Question

Find the angle between the straight line 5x+2y+1 and 3x-4y+3

Answer 1

Answer:

8x-2y+4

Step-by-step explanation:

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Answer 2

If slope of two line are m₁ and m₂ , then angle between them :-

$\boxed{ \boxed{ \rm \large tan \theta = \bigg | \frac{m_1 - m_2 }{1 +m_1 m_2 } \bigg | }} \\ \\ \\ \rm where \: \bold{\theta} \: is \: \: acute \: angle \: between \: two \: lines.$

Now we will find slope of line 5x+2y+1=0

➝ 5x+2y+1=0

➝ 2y = -5x -1

➝ y = (-5/2)x - (1/2)

We have transformed 5x+2y+1=0 in the form of y = mx+c.

Therefore, slope of line 5x+2y+1=0 :- -5/2

Now we will find slope of line 5x+2y+1=0

➝ 5x+2y+1=0

➝ 2y = -5x -1

➝ y = (-5/2)x - (1/2)

We have transformed 5x+2y+1=0 in the form of y = mx+c.

Therefore, slope of line 5x+2y+1=0 :- -5/2

Now we will find slope of line 3x-4y+3=0

➝ -4y= -3-3x

➝ 4y= 3+ 3x

➝ y= (3/4)+ (3/4)x

➝ y= (3/4)x + (3/4)

We have transformed 3x-4y+3=0 in the form of y = mx+c.

Therefore, slope of line 3x-4y+3=0 :- 3/4

$\rm \longrightarrow\large tan \theta = \bigg | \dfrac{m_1 - m_2 }{1 +m_1 m_2 } \bigg | \\ \\ \\ \rm \: put \: \: m_1 = \dfrac{ - 5}{2} \: and \: m_2 = \dfrac{3}{4} \\ \\ \\ \rm \longrightarrow\large tan \theta = \bigg | \dfrac{ \frac{ - 5}{2} - \frac{3}{4} }{1 +( \frac{ - 5}{2} \times\frac{3}{4})} \bigg | \\ \\ \\ \rm \longrightarrow\large tan \theta = \bigg | \dfrac{ \frac{ - 13}{4} }{( \frac{ - 7}{8})} \bigg |\\ \\ \\ \rm \longrightarrow\large tan \theta = \bigg | \dfrac{ - 13 \times 8 }{ - 7 \times 4} \bigg |\\ \\ \\ \rm \longrightarrow\large tan \theta = \bigg | \dfrac{ 13 \times 2 }{ 7 \times 1} \bigg |\\ \\ \\ \rm \longrightarrow\large tan \theta = \dfrac{ 26 }{ 7 } \\ \\ \\ \rm \longrightarrow\large \theta = {tan}^{ - 1} \dfrac{ 26 }{ 7 }$