Question

the acute angle between the lines 7X-4y=0 and 3x-11y+5=0 is _______​

Answer 1

Answer:

The above picture is your answer

Step-by-step explanation:

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Answer 2

answer : angle between lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is 45°

lines are ; 7x - 4y = 0 and 3x - 11y + 5 = 0

slope of line 7x - 4y = 0 , m₁ = 7/4

slope of line 3x - 11y + 5 = 0, m₂ = 3/11

using formula, tanθ = |m₁ - m₂|/|1 + m₁.m₂|

= |7/4 - 3/11|/|1 + 7/4 × 3/11|

= |7 × 11 - 3 × 4|/|4 × 11 + 7 × 3 |

= |77 - 12|/|44 + 21|

= |65/65|

= 1

so, tanθ = 1 = tan45° ⇒θ = 45°

angle between lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is 45°

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