Let point (x1,y1) be on the line 3x+4y=5. ∴‌‌3x1+4y1=5 . . . (i) Also, ‌‌(x1−1)2+(y1−2)2=(x1−3)2+(y1−4)2 ⇒x12+y12−2x1−4y1+5=x12+y12−6x1−8y1+25 ⇒‌‌4x1+4y1=20 . . . (ii) On solving Eqs. (i) and (ii), we get x1=15,y1=−10