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Two lines L1 and L2 pass through the point of intersection of 2y=x-13 and 3y+x+12=0. L1 passes through P(-4,-7) and L2 is perpendicular to 2x-5y=4.find the acute angle between L1 andL2

Sagot :

Answer:

84°

Step-by-step explanation:

1. point of intersection of 2y=x-13 and 3y+x+12=0

x = 2y + 13   ==> 3y + (2y+13) + 12 = 0   ==>  5y + 25 = 0  ==> y = -5

x = 2*(-5) +13 = 3

point of intersection: (3 , -5)        L1: pass (3 , -5) and (-4 , -7)

slope of L1 = (-7 - -5)/(-4-3) = -2 / -7 = 2/7

L2 pass (3 , -5) perpendicular to 2x-5y=4

2x-5y=4  ==> y = 2/5 x - 4  slope = 2/5  

so slope of L2 = -5/2

angle Θ between two slopes: tan Θ = | (m2-m1) / (1 + m1*m2)|

==> = | (-5/2 - 2/7) / (1 + -5/2*2/7) | = |(-39/14) / (4/14) | = |- 39/4| = 39/4 = 9.75

Θ = 84°

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