Question -
Answer -
(i)┬а(1, 4), (2, -3) and(-1, -2)
Given:
Points A (1, 4), B (2,-3) and C (-1, -2).
Let us assume,
m1,┬аm2,┬аandm3┬аbe the slope of the sides AB, BC and CA, respectively.
So,
The equation of theline passing through the two points┬а(x1, y1)and┬а(x2, y2).
Then,
m1 =┬а-7,┬аm2┬а=-1/3┬аand m3┬а= 3
So, the equation ofthe sides AB, BC and CA are
By using the formula,
y тАУ y1= m(x тАУ x1)
=> y тАУ 4 = -7 (x тАУ1)
y тАУ 4 = -7x + 7
y + 7x = 11,
=> y + 3 = (-1/3)(x тАУ 2)
3y + 9 = -x + 2
3y + x = тАУ 7
x + 3y + 7 = 0 and
=> y + 2 = 3(x+1)
y + 2 = 3x + 3
y тАУ 3x = 1
So, we get
y + 7x =11, x+ 3y + 7=0 and y тАУ 3x = 1
тИ┤ The equation of sidesare y + 7x =11, x+ 3y + 7 =0 and y тАУ 3x = 1
(ii)┬а(0, 1), (2, 0) and(-1, -2)
Given:
Points A (0, 1), B (2,0) and C (-1, -2).
Let us assume,
m1,┬аm2,┬аandm3┬аbe the slope of the sides AB, BC and CA, respectively.
So,
The equation of theline passing through the two points┬а(x1, y1)and┬а(x2, y2).
Then,
m1┬а=-1/2, m2┬а= -2/3┬аand m3= 3
So, the equation ofthe sides AB, BC and CA are
By using the formula,
y тАУ y1= m(x тАУ x1)
=> y тАУ 1 = (-1/2)(x тАУ 0)
2y тАУ 2 = -x
x + 2y = 2
=> y тАУ 0 = (-2/3)(x тАУ 2)
3y = -2x + 4
2x тАУ 3y = 4
=> y + 2 = 3(x+1)
y + 2 = 3x + 3
y тАУ 3x = 1
So, we get
x + 2y = 2, 2x тАУ 3y =4and y тАУ 3x = 1
тИ┤ The equation of sidesare x + 2y = 2, 2x тАУ 3y =4 and y тАУ 3x = 1