Question

Find the equation of the bisector of the pair of acute angles formed by the lines 4x+2y=9 and 2x-y=8

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The Bisector of the pair of acute angles formed by the line 8x-25=0

Step-by-step explanation:

$\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2} } }=$± ($\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2} } }$ )

4x+2y-9=0($a_{1} +b_{1} +c_{1}$=0)

2x-y-8=0($a_{2}+b_{2}+c_{2}$)=0)

$a_{1} a_{2} +b_{1}b_{2}$=4(2)+2(-1))

8-1=7>0

$\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2} } }=$± ($\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2} } }$ )

+ will give obtuse angle bisector

- will give acute angle bisector

$\frac{4x+2{}y-9{}}{\sqrt{4{}^{2}+2_{1}^{2} } }= -$($\frac{2{}x+-1_{}y-8_{}}{\sqrt{2{}^{2}+(-1)_{}^{2} } }$

(4x+2y-9)/√16+4=-(2x-y-8/(√4+1))

4x+2y-9/√20=-(2x-y-8/√5)

4x+2y-9/2√5=-(2x-y-8)/√5

√5 gets cancel

4x+2y-9/2=-(2x-y-8)

4x+2y-9=-2(2x-y-8)

4x+2y-9=-4x+2y+16

4x+4x+2y-2y-9-16=0

8x-25=0