A line passes through a point A(1,2) and makes an angle of 60degree with x-axis and intersects line x+y=6 at point P.find AP.

Suppose the equation of line AP is y = mx + c
Since this line makes an angle of 60 degree with x-axis;
So the slope of line AP; m = tan 60° = 3
So the equation of line AP is y=3x+c
And this line passes through the point A(1, 2), so we have;
2 = 1×3+cc = 2-3
Therefore the equation of the line AP is y = 3x+2-3...(i)
The line AP intersects the line x + y = 6 i.e. y = 6 - x...(ii)
solving (i) and (ii) we get;
x = 4+33+1 and y = 53+23+1
So intersection point is P4+33+1,53+23+1.
Using the distance formula to find the distance between the point A(1, 2)  and P4+33+1,53+23+1, we have;
AP = 4+33+1-12+53+23+1-22= 4+3-3-13+12+53+2-23-23+12= 33+12+333+12= 93+12+273+12= 9+273+12= 363+12= 63+1
Therefore AP = 63+1


 

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