NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.3
Area of a Triangle in Coordinate Geometry
If coordinates of the vertices of a triangle ABC are
\( A(x_1, \;y_1), \;B(x_2, \;y_2) \) and \( C(x_3, \;y_3) \),
then area of the triangle w.r.t its co-ordinates of the vertices is:
Formula for area of a triangle in Coordinate Geometry
\[\frac12 \times {\color{red} \{}
x_1\times(y_2 - y_3) \boldsymbol{\color{red} +} x_2\times(y_3 - y_1) \boldsymbol{\color{red} +} x_3\times(y_1 - y_2)
{\color{red} \}} \]
Q. 1(i): Find the area of the triangle whose vertices are:
\((2, \;3), (-1, \;0)\), and \((2, \;-4)\).
Solution
The given vertices are
\begin{align}
& x_1=2,\; y_1=3, \\
& x_2= -1,\; y_2=0,\; x_3=2,\; y_3=-4 \\
\end{align}
Area of a triangle w.r.t its co-ordinates of the vertices is:
\[\frac12 \times {\color{red} \{}
x_1\times(y_2 - y_3) \boldsymbol{\color{red} +} x_2\times(y_3 - y_1) \boldsymbol{\color{red} +} x_3\times(y_1 - y_2)
{\color{red} \}} \]
\[\frac12 \times {\color{red} \{}
2\times(0 - (-4)) \boldsymbol{\color{red} +} (-1)\times(-4 - 3) \boldsymbol{\color{red} +} 2\times(3 - 0)
{\color{red} \}} \]
\[\frac12 \times \{
2 \times 4 + (-1) \times (-7) + 2 \times 3
\} \]
\[= \frac12 \times \{8+7+6\}\]
\[ = \frac12 \times 21 \]
\[ = 10.5 \text{ sq. unit} \]
So, the required area of the given triangle
= 10.5 sq. unit Ans.
Following is the graph of the given triangle:
Q. 1(ii): Find the area of the triangle whose vertices are:
\( (-5, \;-1),\, (3, \;-5),\) and \((5, \;2)\).
Solution
The given vertices are
\begin{align}
& x_1=-5,\; y_1=-1, \\
& x_2= 3,\; y_2=-5,\; x_3=5,\; y_3=2 \\
\end{align}
Area of a triangle w.r.t its co-ordinates of the vertices is:
\[\frac12 \times {\color{red} \{}
x_1\times(y_2 - y_3) \boldsymbol{\color{red} +} x_2\times(y_3 - y_1) \boldsymbol{\color{red} +} x_3\times(y_1 - y_2)
{\color{red} \}} \]
Substituting \( x_1, \,y_1, \,x_2, \,y_2, \,x_3, \,y_3 \) with the given values,
area of the triangle
\[\frac12 \times {\color{red} \{}
-5\times(-5 - 2) \boldsymbol{\color{red} +} 3\times(2 - (-1)) \boldsymbol{\color{red} +} 5\times(-1 - (-5))
{\color{red} \}} \]
\[\frac12 \times \{
-5 \times (-7) + 3 \times 3 + 5 \times 4
\} \]
\[\frac12 \times \{35 + 9 + 20 \}\]
\[ = \frac12 \times 64 \]
\[ = 32 \text{ sq. unit } \]
So, the required area of the given triangle
= 32 sq. unit Ans.
Following is the graph of the given triangle:
Q. 2(i): Find the value of 'k' for which the points
\( (7, \;-2), (5, \;1), (3, \;k) \) are collinear.
Solution
When three points are collinear, they actually cannot form
any triangle, or we can say that the area formed by
the triangle is zero.
The given vertices are
\begin{align}
& x_1=7,\; y_1=-2, \\
& x_2= 5,\; y_2=1,\; x_3=3,\; y_3=k \\
\end{align}
Area of the triangle formed by those vertices = 0.
\[\frac12 \times {\color{red} \{}
x_1\times(y_2 - y_3) \boldsymbol{\color{red} +} x_2\times(y_3 - y_1) \boldsymbol{\color{red} +} x_3\times(y_1 - y_2)
{\color{red} \}} \]
Substituting \( x_1,\; y_1,\; x_2,\; y_2,\; x_3,\; y_3 \) with the given values,
\[\frac12 \times {\color{red} \{}
7\times(1 - k) \boldsymbol{\color{red} +} 5\times(k - (-2)) \boldsymbol{\color{red} +} 3\times(-2 - 1)
{\color{red} \}} \]
\[\frac12 \times {\color{red} \{}
7\times((k+2) - (-3)) \boldsymbol{\color{red} +} 5\times(-3 - (1-k)) \boldsymbol{\color{red} +} 3\times((1-k) - (k+2))
{\color{red} \}} \]
\[ \Rightarrow \frac12 \times (7-7k+5k+10-9) = 0 \]
\[ \Rightarrow \frac12 \times (8-2k) = 0 \]
\[ \Rightarrow \frac{ 2 \times(4-k)}{2} = 0 \]
\[ \Rightarrow 4-k = 0 \]
\[ \therefore k = 4 \]
So, the vertices are collinear when
k = 4 Ans.
Following is the graph of the given three coordinates:
Q. 2(ii): Find the value of 'k' for which the points
\( (8, \;1), (k, \;-4), (2, \;-5) \) are collinear.
Solution
When three points are collinear, they actually cannot form
any triangle, or we can say that the area formed by
the triangle is zero.
The given vertices are
\begin{align}
& x_1=8,\; y_1=1, \\
& x_2= k,\; y_2=-4,\; x_3=2,\; y_3=-5 \\
\end{align}
Area of the triangle formed by those vertices = 0.
\[\frac12 \times {\color{red} \{}
x_1\times(y_2 - y_3) \boldsymbol{\color{red} +} x_2\times(y_3 - y_1) \boldsymbol{\color{red} +} x_3\times(y_1 - y_2)
{\color{red} \}} \]
Substituting \( x_1, \;y_1, \;x_2, \;y_2, \;x_3, \;y_3 \) with the given values,
\[\frac12 \times {\color{red} \{}
8\times(-4 - (-5)) \boldsymbol{\color{red} +} k\times(-5 - 1) \boldsymbol{\color{red} +} 2\times(1 - (-4))
{\color{red} \}} \]
\[\frac12 \times \{
8 \times (-4+5) + k \times (-6) + 2 \times (1+4)
\} \]
\[\frac12 \times \{
8 \times 1 + k \times (-6) + 2 \times 5
\} \]
\[ \Rightarrow \frac12 \times (8-6k+10) = 0 \]
\[ \Rightarrow \frac12 \times (18-6k) = 0 \]
\[ \Rightarrow \frac{ 2 \times(9-3k)}{2} = 0 \]
\[ \Rightarrow 9-3k = 0 \]
\[ \Rightarrow -3k = -9 \]
\[ \therefore k = 3 \]
So, the vertices are collinear when
k = 3 Ans.
Following is the graph of the given three coordinates:
Q. 3: Find the area of the triangle formed by joining
the mid-points of the sides of the triangle whose vertices are
\((0, \;-1),\; (2, \;1),\; (0, \;3)\).
Find the ratio of this area to the
area of the given triangle.
Solution
Let ΔABC be the given triangle whose vertices are
\(A(0, \;-1),\; B(2, \;1)\) and \(C(0, \;3)\). What are the tasks here?
(1) Join the given vertices to form ΔABC.
(2) Find the area of ΔABC with given coordinates.
(3) Join the midpoints of AB, BC and CA to form inner ΔPQR.
(4) Find the coordinates of point P, Q and R with
midpoint formula.
(5) Find the area of ΔPQR.
(6) Calculate the ratio of the area of ΔPQR to ΔABC.
Graph in box:
Here, ABC is the given triangle with vertices:
\begin{align}
& x_1=0,\; y_1=-1, \\
& x_2= 2,\; y_2=1,\; x_3=0,\; y_3=3 \\
\end{align}
\(\therefore \) Area of triangle ABC:
\[\frac12 \times {\color{red} \{}
0\times(1 - 3) \boldsymbol{\color{red} +} 2\times(3 - (-1)) \boldsymbol{\color{red} +} 0\times(-1 - 1)
{\color{red} \}} \]
\[\frac12 \times \{
0 \times (-2) + 2 \times 4 + 0 \times (-2)
\} \]
\[ = \frac12 \times (0+8+0) \]
\[ = \frac12 \times 8 \]
\[ = 4 \text{ sq. unit} \]
Let \(P\) be the midpoint of \(AB\),
\(Q\) be the midpoint of \(BC\),
and \(R\) be the midpoint of \(CA\).
The formula to find the midpoint of a straight line is:
\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \, \right)\]
Let us find coordinates of midpoint \(P\) of \( A(0, \;-1),\; B(2, \;1) \) where
\[x_1= 0,\, y_1= -1,\, x_2= 2,\, y_2= 1\]
\[\left(\frac{0+2}{2}, \frac{(-1)+1}{2} \, \right)\]
\[\left(\frac{2}{2},\, \frac{0}{2} \, \right)\]
\[ = (1, \;0) \]
We find coordinates of midpoint \(Q\) of \( B(2, 1),\; C(0, 3) \) where
\[x_1= 2,\, y_1= 1,\, x_2= 0,\, y_2= 3\]
\[\left(\frac{2+0}{2}, \frac{1+3}{2} \, \right)\]
\[\left(\frac{2}{2},\, \frac{4}{2} \, \right)\]
\[ = (1, \;2) \]
and coordinates of midpoint \(R\) of \( C(0, \;3),\; A(0, \;-1) \) where
\[x_1= 0,\, y_1= 3,\, x_2= 0,\, y_2= -1\]
\[\left(\frac{0+0}{2}, \frac{3+(-1)}{2} \, \right)\]
\[\left(\frac{0}{2},\, \frac{2}{2} \, \right)\]
\[ = (0, \;1) \]
Now we know the coordinates of points \(P, Q, R\) as \((1, \;0),\; (1, \;2), (0, \;1)\).
For triangle PQR,
\begin{align}
& x_1=1,\; y_1=0, \\
& x_2= 1,\; y_2=2,\; x_3=0,\; y_3=1 \\
\end{align}
\(\therefore \) Area of triangle PQR:
\[\frac12 \times {\color{red} \{}
1\times(2 - 1) \boldsymbol{\color{red} +} 1\times(1 - 0) \boldsymbol{\color{red} +} 0\times(0 - 2)
{\color{red} \}} \]
\[\frac12 \times \{
1 \times 1 + 1 \times 1 + 0 \times (-2)
\} \]
\[ = \frac12 \times (1+1+0) \]
\[ = \frac12 \times 2 \]
\[ = 1 \text{ sq. unit} \]
So area of ΔPQR = 1 sq. unit,
area of ΔABC = 4 sq. unit.
\(\therefore \) Ratio of areas \(ΔPQR : ΔABC = 1:4\) Ans.
Following is the solved graph:
Q. 4: Find the area of the quadrilateral whose vertices,
in order, are \((-4, \;-2),(-3, \;-5),(3, \;-2)\) and \((2, \;3)\).
Solution
A diagonal inside the quadrilateral will divide it in two
triangles. We find the areas of these two triangles and add them
to find the area of the whole quadrilateral.
Let the vertices be \(A(-4, \;-2),\; B(-3, \;-5),\; C(3, \;-2),\; D(2, \;3)\).
We draw the diagonal \(BD\) and get two triangles Δ\(ABD\) and Δ\(BCD\)
as per the following graph.
Now, area of the quadrilateral ABCD
= area of ΔABD + area of ΔBCD
\[ = \frac{23}{2} + \frac{33}{2} \]
\[ = \frac{23+33}{2} \]
\[ = \frac{56}{2} \]
\[ = 28 \text{ sq. unit} \color{green}{\text{ Ans.}} \]
Q. 5:
You have studied in Class IX that a median
of a triangle divides it into two triangles of equal areas.
Verify this result for ΔABC whose vertices
are \( A(4, \;-6),\; B(3, \;-2)\) and \(C(5, \;2)\).
Solution
Let the given triangle be ABC with D midpoint of AC.
So, BD is the median of the ΔABC which divides it into
two smaller ΔABD and ΔBCD.
Let us find the coordinates of D (which is the midpoint of AC).
By midpoint formula,
\[x_1= 4,\, y_1= -6,\, x_2= 5,\, y_2= 2\]
\[\left(\frac{4+5}{2}, \frac{(-6)+2}{2} \, \right)\]
\[\left(\frac{9}{2},\, \frac{-4}{2} \, \right)\]
\[ = \left( \frac92, -2 \right) \]
\[ = ( 4.5, \;-2) \]
So, for ΔABD, coordinates of the vertices are
\begin{align}
& x_1=4,\; y_1=-6, \\
& x_2= 3,\; y_2=-2,\; x_3=4.5,\; y_3=-2 \\
\end{align}
So, area of ΔABD :
\[\frac12 \times {\color{red} \{}
4\times(-2 - (-2)) \boldsymbol{\color{red} +} 3\times(-2 - (-6)) \boldsymbol{\color{red} +} 4.5\times(-6 - (-2))
{\color{red} \}} \]
\[\frac12 \times \{
4 \times (-2+2) + 3 \times (-2+6) + 4.5 \times (-6+2)
\} \]
\[\frac12 \times \{
4 \times 0 + 3 \times 4 + 4.5 \times (-4)
\} \]
\[ = \frac12 \times (0+12-4.5 \times4) \]
\[ = \frac12 \times (12-18) \]
\[ = \frac12 \times (-6) \]
\[ = -3 \]
Note: In Coordinate Geometry, if an area is found to be negative,
we should take its absolute value, i.e.,
the positive value.
So area of ΔABD = 3 sq. unit.
Again, for ΔBCD, coordinates of the vertices are
\begin{align}
& x_1=3,\; y_1=-2, \\
& x_2= 5,\; y_2=2,\; x_3=4.5,\; y_3=-2 \\
\end{align}
So, area of ΔBCD :
\[\frac12 \times {\color{red} \{}
3\times(2 - (-2)) \boldsymbol{\color{red} +} 5\times(-2 - (-2)) \boldsymbol{\color{red} +} 4.5\times(-2 - 2)
{\color{red} \}} \]
\[\frac12 \times \{
3 \times (2+2) + 5 \times (-2+2) + 4.5 \times (-4)
\} \]
\[\frac12 \times \{
3 \times 4 + 5 \times 0 + 4.5 \times (-4)
\} \]
\[ = \frac12 \times (12+0-18) \]
\[ = \frac12 \times (12-18) \]
\[ = \frac12 \times (-6) \]
\[ = -3 \]
As per above note, take the absolute value of -3.
So area of ΔBCD = 3 sq. unit
So both the ΔABD and ΔBCD have equal areas of 3 sq. unit
Verified.
