)}{2}$ ) = ($\dfrac{(1+x)}{2}$ , 4)
Midpoint of BD = ($\dfrac{(4+3)}{2}$ , $\dfrac{(5+y)}{2}$ ) = ($\dfrac{7}{2}$ , $\dfrac{(5+y)}{2}$)
Midpoint of AC and BD are the same; this implies
$\dfrac{(1+x)}{2}$ = $\dfrac{7}{2}$ and 4 = $\dfrac{(5+y)}{2}$
x + 1 = 7 and 5 + y = 8
x = 6 and y = 3
Let the coordinates of point A be (x, y).
Midpoint of AB is (2, – 3), which is the center of the circle.
Coordinate of B = (1, 4)
(2, -3) =($\dfrac{(x+1)}{2}$ ,$\dfrac{ (y+4)}{2}$)
$\dfrac{(x+1)}{2}$ = 2 and $\dfrac{(y+4)}{2}$ = -3
x + 1 = 4 and y + 4 = -6
x = 3 and y = -10
The coordinates of A(3,-10).
The coordinates of point A and B are (-2,-2) and (2,-4), respectively.
Since AP =$\dfrac{ 3}{7}$ AB
Therefore, AP: PB = 3:4
Point P divides the line segment AB in the ratio 3:4.
Coordinate of P=($\dfrac{3(2)+4(2)}{3+4}$,$\dfrac{3(-4)+4(-2)}{3+4}$) =($\dfrac{6-8}{7}$,$\dfrac{-12-8}{7}$) =(-$\dfrac{2}{7}$,-$\dfrac{20}{7}$)which is the required answer.
The coordinates of point A and B are (-2,-2) and (2,-4), respectively.
Since AP =$\dfrac{ 3}{7}$ AB
Therefore, AP: PB = 3:4
Point P divides the line segment AB in the ratio 3:4.
Coordinate of P=($\dfrac{3(2)+4(2)}{3+4}$,$\dfrac{3(-4)+4(-2)}{3+4}$) =($\dfrac{6-8}{7}$,$\dfrac{-12-8}{7}$) =(-$\dfrac{2}{7}$,-$\dfrac{20}{7}$)which is the required answer.
Draw a figure, line dividing by 4 points.
From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1, respectively.
Coordinates of X=($\dfrac{1(2)+3(-2)}{1+3}$,$\dfrac{1(8)+3(2)}{1+3}$)=(-1,$\dfrac{7}{2}$)
Coordinates of Y=($\dfrac{2(1)-2(1)}{1+1}$,$\dfrac{2(1)+8(1)}{1+1}$)=(0,5)
Coordinates of Z=($\dfrac{3(2)+1(-2)}{1+3}$,$\dfrac{3(8)+1(2)}{1+3}$)=(1,$\dfrac{13}{2}$)