Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

Coordinate Geometry : Exercise - 7.3 (Mathematics NCERT Class 10th)

Q.1     Find the area of the triangle whose vertices are :
(i) (2, 3), (–1, 0), (2, – 4)             (ii) (– 5,–1), (3, – 5), (5, 2)
Sol.      (i) Let$A = \left( {{x_1},{y_1}} \right) = \left( {2,3} \right),B = \left( {{x_2},{y_2}} \right) = \left( { - 1,0} \right)$
and $C = \left( {{x_3},{y_3}} \right) = \left( {2, - 4} \right)$
Area of $\Delta ABC$
$= {1 \over 2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
$= {1 \over 2}\left[ {2\left( {0 + 4} \right) + \left( { - 1} \right)\left( { - 4 - 3} \right) + 2\left( {3 - 0} \right)} \right]$
$= {1 \over 2}\left( {8 + 7 + 6} \right) = {{21} \over 2}sq.units$

(ii) Let $A = \left( {{x_1},{y_1}} \right) = \left( { - 5, - 1} \right),B = \left( {{x_2},{y_2}} \right) = \left( {3, - 5} \right)$
and $C = \left( {{x_3},{y_3}} \right) = \left( {5,\,\,2} \right)$

Area of $\Delta ABC$
$= {1 \over 2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
$= {1 \over 2}\left[ { - 5\left( { - 5 - 2} \right) + 3\left( {2 + 1} \right) + 5\left( { - 1 + 5} \right)} \right]$
$= {1 \over 2}\left( {35 + 9 + 20} \right) = {1 \over 2} \times 64$
= 32 sq. units

Q.2     In each of the following find the value of 'k' for which the points are collinear :
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, – 5)
Sol.      (i) Let the given points be $A = \left( {{x_1},{y_1}} \right) = \left( {7, - 2} \right),B = \left( {{x_2},{y_2}} \right) = \left( {5,\,\,1} \right)$ and $C = \left( {{x_3},{y_3}} \right) = \left( {3,\,k} \right)$
These points lie on a line if
Area $\left( {\Delta ABC} \right) = 0$
$\Rightarrow {x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) = 0$
$\Rightarrow 7\left( {1 - k} \right) + 5\left( {k + 2} \right) + 3\left( { - 2 - 1} \right) = 0$
$\Rightarrow 7 - 7k + 5k + 10 - 9 = 0$
$\Rightarrow 8 - 2k = 0$
$\Rightarrow$ 2k = 8
$\Rightarrow$ k = 4
Hence, the given points are collinear for k = 4

(ii) Let the given points be $A = \left( {{x_1},{y_1}} \right) = \left( {8,1} \right),B = \left( {{x_2},{y_2}} \right) = \left( {k, - 4} \right) and\,C = \left( {{x_3},{y_3}} \right) = \left( {2,{\mkern 1mu} - 5} \right).$
If the given points are collinear, then
$\Rightarrow {x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) = 0$
$\Rightarrow 8\left( { - 4 + 5} \right) + k\left( { - 5 - 1} \right) + 2\left( {1 + 4} \right) = 0$
$\Rightarrow 8 - 6k + 10 = 0$
$\Rightarrow$ – 6k = – 18
$\Rightarrow$ k = 3
Hence, the given points are collinear for k = 3.

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), and (0, 3). Find the ratio of this area to the area of the  given triangle.

Sol.     Let $A = \left( {{x_1},{y_1}} \right) = \left( {0,\,\, - 1} \right),B = \left( {{x_2},{y_2}} \right) = \left( {2,\,\,1} \right)$
and $C = \left( {{x_3},{y_3}} \right) = \left( {0,\,3} \right)$ be the vertices of ${\Delta ABC}$.
Area $\left( {\Delta PQR} \right)$
$= {1 \over 2}\left[ {1\left( {0 - 1} \right) + 1\left( {1 - 2} \right) + 0\left( {2 - 0} \right)} \right]$
$= {1 \over 2}\left( { - 1 - 1 + 0} \right) = - 1$
= 1 sq. unit (numerically) Area $\left( {\Delta ABC} \right)$
= ${1 \over 2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
$= {1 \over 2}\left[ {0\left( {1 - 3} \right) + 2\left( {3 + 1} \right) + 0\left( { - 1 - 1} \right)} \right]$
$= {1 \over 2}\left( {0 + 8 + 0} \right) = 4$ sq. units
Let $P\left( {{{0 + 2} \over 2},{{3 + 1} \over 2}} \right)$,i.e., (1, 2), $Q\left( {{{2 + 0} \over 2},{{1 - 1} \over 2}} \right)$, i.e.,  (1, 0) and $R\left( {{{0 + 0} \over 2},{{3 - 1} \over 2}} \right)$, i.e., (0, 1) are the vertices
of ${\Delta PQR}$  formed by joining the mid-points of the sides of ${\Delta ABC}$.

Ratio of the area $\left( {\Delta PQR} \right)$ to the area $\left( {\Delta ABC} \right)$ = 1 : 4.

Q.4     Find the area of the quadrilateral whose vertices, taken in order, are (– 4, –2), (– 3, –5), (3, –2) and (2, 3).
Sol.     Let A(– 4, –2), B (– 3, –5), C (3, –2) and D (2, 3) be the vertices of the quadrilateral ABCD.
= Area of $\Delta$ ABC + Area of $\Delta$ ACD
$= {1 \over 2}\left[ { - 4\left( { - 5 + 2} \right) - 3\left( { - 2 + 2} \right) + 3\left( { - 2 + 5} \right)} \right]$
$+ {1 \over 2}\left[ { - 4\left( { - 2 - 3} \right) + 3\left( {3 + 2} \right) + 2\left( { - 2 + 2} \right)} \right]$
$= {1 \over 2}\left( {12 - 0 + 9} \right) + {1 \over 2}\left( {20 + 15 + 0} \right)$
$= {1 \over 2}\left( {21 + 35} \right) = {1 \over 2} \times 56 = 28$ sq. units

Q.5     You have studied in class IX (Chapter 9 Example 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta$ ABC whose vertices are A(4, –6), B(3, –2) and C(5, 2).
Sol.     Since AD is the median of ${\Delta ABC}$, therefore, D is the mid-point of BC. Coordinates of D are
$\left( {{{3 + 5} \over 2},{{2 + 2} \over 2}} \right)$,i.e., (4, 0). Area of $\Delta ADC$
$= {1 \over 2}\left[ {4\left( {0 - 2} \right) + 4\left( {2 + 6} \right) + 5\left( { - 6 - 0} \right)} \right]$
$= {1 \over 2}\left( { - 8 + 32 - 30} \right) = {1 \over 2} \times - 6 = - 3$
= 3sq. units (numerically)
$= {1 \over 2}\left[ {4\left( { - 2 - 0} \right) + 3\left( {0 + 6} \right) + 4\left( { - 6 + 2} \right)} \right]$
$= {1 \over 2}\left( { - 8 + 18 - 16} \right) = {1 \over 2}\left( { - 6} \right) = - 3$
= 3 sq. units (numerically)
Clearly, area $\left( {\Delta ADC} \right)$ = area $\left( {\Delta ABD} \right)$
Hence, the median of the triangle divides it into two triangles of equal areas.

• Pintu from muzaffarpur srt school

• Anonymous

Question 3 mein kaun sa formula use hua hai

• Bro dono triangle ka area nikala hai and then compare kiya. 1st ka answer 1 sq.unit ka ya and 2nd ka answer 4 sq.units aaya that is why 1:4.
Formula: area of triangle

• Hemanshi

Awsm app thank you from study

• very good

• yash kumar

nice

• Yash kumar

this app is good

• Anonymous

Thanks to you

• hamshavardhini

Easy explanation!!

• Riti Chaudhary

Thanks for your help .

• D.Raviteja

it was very helpful to me thnx

• Its helped me a lot so thnx

• Prakashjaanu

Thanks to app and u r doing a good job for helping me and every one..........and continue u r job for helping to everyone and it is very useful to everyone and it helps me in exams time

• Thrisha deepthi

Yah !

Its good& very useful .Thanks a lot

• Anonymous

Nice

• Anonymous

• Krishna CHOUDHAR¥

Bawall app h bhai mast Cong...

• Devendra

Fantastic

• Lekhivi

Its really helpful for reference and understanding......so im grateful TBH

• Aakash

Good one...

• yash slps

It is very helpful to me at the time of study and at time of exam

• yash slps

It is very helpful to me at the time of exams as it has all the answers of ncert book and related

• Harshita

This app is superb ........here we get all the NCERT questions with answer of each subject ....it also contains many test papers ............its an ossum app...... thanku dornstudy

• ye app muje bhi achhi lagi yaha par har sum ka answer mil jata hai

• Himanshu

This app is one of the best app .as all the ncert questions are also available here and tests also.......thanks dronstudy

• Rahul Roy

Yah app hame bahut Achha laga Kuyki sbhi subject ka question paper hai free test jai, or ha sabhi subject ka solve question Kai OK

• Generation Next

Thank you Rahul Roy

• Its the easiest way to understand this chapter or the other chapters. I'm thankful for this app.
Nowadays sites and apps go on for help in exchange of money transactions but here all of the tests and solved up questions are free which is indeed helpful and generous.
Thanks