Question

Find the angle between the lines 4B and BC
where i) 40,-1), B(2, 1), C (0,3).
ii) A(0,0), B(√3,3), C(-√3,3)​

Answer 1

Answer:

A(0,-1), B(2,1)

So slope of AB = m1 =( y2-y1)/x2-x1)

={ 1-(-1)}/(2–0) = 2/2 = 1

B(2,1) C (0,3)

Slope of BC = m2 = (y2-y1)/(x2-x1)

= (3–1)/(0–2) = 2/-2 = -1

Here m1×m2 = 1×(-1) = -1.

So the line are perpendicular.

ie.angle between AB and BC is 90°