Question -
Answer -
(i) x + y – 4 = 0, 2x – y+ 3 0 and x – 3y + 2 = 0
Given:
x + y − 4 = 0, 2x − y+ 3 = 0 and x − 3y + 2 = 0
Let us find the pointof intersection of pair of lines.
x + y − 4 = 0 … (1)
2x − y + 3 = 0 … (2)
x − 3y + 2 = 0 … (3)
By solving (1) and (2)using cross – multiplication method, we get
x = 1/3, y = 11/3
Solving (1) and (3)using cross – multiplication method, we get
x = 5/2, y = 3/2
Similarly, solving (2)and (3) using cross – multiplication method, we get
x = – 7/5, y = 1/5
∴ The coordinatesof the vertices of the triangle are (1/3, 11/3), (5/2, 3/2) and (-7/5, 1/5)
(ii) y (t1 +t2) = 2x + 2at1t2, y (t2 + t3)= 2x + 2at2t3 and, y(t3 + t1)= 2x + 2at1t3.
Given:
y (t1 +t2) = 2x + 2a t1t2, y (t2 + t3)= 2x + 2a t2t3 and y (t3 + t1)= 2x + 2a t1t3
Let us find the pointof intersection of pair of lines.
2x − y (t1 +t2) + 2a t1t2 = 0 … (1)
2x − y (t2 +t3) + 2a t2t3 = 0 … (2)
2x − y (t3 +t1) + 2a t1t3 = 0 … (3)
By solving (1) and (2)using cross – multiplication method, we get
Solving (1) and (3)using cross – multiplication method, we get
Solving (2) and (3)using cross – multiplication method, we get
∴ The coordinatesof the vertices of the triangle are (at21, 2at1), (at22,2at2) and (at23, 2at3).