xsquareroot ÷(a+1) x+b are - 2 & 3
Answers
Step-by-step explanation:
y - \sqrt{3} x - 5 = 0 ---- (1)y−
3
x−5=0−−−−(1)
y = \sqrt{3} x + 5y=
3
x+5
It is in the form of y1 = m1x1 + c.
m1 = \sqrt{3}m1=
3
\sqrt{3} y - x + 6 = 0
3
y−x+6=0 ------ (2)
\sqrt{3} y = x - 6
3
y=x−6
y = \frac{1}{ \sqrt{3} } x - \frac{6}{ \sqrt{3} }y=
3
1
x−
3
6
y = \frac{1}{ \sqrt{3} } x - 2 \sqrt{3}y=
3
1
x−2
3
It is in the form of y2 = m2x2 + c.
m2 = \frac{1}{ \sqrt{3} }m2=
3
1
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}tanA=
1+m1m2
m1−m2
= \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} * \frac{1}{ \sqrt{3} } }=
1+
3
∗
3
1
3
−
3
1
= \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + 1}=
1+1
3
−
3
1
= \frac{2}{ \frac{ \sqrt{3} }{2} }=
2
3
2
= \frac{2}{2 \sqrt{3} }=
2
3
2
= \frac{1}{ \sqrt{3} }=
3
1
Tan A = \frac{1}{ \sqrt{3} }TanA=
3
1
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!