Question -
Answer -
(i) (1, 4), (2, -3) and(-1, -2)
Given:
Points A (1, 4), B (2,-3) and C (-1, -2).
Let us assume,
m1, m2, andm3 be the slope of the sides AB, BC and CA, respectively.
So,
The equation of theline passing through the two points (x1, y1)and (x2, y2).
Then,
m1 = -7, m2 =-1/3 and m3 = 3
So, the equation ofthe sides AB, BC and CA are
By using the formula,
y – y1= m(x – x1)
=> y – 4 = -7 (x –1)
y – 4 = -7x + 7
y + 7x = 11,
=> y + 3 = (-1/3)(x – 2)
3y + 9 = -x + 2
3y + x = – 7
x + 3y + 7 = 0 and
=> y + 2 = 3(x+1)
y + 2 = 3x + 3
y – 3x = 1
So, we get
y + 7x =11, x+ 3y + 7=0 and y – 3x = 1
∴ The equation of sidesare y + 7x =11, x+ 3y + 7 =0 and y – 3x = 1
(ii) (0, 1), (2, 0) and(-1, -2)
Given:
Points A (0, 1), B (2,0) and C (-1, -2).
Let us assume,
m1, m2, andm3 be the slope of the sides AB, BC and CA, respectively.
So,
The equation of theline passing through the two points (x1, y1)and (x2, y2).
Then,
m1 =-1/2, m2 = -2/3 and m3= 3
So, the equation ofthe sides AB, BC and CA are
By using the formula,
y – y1= m(x – x1)
=> y – 1 = (-1/2)(x – 0)
2y – 2 = -x
x + 2y = 2
=> y – 0 = (-2/3)(x – 2)
3y = -2x + 4
2x – 3y = 4
=> y + 2 = 3(x+1)
y + 2 = 3x + 3
y – 3x = 1
So, we get
x + 2y = 2, 2x – 3y =4and y – 3x = 1
∴ The equation of sidesare x + 2y = 2, 2x – 3y =4 and y – 3x = 1