The combined equation of the angle bisectors of the lines represented by 2x2+xy-y2-10x+2y+8=0is
Answers
Given : lines represented by 2x²+xy-y²-10x+2y+8=0
To Find : combined equation of the angle bisectors of the lines
Solution:
2x²+xy-y²-10x+2y+8=0
(2x - y - 2)(x + y - 4) = 0
2x - y - 2 = 0
=> y = 2x - 2
m₁ = 2
x + y - 4 = 0
=> y = -x + 4
m₂ = - 1
slope of angle bisector = m
| (m - m₁)/(1 + m.m₁) | = | (m - m₂)/(1 + m.m₂) |
=> | (m - 2)/(1 + 2m) | = | (m+1)/(1 - m) |
case 1:
(m - 2)/(1 + 2m) = (m+1)/(1 - m)
=> -m² + 3m - 2 = 2m² + 3m + 1
=> 3m² = - 3
=> m² = -1 ( not possible)
case 2
(m - 2)/(1 + 2m) = -(m+1)/(1 - m)
=> -m² + 3m - 2 = -2m² - 3m - 1
=> m² +6m - 1 = 0
m =( -6 ± √40)/2
= - 3 ± √10
2x - y - 2 = 0 and x + y - 4 = 0 intersection
=> x = 2 , 2
y - 2 = (- 3 ± √10)(x - 2)
Equations of angle bisectors
y + (3 - √10)x - 8 + 2√10 = 0
y + (3 + √10)x -8 - 2√10 = 0
(y + (3 - √10)x - 8 + 2√10)(y + (3 + √10)x -8 - 2√10) = 0
=> y² -16y +6xy -x² -8x + 24 = 0
y² -16y +6xy -x² -8x + 24 = 0 is the combined equation of the angle bisectors of the lines represented by 2x²+xy-y²-10x+2y+8=0
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