Question

7. The side BC of an equilateral 3ABC is parallel to x-axis. Find the slope of AB and
the slope of BC

Answer 1

side BC is parallel to x - axis. means, angle made by BC with x - axis is zero.
hence, slope of BC = tan0° = 0

we know, angle between two sides of an equilateral triangle equals 60°.
so, angle between side BC and side AB = 60°
let slope of AB = m and slope of BC = 0

now, we know
$\bold{\frac{|m_1-m_2|}{|1+m_1m_2|}=tan\theta}$
here m1 is the slope of first line and m2 is the slope of 2nd line . theta is angle between two lines (or their slopes ).
put m1 = m and m2 = 0 and theta= 60°
then,
$tan{ \theta} = \frac{m }{1 + 0}$
=> tan60° = m
=> m = tan60° = √3

hence, slope of AB = √3
slope of BC = 0

Answer 2

Solution :

It is given that ,

∆ABC is an equilateral triangle.

BC is parallel to X - axis .

Inclination of AB = x° = 60°

Slope of AB = m

m = tan x°

=> m = tan 60°

m = √3

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