- ● Section Formula
- ● Midpoint Formula
- ● Question 1
- ● Question 2
- ● Question 3
- ● Question 4
- ● Question 5
- ● Question 6
- ● Question 7
- ● Question 8
- ● Question 9
- ● Question 10

If there is a line AB where point \(A\) is at \((x_1, \;y_1)\) and point \(B\) is at \((x_2, \;y_2)\), and if a point \(P(x, \;y)\) divides the line in the ratio of \(m : n\), then \(P(x, \;y)\) may be defined as:

The Section Formula \[ P(x, \;y) = \left( \frac{m \cdot x_2 + n \cdot x_1}{m+n}, \; \frac{m \cdot y_2 + n \cdot y_1}{m+n} \right) \]

This is known as Section formula or Internal Section formula in the world of Coordinate Geometry.

When \(m:n = 1:1\), the point \(P(x, \;y)\) becomes the midpoint of line \(AB\). So, putting \(m=1\) and \(n=1\), the formula becomes \[ \left(\, \frac { 1 \cdot x_2 + 1 \cdot x_1 } { 1 + 1}, \frac { 1 \cdot y_2 + 1 \cdot y_1 } { 1 + 1} \, \right) \]

The Midpoint Formula \[ M(x, \;y) = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right) \]

This is the midpoint formula in Coordinate Geometry.

- ● Section Formula
- ● Midpoint Formula
- ● Question 1
- ● Question 2
- ● Question 3
- ● Question 4
- ● Question 5
- ● Question 6
- ● Question 7
- ● Question 8
- ● Question 9
- ● Question 10

Q. 1: Find the coordinates of the point which divides the join of \((−1, \;7)\) and \((4, \;−3)\) in the ratio 2 : 3.

**Solution**

Let point \(M(x,y)\) divide the line in ratio of \(2:3\).

In this problem, to find \(M(x, \;y)\), we have to apply the section formula.
\begin{align}
& m=2,\; n=3, \\
& x_1= -1,\; y_1=7,\; x_2=4,\; y_2=-3 \\
\end{align}
\[ \left(\,
\frac { m \cdot x_2 + n \cdot x_1 } { m + n},
\frac { m \cdot y_2 + n \cdot y_1 } { m + n} \,
\right)
\]
\[ \left(\,
\frac { 2 \cdot 4 + 3 \cdot (-1) } { 2 + 3},
\frac { 2 \cdot (-3) + 3 \cdot 7 } { 2 + 3} \,
\right)
\]
\[\left(\frac{8-3}{5},\, \frac{-6+21}{5} \, \right)\]
\[\left(\frac{5}{5},\, \frac{15}{5} \, \right)\]
\[ = ( 1, 3 ) \]

So, \((1, 3)\) is the point which divides the given line AB in \(2:3\) ratio.
Ans.

Following is the graph of the given scenario:

Q. 2: Find the coordinates of the points of trisection of the line segment joining \((4,−1)\) and \((−2,−3)\).

**Solution**

When a line is trisected, two collinear points divide the line into three equal parts.
Let the end points of the line be \(A(4,−1)\) and \(B(−2,−3)\).
Let two points \(P(x, y)\) and \(Q(x, y)\) trisect \(AB\).

We have to find the coordinates of points \(P\) and \(Q\).

Step #1: We find \(m:n\) at point \(P\).

Because AB is trisected by points \(P\) and \(Q\), the following observations can be made:

\begin{align} &\quad AP = PQ = BQ \quad \color{gray} \text{[AB is trisected]} \\[4pt] &\Rightarrow BP = PQ + BQ \quad \color{gray} \text{[Q is the midpoint of BP]} \\[4pt] &\Rightarrow BP = PQ + PQ \quad \color{gray} \text{[BQ = PQ]} \\[4pt] &\Rightarrow BP = 2 \times PQ \\[4pt] &\Rightarrow BP = 2 \times AP \quad \color{gray} \text{[PQ = AP]} \\[4pt] &\Rightarrow \frac{BP}{AP} = 2 \quad \color{gray} \text{[dividing both sides by AP]} \\[4pt] &\Rightarrow \frac{AP}{BP} = \frac12 \quad \color{gray} \text{[reciprocals]} \\[4pt] &\Rightarrow AP : BP = 1 : 2 \\[4pt] &\therefore m:n = 1:2 \end{align}

Step #2: Ratio \(m:n\) at \(P\) calculated, find the coordinates of \(P\).

\begin{align} & m=1,\; n=2, \\ & x_1= 4,\; y_1=-1,\; x_2=-2,\; y_2=-3 \\ \end{align} \[ \left(\, \frac { 1 \cdot (-2) + 2 \cdot 4 } { 1 + 2}, \frac { 1 \cdot (-3) + 2 \cdot (-1) } { 1 + 2} \, \right) \] \[\left(\frac{-2+8}{3},\, \frac{-3-2}{3} \, \right)\] \[\left(\frac{6}{3},\, \frac{-5}{3} \, \right)\] \[\left(2,\, \frac{-5}{3} \, \right)\]

Step #3: To find \(m:n\) at point \(Q\), follow Step #1

\[AQ = 2 \times BQ \Rightarrow AQ:BQ=2:1,\; \therefore m:n=2:1\]

Step #4: Ratio \(m:n\) at \(Q\) calculated, find the coordinates of \(Q\).

\begin{align}
& m=2,\; n=1, \\
& x_1= 4,\; y_1=-1,\; x_2=-2,\; y_2=-3 \\
\end{align}
\[ \left(\,
\frac { 2 \cdot (-2) + 1 \cdot 4 } { 2 + 1},
\frac { 2 \cdot (-3) + 1 \cdot (-1) } { 2 + 1} \,
\right)
\]
\[\left(\frac{-4+4}{3},\, \frac{-6-1}{3} \, \right)\]
\[\left(\frac{0}{3},\, \frac{-7}{3} \, \right)\]
\[ = \left( 0, \frac{-7}{3} \right) \]
So, the trisector points of the given line are:

\begin{align}\\
\left( 2,\,\frac{-5}{3} \right) \;\text{ and }
\left( 0,\,\frac{-7}{3} \right) \;\color{green}\text{ Ans.}
\end{align}

The graph is given below.

Q. 3: To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs \(\frac14\)th the distance AD on the 2nd line and posts a green flag. Preet runs \(\frac15\)th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag?

**Solution**

100 flower pots have been placed at a distance of 1m.
So total distance is 100m.

Step #1: Distances covered by Niharika and Preet.

Niharika posts her green flag in 2nd row at
\[ \left( \frac14 \times 100 \right) m = 25m \]
So her location is \( A(2, 25) \)

Preet posts her red flag in 8th row at
\[ \left( \frac15 \times 100 \right) m = 20m \]
So her location is \( B(8, 20) \).

Step #2: We find distance between the green and red flag

with the help of the
distance formula.

\[x_1= 2,\, y_1= 25,\, x_2= 8,\, y_2= 20\] \[\sqrt {( x_2 - x_1 )^2 + (y_2 - y_1)^2 }\] \[\sqrt {( 8 - 2 )^2 + (20 - 25)^2 }\] \[ = \sqrt{ 6^2 + (-5)^2} \] \[ = \sqrt{ 36 + 25} \] \[ = \sqrt{61}\,m \]

Let \( M(x, y) \) be the midpoint of AB where Rashmi has to post her
blue flag.

Step #3: Find the midpoint of \(AB [M(x, y)] \)

\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \, \right)\]
\[\left(\frac{2+8}{2}, \frac{25+20}{2} \, \right)\]
\[ = \left( \frac{10}{2},\, \frac{45}{2} \right) \]
\[ = \left( 5, 22.5 \right) \]

So, the distance between red flag and green flag = \( \sqrt{61}m \).

Rashmi has to post her blue flag at \((5,\,22.5)\),

i.e. at row \(5\) and vertically at \(22.5m\) Ans.

Following is the modified image of the scenario:

Q. 4: Find the ratio in which the line segment joining the points \((−3, 10)\) and \((6, −8)\) is divided by \((−1, 6)\).

**Solution**

Let the point \(P(-1, 6)\) divide the line between
\(A(-3, 10)\) and \(B(6, -8)\) in the ratio \(k:1\), or \(m:n=k:1\).

Here, \(x\) coordinate of \(P\) is \(-1\). By applying the section formula,

\[\text{x-coordinate of point P} \;=\; -1 \\\\[5pt] \] \[ \left(\, \frac { m \cdot x_2 + n \cdot x_1 } { m + n}, \frac { m \cdot 20 + n \cdot 25 } { m + n} \, \right) \] \[ \left(\, \frac { k \cdot 6 + 1 \cdot (-3) } { k + 1}, \frac { k \cdot 20 + 1 \cdot 25 } { k + 1} \, \right) \] \[ \Rightarrow \;\frac{6k - 3}{k + 1} \;=\; -1 \\[6pt] \] \[ \Rightarrow \;6k - 3 \;=\; -k - 1 \\[4pt] \] \[ \Rightarrow \;6k+k \;=\; -1 + 3 \\[4pt] \] \[ \Rightarrow \;7k \;=\; 2 \\[4pt] \] \[ \therefore \; k \;=\; \frac27 \] \[ \therefore \; m:n \;=\; \frac27 : 1 \] \[ \therefore \; m:n \;=\; 2 : 7 \]

So, point \(\;(−1, 6)\) divides the line between \(\;A(-3, 10)\) and
\(\;B(6, -8)\) in the ratio 2 : 7
Ans.

Q. 5: Find the ratio in which the line segment joining \(A(1, −5)\) and \(B(−4, 5)\) is divided by the x-axis. Also find the coordinates of the point of division.

**Solution**

If any point is on the X-axis, its y-coordinate is always 0.

Because the divider point lies on x-axis,
the y coordinate of the point is 0.
Let point \(P(x, 0)\) divide the given line \(AB\) in the ratio of

\(1:1, \; or \; m:n=k:1\)

By applying the section formula,

\[\text{x-coordinate of point P} \;=\; -1 \] \[ \left(\, \frac { m \cdot y_2 + n \cdot y_1 } { m + n}, \frac { m \cdot 20 + n \cdot 25 } { m + n} \, \right) \] \[ \left(\, \frac { k \cdot 5 + 1 \cdot (-5) } { k + 1}, \frac { k \cdot 20 + 1 \cdot 25 } { k + 1} \, \right) \] \[ \Rightarrow \;\frac{5k - 5}{k + 1} \;=\; 0 \] \[ \Rightarrow \;5k - 5 \;=\; 0 \\[4pt] \] \[ \Rightarrow \;5k \;=\; 5 \\[4pt] \] \[ \therefore \; k \;=\; 1 \] \[ \therefore \; m:n \;=\; 1 : 1 \]

Step #2: Find the x-coordinate of point \(P\)

\[\text{x-coordinate of point P} \]
\[ \left(\,
\frac { m \cdot x_2 + n \cdot x_1 } { m + n},
\frac { m \cdot 20 + n \cdot 25 } { m + n} \,
\right)
\]
\[ \left(\,
\frac { 1 \cdot (-4) + 1 \cdot 1 } { 1 + 1},
\frac { 1 \cdot 20 + 1 \cdot 25 } { 1 + 1} \,
\right)
\]
\[ =\; \frac{-4+1}{2} \]
\[ =\; \frac{-3}{2} \]
So, the required coordinates of P are
\[ \left( \frac{-3}{2}, 0 \right) \]
and it divides \(AB\) in the ratio of \(1:1\)
Ans.

Q. 6: If \((1,2), \; (4,y), \; (x,6)\) and \((3,5)\) are the vertices of a parallelogram taken in order, find \(x\) and \(y\).

**Solution**

Let the vertices of the given parallelogram are:

\(A(1, 2), \; B(4, y), \; C(x, 6),\) and \(D(3, 5)\).

Let the diagonals AC and BD intersect at point \(O(p, q)\).

Step #1: Find \(\,m : n\) for both \(\,AC\) and \(\,BD\)

Note: In a parallelogram, the diagonals bisect each other.

So, O is the midpoint of AC as well as BD.

In both the cases, \(m : n = 1 : 1\)

Step #2: Find coordinates of point \(\,O\) w.r.t. \(\,AC\)

When O is the midpoint of AC, \(A(1, 2),\; C(x, 6)\)

\[x_1= 1,\, y_1= 2,\, x_2= x,\, y_2= 6\]
\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \, \right)\]
\[\left(\frac{1+x}{2}, \frac{2+6}{2} \, \right)\]
\[ = \left( \frac{1+x}{2}\,, \frac{8}{2} \right) \]
\[ = \left( \frac{1+x}{2}\,, 4 \right) ...(i) \]

Step #3: Find coordinates of point \(\,O\) w.r.t. \(\,BD\)

When O is the midpoint of BD, \(B(4, y),\; D(3, 5)\)

\[x_1= 4,\, y_1= y,\, x_2= 3,\, y_2= 5\]
\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \, \right)\]
\[\left(\frac{4+3}{2}, \frac{y+5}{2} \, \right)\]
\[ = \left( \frac{7}{2}\,, \frac{y+5}{2} \right) ...(ii) \]

Step #4: Find the values of \(x\) and \(y\) from (i) and (ii)

Since (i) and (ii) represent the same point O,

\[ \therefore (i)=(ii) \]
\[ \Rightarrow \frac{1+x}{2} = \frac72 \]
\[ \quad\color{gray}\text{[multiply both sides by 2]} \]
\[ \Rightarrow 1+x = 7 \]
\[ \therefore x = 7-1 = 6 \]

\[ \text{Again, } 4 = \frac{y+5}{2} \]
\[ \quad\color{gray}\text{[by cross-multiplication]}\]
\[ \Rightarrow y+5 = 8 \]
\[ \therefore y = 8-5 = 3 \]
So, the required values are \(x=6,\; y=3 \) Ans.

Q. 7: Find the coordinates of a point A, where AB is the diameter of circle whose center is (2,−3) and B is (1,4)

**Solution**

Let coordinates of \(A\) be \((x, y)\) & center point be \(C\).

\(C(2, -3)\) is the mid-point of diameter \(AB\).

So we have endpoints \(A, B\) and midpoint \(C\).

Here,
\[x_1= x,\, y_1= y,\, x_2= 1,\, y_2= 4\]
\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \, \right)\]
\[\left(\frac{x+1}{2}, \frac{y+4}{2} \, \right)\]
We have got two equations here.

\begin{align}
&\frac{x+1}2 = 2, \\[6pt]
&\Rightarrow x+1 = 4, \\[6pt]
&\therefore x = 3, \\[6pt]
&\frac{y+4}2 = -3 \\[6pt]
&\Rightarrow y+4 = −6 \\[6pt]
&\Rightarrow y = -6-4 \\[6pt]
&\therefore y = -10
\end{align}
So , the coordinates of point \(\;A\) are \(\;(3,-10)\) Ans.

Q. 8: If A and B are \((−2,−2)\) and \((2,−4)\), respectively, find the coordinates of P such that \( AP=\frac37 AB \) and P lies on the line segment AB.

**Solution**

Step #1: Given \(AP = 3/7 \times AB\), find \(m : n\).

Given \( \quad AP = \frac37 \times AB \), and
\[x_1= -2,\, y_1= -2,\, x_2= 2,\, y_2= -4\]
we find the \(m:n\)

We know,
\begin{align}
BP &= AB - AP \\[6pt]
& = AB - \frac{3AB}7 \\[6pt]
& = \frac{7AB - 3AB}7 \\[6pt]
& = \frac47 AB \\[6pt]
\therefore\quad \frac{AP}{BP} & = \,\,\,{\frac37 AB \over \frac47 AB} \,\,\,
= {\frac37 \over \frac47} \,\,
= \frac34 \\[6pt]
\therefore\quad m : n &= 3 : 4 \\
\end{align}

Step #2: find coordinates of \(P\) while \(m:n = 3:4\)

\begin{align}
& m=3,\; n=4, \\
& x_1= -2,\; y_1=-2,\; x_2=2,\; y_2=-4 \\
\end{align}
\[ \left(\,
\frac { 3 \cdot 2 + 4 \cdot (-2) } { 3 + 4},
\frac { 3 \cdot (-4) + 4 \cdot (-2) } { 3 + 4} \,
\right)
\]
\[\left(\frac{6-8}{7},\, \frac{-12-8}{7} \, \right)\]
\[\left(\frac{-2}{7},\, \frac{-20}{7} \, \right)\]
So , the coordinates of point P are
\[ \left(\frac{-2}{7}, \frac{-20}{7}\right) \color{green}{\text{ Ans.}} \]
The straight line is as in the following:

Q. 9: Find the coordinates of the points which divide
the line segment joining \(A(−2,2)\) and \(B(2,8)\) into four equal parts.

**Solution**

In the above figure, a/q, the points P, Q and R divide
the line AB in equal parts.

So, \(AP = PQ = QR = BR\), and

Q is the mid-point of AB --- (i)

P is the mid-point of AQ --- (ii)

R is the mid-point of BQ --- (iii)

We calculate mid-point of two points with the following formula:

\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \, \right)\]
Q is the midpoint of \(A(-2, 2)\) and \(B(2, 8)\)
\[x_1= -2,\, y_1= 2,\, x_2= 2,\, y_2= 8\]
\[\left(\frac{(-2)+2}{2}, \frac{2+8}{2} \, \right)\]
\[ = \left( \frac{0}{2}\,, \frac{10}{2} \right) \]
\[ = ( 0, 5) \]
P is the midpoint of \(A(-2, 2)\) and \(Q(0, 5)\)
\[x_1= -2,\, y_1= 2,\, x_2= 0,\, y_2= 5\]
\[\left(\frac{(-2)+0}{2}, \frac{2+5}{2} \, \right)\]
\[ = \left( -1\,, \frac{7}{2} \right) \]
R is the midpoint of \(Q(0, 5)\) and \(B(2, 8)\)
\[x_1= 0,\, y_1= 5,\, x_2= 2,\, y_2= 8\]
\[\left(\frac{0+2}{2}, \frac{5+8}{2} \, \right)\]
\[ = \left( 1\,, \frac{13}{2} \right) \]
So , the coordinates of the required points serially from left are:

\[ (0, 5), \left(-1, \frac72 \right) \;and\; \left(1, \frac{13}2 \right) \color{green}{\text{ Ans.}} \]
The straight line is as in the following:

Q. 10: Find the area of a rhombus if its vertices are
\((3,0), \, (4,5), \, (-1,4) \, and \, (-2,-1)\) taken in order.

[Hint: Area of a rhombus = \( \frac12 \times \) (product of its diagonals)]

**Solution**

Let ABCD be the rhombus so that AC and BD are its diagonals.

We find the lengths of the two diagonals
with the formula of distance between two points :

\[\sqrt {( x_2 - x_1 )^2 + (y_2 - y_1)^2 }\]
For distance A to C,
\[x_1= 3,\, y_1= 0,\, x_2= -1,\, y_2= 4\]
\[\sqrt {( (-1) - 3 )^2 + (4 - 0)^2 }\]
\[ = \sqrt{(-4)^2 + 4^2} \]
\[ = \sqrt{16 + 16} \]
\[ = \sqrt{32} \]
\[ = \sqrt{4^2 \times 2} \]
\[ = 4\sqrt{2} \]

For distance B to D,
\[x_1= 4,\, y_1= 5,\, x_2= -2,\, y_2= -1\]
\[\sqrt {( (-2) - 4 )^2 + ((-1) - 5)^2 }\]
\[ = \sqrt{(-6)^2 + (-6)^2} \]
\[ = \sqrt{36 + 36} \]
\[ = \sqrt{72} \]
\[ = \sqrt{6^2 \times 2} \]
\[ = 6\sqrt{2} \]
*Area of the rhombus ABCD *

\[ \begin{align}
&= \frac12 \times \text{(product of the diagonals)} \\\\[5pt]
&= \frac12 \times {AC} \times {BD} \\\\[5pt]
&= \frac{1}{2} \times 4\sqrt2 \times 6\sqrt2 \\[6pt]
&= \frac{4 \times 6 \times \sqrt2 \times \sqrt2}{2} \\\\[5pt]
&= \frac{4 \times 6 \times 2}{2} \\\\[5pt]
&= 24 \; \text{sq. unit}
\end{align} \]

So the area of the rhombus = 24 Sq. Unit
Ans.

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