Question

find the value of k if the angle between the straight line 4x-y+7=0 ,kx-5y-9=0 is 45 degrees

Answer 1

Answer:

3 or - 25/3

Step-by-step-solution:

Using y = mx + c, slopes of given lines can be given.

Slope of 4x - y + 7 = 0 is 4.

Slope of kx - 5y + 9 = 0 is k/5

Using,

tanA = | (m-n)/(1+mn) |, where m and n are slopes of two lines and A is angle between them.

Here, angle between them is 45°

= > tan45° = | (k/5 - 4)/(1 + 4*k/5) |

= > 1 = | (k/5 - 4)/(1 + 4k/5) |

= > 1 = | {(k - 20)/5}/{(5 + 4k)/5} |

= > 1 = | ( k - 20 )/( 5 + 4k )

= > ± 1 = (k - 20) / (5 + 4k)

Case 1:

→ 1 = (k - 20)/(5 + 4k)

→ 4k + 5 = k - 20

→ 4k - k = - 20 - 5

→ k = -25/3

Case 2:

→ - 1 = (k - 20)/(5 + 4k)

→ - 5 - 4k = k - 20

→ 20 - 5 = 4k + k

→ 3 = k

Answer 2

Answer:

3 or - 25/3

Step-by-step explanation:

y = mx + c, slopes of given lines can be derived which are

Slope of 4x - y + 7 = 0 is 4.

Slope of kx - 5y + 9 = 0 is k/5

as we know , tanA = | (m-n)/(1+mn) |, (m,n are the slopes and A is the angle between the lines)

 

= > tan45° = | (k/5 - 4)/(1 + 4*k/5) |

= > 1 = | (k/5 - 4)/(1 + 4k/5) |

= > 1 = | {(k - 20)/5}/{(5 + 4k)/5} |

= > 1 = | ( k - 20 )/( 5 + 4k )

= > ± 1 = (k - 20) / (5 + 4k)

Case 1: if 1 is positive  

1 = (k - 20)/(5 + 4k)

4k + 5 = k - 20

4k - k = - 20 - 5

k = -25/3

Case 2: if 1 is negative  

- 1 = (k - 20)/(5 + 4k)

- 5 - 4k = k - 20

20 - 5 = 4k + k

3 = k