On a graph paper draw two straight lines which represent x -3y + 1 =0 and 2x - 3y - 4 = 0. Also find the point of intersection of the two lines on the graph
Answers
EXPLANATION.
Graph of straight lines.
⇒ x - 3y + 1 = 0. - - - - - (1).
⇒ 2x - 3y - 4 = 0. - - - - - (2).
As we know that,
From equation (1), we get.
⇒ x - 3y + 1 = 0. - - - - - (1).
Put the value of x = 0 in the equation, we get.
⇒ (0) - 3y + 1 = 0.
⇒ - 3y + 1 = 0.
⇒ - 3y = - 1.
⇒ 3y = 1.
⇒ y = 1/3.
⇒ y = 0.33.
Their Co-ordinates = (0,0.33).
Put the value of y = 0 in the equation, we get.
x - 3(0) + 1 = 0.
x + 1 = 0.
x = - 1.
Their Co-ordinates = (-1,0).
From equation (2), we get.
2x - 3y - 4 = 0. - - - - - (2).
Put the value of x = 0 in the equation, we get.
⇒ 2(0) - 3y - 4 = 0.
⇒ - 3y - 4 = 0.
⇒ - 3y = 4.
⇒ y = - 4/3.
⇒ y = - 1.33.
Their Co-ordinates = (0,-1.33).
Put the value of y = 0 in the equation, we get.
⇒ 2x - 3(0) - 4 = 0.
⇒ 2x - 4 = 0.
⇒ 2x = 4.
⇒ x = 2.
Their Co-ordinates = (2,0).
Both curves intersects at a point = (5,2).
Answer:
Required Answer :
On straight lines,
➙ x - 3y + 1 = 0 ----- equation [1]
➙ 2x -3y -4 = 0 ----- equation [2]
Now, from equation [1]
➙ x - 3y + 1 = 0 ----- equation [1]
putting value of x = 0 in the equation,
➙ (0 )- 3y + 1 = 0
➙ - 3y + 1 = 0
➙ - 3y = - 1
➙ 3y = 1 ( take of - )
➙ 1/3 = y
➙ y = 0.33
- So, Coordinates is 0 & 0.33
putting the value y - 0 in the equation,
➙ x - 3(0) + 1 - 0
➙ x + 1 = 0
➙ x = -1
- So, coordinates is - 1 & 0
Now, from equation [2]
➙ 2x -3y -4 = 0 ----- equation [2]
putting the value x = 0 in the equation
➙ 2(0) -3y -4 = 0
➙ - 3y -4 = 0
➙ -3y = 4
➙ y = -4/3 (division)
➙ y = - 1.33
- So, coordinates is 1.33 & 0
putting value of y = 0 in the equation
➙ 2x -3(0) -4 = 0
➙ 2x -4 = 0
➙ 2x = 4
➙ x = 2
- So, coordinates is 2 & 0