find the equation to the pair of straight lines passing through the origin and making an acute angle alpha with the straight line x+y+5=0
Answers
Given : pair of straight lines passing through the origin and making an acute angle alpha with the straight line x+y+5=0
To Find : equation to the pair of straight lines
Solution:
straight line x+y+5=0
=> y = - x - 5
Slope = - 1
Slope of lines = m which line makes an acute angle alpha with the straight line x+y+5=0
tanα = | (m - (-1) )/ ( 1 + m(-1)) |
=> tanα = | (m + 1)/( 1 - m) |
(m + 1)/( 1 - m) = ± tanα
(m + 1)/( 1 - m) = tanα
=> m + 1 = tanα - m tanα
=> m(1 + tanα) = tanα - 1
=> m = ( tanα - 1)/(1 + tanα)
(m + 1)/( 1 - m) = - tanα
=> m + 1 = -tanα + m tanα
=> m(1 - tanα) = -tanα - 1
=> m = (1 + tanα)/( tanα - 1)
Equation of line passing through origin with slope m
y = mx
y = ( tanα - 1)x/(1 + tanα) => y ( 1 + tanα) = x ( tanα - 1)
=> x ( tanα - 1) - y ( 1 + tanα) = 0
y = (1 + tanα)x/( tanα - 1) => y ( tanα - 1) = x ( 1 + tanα)
=> x ( 1 + tanα) - y ( tanα - 1) = 0
Equation of lines
(x ( tanα - 1) - y ( 1 + tanα) )(x ( 1 + tanα) - y ( tanα - 1) ) = 0
=> x²(tan²α - 1) + y²(tan²α - 1) - xy ((tan α - 1)² + (1 + tanα)²) = 0
=> x²(tan²α - 1) + y²(tan²α - 1) - 2xy (tan²α+ 1 ) = 0
=> x²(tan²α - 1) + y²(tan²α - 1) - 2xy (sec²α) = 0
x²(tan²45 - 1) + y²(tan²45 - 1) - 2xy (sec²45) = 0
equation to the pair of straight lines is :
x²(tan²45 - 1) + y²(tan²45 - 1) - 2xy (sec²45) = 0
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