determine graphically the vertices of the triangle formed by the lines 3x-y=3, 2x-2y=2, x+2y=8
Answers
Step-by-step explanation:
Given linear equations are
3x−y=3 ___(i)
2x−3y=2 ___(ii)
x+2y=8 ___(iii)
Let the intersecting points of lines (i) and (ii) is A, and of lines (ii) and (iii) is B and that of lines (iii) and (i) is C.
The intersecting point of (ii) and (i) can be find out by solving (i) and (ii) for (x, y).
3x−y=3 [From (i)]
2x−3y=2 [From (ii)]
9x−3y=9 __(iv) [multiplying eqn. (i) by 3]
2x−3y=2 [from (ii)]
− + −
7x=7
[By subtracting (ii) from (iv)]
⇒x=
7
7
⇒x=1
Now, 3x−y=3 [From (i)]
⇒3(1)−y=3[x=1]
⇒−y=3−3
⇒−y=0
⇒y=0
So, intersecting point of eqns.(i) and (ii) is A(1,0).
Similarly, intersecting point B of eqns. (ii) and (iii) can be find out as follows:
2x−3y=2 [from (ii)]
x+2y=8 [from (iii)]
2x−3y=2 [From (ii)]
2x+4y=16 __(v) [By multiplying (iii) by 2]
− − −
−7y=−14
[Subtracting (v) from (ii)]
⇒y=
7
14
⇒y=2
Now, x+2y=8 [From (iii)]
⇒x+2(2)=8
⇒x=8−4
⇒x=4
So, the coordinates of B are (4,2)
Similarly, for intersecting point of C of eqns. (i) and (iii), we have
3x−y=3 [From (i)]
x+2y=8 [From (iii)]
Multiplying (i) by 2, we get
6x−2y=6 ___(vi)
x+2y=8 [From (iii)]
7x=14
[Adding (vi) and (iii)]
⇒x=
7
14
⇒x=2
Now, 3x−y=3 [from (i)]
⇒3(2)−y=3
⇒−y=3−6
−y=−3⇒y=3
So, point C is (2,3)
Hence, the vertices of ΔABC formed by given three linear equations are A(1,0),B(4,2) and C(2,3)