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# Pls help asap

0
17
7
+101

The line y=mx bisects the angle between the two lines shown below. Find m.

Jan 26, 2024

#1
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i am finding answer of this question.

mcdvoice

Jan 26, 2024
#2
+129820
+1

m  =  tan ( [ arctan (1)  + arctan (1/7) ]  /   2 )  =    .5

Corrected

Jan 26, 2024
edited by CPhill  Jan 27, 2024
#3
+101
0

Thanks, but that answer is incorrect. I haven't learned tan and arctan yet, do you have another solution? Thanks again.

Vxritate  Jan 26, 2024
#4
+129820
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Sorry......punched it into  the  calculator incorrectly.....see my corrected answer

CPhill  Jan 27, 2024
#5
+129820
+1

Another  way

Construct  a circle at the origin with a radius of 1

Equation   is     x^2 + y^2  = 1   (1)

Slope  of line   y = x    =  1

So

x = y

Sub this into  (1)  for y

x^2 + x^2  = 1

2x^2 = 1

x^2 = 1/2

x= 1/sqrt 2      and y =1/sqrt 2

Construct a perpendicular  line to y = x    passing through  (1/sqrt 2, 1/sqrt 2)

The equation is

y = (-1) (x - 1/sqrt 2) + 1/sqrt2 =

y =  -x + 2/sqrt2

y = -x + sqrt 2       (2)

Now consider the line y = (1/7)x

Sub this into (1) for y

x^2 + (1/7 x )^2   =1

x^2 + 1/49 x^2   = 1

(50 /49) x^2   =1

x^2  = 49/50

x = 7  /sqrt  (50)

y = (1/7) (7/sqrt (50)  = 1 /sqrt (50)

Now construct a perpendicular line  to y =(1/7)x    passing through  ( 7/sqrt 50 , 1/sqrt 50)

The equation  is

y = (-7)(x - 7/sqrt50) + 1/sqrt (50)

y = -7x + 50/sqrt50

y = -7x + sqrt (50)     ( 3)

Find  the  x intersection of   (2)  and (3)

-x + sqrt 2 =  -7x + sqrt 50

6x = sqrt (50)  - sqrt 2

6x  = 5sqrt (2) - sqrt 2

6x = 4sqrt 2

x = (2/3)sqrt 2

And  y =  - (2/3)sqrt 2  + sqrt 2  =   (1/3)sqrt 2

The intersection  of these two lines  is   [ (2/3)sqrt 2  ,  (1/3)sqrt 2) ]

The slope of the line we  want (through BD)  =   [ (1/3)sqrt 2  ]  / [(2/3)sqrt 2 ]  =   1/2   =  .5 =  m

(Which we found by calculator  !!!)

Jan 27, 2024
#6
+101
0

Thanks CPhill i understand now!

Vxritate  Jan 27, 2024
#7
+758
0

I haven't even learnt arctan yet :( I'm still learning tan.

history  Jan 27, 2024