Determinants NCERT Solutions
NCERT Exercise Solution 4.3
Question:1 .Find area of the triangle with vertices at the point given in each of the following:
(i)(1,0)(6,0),(4,3)
(ii)(2,7),(1,1),(10,8)
(iii)(-2,-3),(3,2),(-1,-8)
Solution:(i)Let A (1,0),B(6,0)and C(4,3)be three vertices of triangle ABC.Then
area of `Delta ABC=1/2|(1, 0, 1), (6, 0, 1), (4, 3, 1)|`
`=1/2 [1(0-3)-0(6-4)+1(18-0)]`
`=1/2[-3+18]=(15)/2= 7.5sq.` units.
(ii)Let A (2,7),B(1,1)and C (10,8)be three vertices of triangle ABC. Then
area of `Delta ABC=1/2|(2, 7, 1), (1, 1, 1), (10, 8, 1)|`
`=1/2[2(1-8)-7(1-10)+1(8-10)]`
`=1/2[-14+63-2]=(47)/2` sq.units
(iii) Let A (-2,-3),B(3,2)and C(-1, -8)be three vertices of triangleABC, Then
aera of `Delta ABC=1/2|(-2, -3, 1), (3, 2, 1), (-1, -8, 1)|`
`=1/2[-2(2+8)+3(3+1)+1(-24+2)]`
`=1/2[-20+12-22]` `=|(-30)/2|=15`sq.units
Question:2 . Show that the points A (a,b+c)B(b,c+a),C(c,a+b)are collinear.
Solution: (i)Let A (a,b+c),B(b,c+a)and C(c,a+b)be three vertices of triangle ABC, Then
area of `Delta ABC= i/2|(a, b+c, 1), (b, c+a, 1),(c, a+b, 1)|`
`=1/2 [a(c+a-a-b)-(b+c)(b-c)` `+1(ab+b^2-c^2-ac)]`
`=1/2[ac-ab-b^2+c^2+ab` `+b^2-c^2-ac)]`
`=1/2xx0=0-3`
Now area of `DeltaABC =0`
Thus A (a, b+c),B (b,c+a)and C(c, a+b)are collinear.
Question:3 .Find the values of Kif areaof triangle is 4sq. units and vertices are
(i)`(k,0),(4,0),(0,2)`,
(ii)`(-2,0),(0,4),(0,K)`
Solution:(i)Let A((k,0),B(4,0)andC(0,2)be three vertices of tringle ABC. Then
area of `Delta ABC =1/2|(k, 0, 1), (4, 0, 1), (0, 2, 1)|`
=`1/2[k(0-2)-0(4-0)+1(8-0)]`
`=1/2[-2k+8]=(-k+4)`sq.units
But area of `Delta ABC` is 4 sq. units,
`:. -+4 +-4`,
`:. -k+4=4 or -k+4=-4-0`
`=>k=0 or k=8`
(ii)Let A (-2, 0),B(0,4),(0,k)be three vertices of triangle ABC.Then
area of `Delta ABC=1/2|(-2, 0, 1), (0, 4, 1), (0, k, 1)|`
`= 1/2 [-2(4-k)` `-0(0-0)+1(0-0)] `
`=1/2[-8+2k]=(k-4)`sq.units
But area of `Delta ABC `is 4sq.units.
`:.k=4=+- 4`
`:. k-4=4 or k-4=-4`
`=>k=4+4 or k=-4+4`
`=> k=8 or k=0`
Question:4 .(i) Find equation of line joining (1,2)and (3,6) using determinanants.
(ii)Find equation of line joining (3,1)and (9,3)using determinants.
Solution:(i)Let A (1,2)and B(3,6)be any two points.Let p(x,y)be any point on line AB.
∴ area of `\ Delta ABP=0`
`1/2|(1, 2, 1), (3, 6, 1),(x, y, 1)|=0`
`:.1/2[1(6-y)-2(3-x)+1` `(3y-6x)]=0`
`:.6-y-6+2x+3y-6x=0`
`=>-4x+2y=0=>-2(2x-y)=0`
`=> 2x=y`
which is the equation of required line.
(ii)Let A(3,1)and B(9,3)be any two points. Let P(x,y)be any point on line AB.
`:.area of Delta ABP=0`
`1/2|(3, 1, 1), (9, 3, 1), (x, y, 1)|=0`
`:.1/2[3(3-y)-1(9-x)+` `1(9y-3x)]=0`
`:.9-3y-9+x+9y-3x=0`
`=>-2x+6y=0` ` =>-2(x-3y)=0`
`=>x-3y=0`
Which is the equation of required line.
Question:5 .If area of tringle is 35 sq. units with vertices (2,-6), (5,4)and (k,4).Then kis
(A) 12
(B)`-2`
(C)`-12, -2`
(D)`12,-2`
Solution:Let A(2,-6), B(5,4) and C (k,4) be there vertices of triangle ABC. Then
area of `Delta ABC=1/2 |(2, -6, 1), (5, 4, 1), (k, 4, 1)|`
`=1/2[2(4-4)+6(5-k)+1(20-4k)]`
`=1/2[0+30-6k+20-4k]`
`=1/2[50-10k]=>` `(25-5k)`sq. unit.
But area of `Delta ABC `is 35 sq.units.
`:. 25-5k=+-35`
`:. 25 - 5K= 35 ` `or 25-5k =-35`
`-5k=35-25` ` or -5k=-35-25`
`-5k=10 or -5k=-60`
`k=-2 or k=12`
Thus answer is (D).
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