We have given a point (2, 3), through which two lines are passing and intersects at an angle of 60°. Let m be the slope of the other line ⇒ tan 60° = $\left|\frac{m-2}{1+2m}\right|$ ⇒ ± $\sqrt{3}$ = $\frac{m-2}{1+2m}$ ⇒ $\sqrt{3}$ = $\frac{m-2}{1+2m}$ ... (i) or - $\sqrt{3}$ = $\frac{m-2}{1+2m}$ ... (ii) ⇒ $\sqrt{3}+2\sqrt{3}m$ = m - 2 or $\sqrt{3}+2\sqrt{3}m$ = 2 - m ⇒ $\sqrt{3}$ + 2 = m - $2\sqrt{3}$ m or $2\sqrt{3}$ m + m = 2 - $\sqrt{3}$ ⇒ m = $\frac{\sqrt{3}+2}{1-2\sqrt{3}}$ or m = $\frac{2-\sqrt{3}}{1+2\sqrt{3}}$ Since the line passes through (2, 3) ∴ Its equation is either y - 3 = $\frac{\sqrt{3}+2}{1-2\sqrt{3}} (x-2)$ ... (iii) or y - 3 = $\frac{2-\sqrt{3}}{1+2\sqrt{3}} (x-2)$ ... (iv) (iii) ⇒ (y − 3)(1 − 2$\sqrt{3}$) = ($\sqrt{3}$ + 2) (x − 2) ⇒ y (1 − 2$\sqrt{3}$) − 3(1 − 2$\sqrt{3}$) = ($\sqrt{3}$ + 2)x − 2($\sqrt{3}$ + 2) ⇒ y (1 − 2$\sqrt{3}$) − (3 − 6 $\sqrt{3}$) = ($\sqrt{3}$ + 2)x − 2 $\sqrt{3}$ −4 ⇒ 2 $\sqrt{3}$ + 4 − 3 + 6 $\sqrt{3}$ =($\sqrt{3}$ + 2)x +(2$\sqrt{3}$ − 1)y ⇒($\sqrt{3}$ + 2)x + (2$\sqrt{3}$ − 1)y− 8 $\sqrt{3}$ − 1 = 0 ... (v) And (iv) ⇒ (y − 3) (1 + 2$\sqrt{3}$) = (2 − $\sqrt{3}$) (x −2) ⇒ y(1 + 2$\sqrt{3}$) − 3 (1 + 2$\sqrt{3}$) = (2 − $\sqrt{3}$)x − 2(2 − $\sqrt{3}$) ⇒ − 3 − 6 $\sqrt{3}$ + 2(2 − $\sqrt{3}$) = (2 − $\sqrt{3}$)x− y(1+ 2 $\sqrt{3}$) ⇒ − 3 − 6 $\sqrt{3}$ +4 − 2 $\sqrt{3}$ =(2 − $\sqrt{3}$)x − y (1 + 2 $\sqrt{3}$) ⇒ (2 − $\sqrt{3}$)x −(1 + 2 $\sqrt{3}$)y + 8$√3$ − 1 = 0 ... (vi) Thus (v) and (vi) are the required equations of the line.