Find Equation of line passing through (1,1) making an angle 45° with the line 2x +3y =10
Answers
Answer: x+5y+4 = 0
Step-by-step explanation:
•Given the point is (1,1) and the angle is 45° with line 2x+3y = 10.
•Let m1 be slope of AB and m2 be slope of CD.
•Given angle between CD &AB is 45°
•Now,Angle between two lines is
Tanθ = |(m2-m1)/(1+m1m2)|
•Putting the angle as 45°,we get
Tan45° = |(m2-m1)/(1+m1m2)|
1 = |(m2-m1)/(1+m1m2)| ...(1)
•Given line 2x+3y = 10.
3y = 10-2x
y = (10/3)-(2/3)x
y = (-2/3)x+(10/3)
•The above equation is in the form of y= mx+c
Where m is the slope
•Therefore,m = (-2/3)= m1
•Substituting m1=(-2/3) in (1),we get
1= |(m2-(-2/3))/(1+(-2/3)m2)|
1= |(m2+(2/3))/(1-(2/3)m2)|
m2+(2/3) = 1-(2/3)m2
m2(1+(2/3)) = 1-(2/3)
(5/3)m2 = (1/3)
5m2 = 1
m2 = (1/5).
•Equation of line passing through(x1,y1) with slope m is
(y-y1) = m(x-x1)
•Equation of line passing through (1,1) with slope m2 is
(y-1)=m2(x-1)
(y-1) =(1/5)(x-1)
x-1 = 5(y-1)
x-1 = 5y-5
x-5y+4 = 0
•Hence,the required equation of line is
x+5y+4 = 0