NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2
The Section formula in Coordinate Geometry
The Section Formula
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.
The Midpoint Formula
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.
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 \(P\).
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.
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,
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.
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.
