This problem can be solved by 2 methods.

Method 1:-

r = -i + s(cosai + sinaj)

or, r = (scosa - 1)i + ssinaj------I

r = i + t(-sinai + cosaj)

or, r = (1 - tsina)i + tcosaj-------II

At the pt. of intersection of st. lines I & II,

(scosa - 1)i + ssinaj = (1 - tsina)i + tcosaj

=>scosa - 1 = 1 - tsina-------III

& ssina = tcosa-------IV

IV=> t = stana-------V

III & V=> scosa - 1 = 1 - s(sin a)^2 * seca

=> s[cosa + (sina)^2/cosa] = 2

=> s = 2cosa-------VI

V & VI=>

t = 2cosa*tana

=> t = 2sina--------VII

I & VI=>

r = [2(cosa)^2 -1]i + 2cosa*sinaj

=> r = cos2ai + sin2aj--------VIII

Therefore, the pt. of intersection is (cos2a,sin2a).

Method 2:-

Put r = xi +yj in both the eqs.

Then, equating the coefficients of i & j for both the eqs. & solving them, we get the eqs. of the 2 st. lines as follows:-

y = tana(x+1)-----A

& y = cota(1-x)-----B

Then, their pt. of intersection can be easily found out.