Question

Find the angle between lines y - √3x - 5 = 0 and √3y - x + 6 = 0.

MATHS | CLASS 11 | CH. STRAIGHT LINES pg. 223 |

Answer 1

slope of line 1,m1 = √3
slope of line 2,m2= 1/√3

Angle b/w two lines is x°, then we know
tan x = (m1- m2)/(1+ m1*m2)
tan x = (√3 -1/√3)/(1+√3*1/√3)
tan x = (3-1)/2*√3
tan x= 1/√3
>> x = 30° Ans

Answer 2

$Given :$

$y - \sqrt{3} x - 5 = 0 ---- (1)$

$y = \sqrt{3} x + 5$

It is in the form of y1 = m1x1 + c.

$m1 = \sqrt{3}$




$\sqrt{3} y - x + 6 = 0$  ------ (2)

$\sqrt{3} y = x - 6$

$y = \frac{1}{ \sqrt{3} } x - \frac{6}{ \sqrt{3} }$

$y = \frac{1}{ \sqrt{3} } x - 2 \sqrt{3}$

It is in the form of y2 = m2x2 + c.

$m2 = \frac{1}{ \sqrt{3} }$


Now,

Note: Here I am writing ∅ as A because it is difficult for me to write ∅ always.

Let A be the angle between the lines.Then

$tan A = \frac{m1 - m2}{1 + m1m2}$

                 $= \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} * \frac{1}{ \sqrt{3} } }$

                 $= \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + 1}$

                 $= \frac{2}{ \frac{ \sqrt{3} }{2} }$

                 $= \frac{2}{2 \sqrt{3} }$

                 $= \frac{1}{ \sqrt{3} }$


$Tan A = \frac{1}{ \sqrt{3} }$

A = 30.  ------- Which is an Acute angle

Thus, the Obtuse angle between the lines = 180 - 30

                                                                       = 150.



Therefore the angle between the lines is 30 or 150.


Hope this helps!