Determine algebraically the vertices of the triangle formed by the lines 3x-y=3 2x-3y=2 and x+2y=8
Answers
The three lines which forms a triangle are :
→ 3 x-y=3 -------(1)
→2 x-3 y=2 -------(2)
→ x+2 y=8 --------(3)
Multiplying Equation (1) by 3 and then subtract equation (2) from it
→9 x - 3 y - 2 x + 3 y = 9 -2
→ 7 x = 7
→ x = 1
Putting the value of x in (1), we get
→ 3 - y = 3
→ y = 0
So, one vertex of triangle is (1,0).
Equation (2) - 2 × Equation (3)
→2 x - 3 y -2 x - 4 y= 2 - 16
→ -7 y = -14
Dividing both sides by 7, we get
→ y = 2
Putting the value of y in equation (3), we get
→ x + 4 = 8
→ x = 8 - 4
→ x = 4
→ The second vertex is (4,2).
2 × equation (1) + equation (3) gives,
→6 x - 2 y + x + 2 y = 6 + 8
→7 x = 14
Dividing both sides by 2, we get
x = 2
putting the value of x in (3), we get
2 + 2y =8
2 y = 8 -2
2 y = 6
y =3
So,the third vertex is (2,3).
The three vertices of triangle are (1,0),(4,2) and (2,3).
We can find the solution graphically also.
Answer:
The vertices formed by the given lines are (2, 3), (1, 0) and (4, 2).
Step-by-step explanation:
The lines are :
3x - y = 3 → (1)
2x - 3y = 2 → (2)
x + 2y = 8 → (3)
To find the first vertex take the lines (1) and (2).
(1) × 3 ⇒ 9x - 3y = 3
(1) - (2) ⇒ 9x - 3y = 9
2x - 3y = 2
--------------------
7x = 7
⇒ x = 7/7
x = 1
∴ 3x - y = 3
⇒ 3 -y = 3
⇒ -y = 3 - 3 = 0
⇒ y = 0
The first vertex is (1,0)
To find the second vertex take lines (2) and (3).
From (3),
x + 2y = 8
⇒ x = 8 - 2y
Substituting value of x in (2),
⇒ 2 (8 - 2y) - 3y = 2
⇒ 16 - 4y -3y -2 = 0
⇒ -7y = -14
⇒ y = 14/7
⇒ y = 2
Substituting value of y in x = 8 - 2y,
x = 8 - 2 × 2
= 8 - 4
x = 4
The second vertex is (4,2)
To find the third vertex take the lines (1) and (3).
From (3),
x + 2y = 8
⇒ x = 8 - 2y
Substituting value of x in (1),
⇒ 3 (8 - 2y) - y = 3
⇒ 24 - 6y -y -3 = 0
⇒ -7y = -21
⇒ y = 21/7
⇒ y = 3
Substituting value of y in x = 8 - 2y,
x = 8 - 2 × 3
= 8 - 6
x = 2
The third vertex is (2,3).
The vertices formed by the given lines are (2, 3), (1, 0) and (4, 2).