Let the variable point be P(h,k). The distance from P to the point (4,3) can be expressed as: √(h−4)2+(k−3)2 The perpendicular distance from P to the line x+2y−1=0 is given by: |‌
h+2k−1
√12+22
|=|‌
h+2k−1
√5
| According to the problem, these two distances are equal: √(h−4)2+(k−3)2=|‌
h+2k−1
√5
| Squaring both sides, we get: (h−4)2+(k−3)2=‌
(h+2k−1)2
5
Expanding both sides: h2−8h+16+k2−6k+9=‌
h2+4hk+4k2−2h−4k+1
5
Multiply through by 5 to eliminate the fraction: 5(h2−8h+16+k2−6k+9)=h2+4hk+4k2−2h−4k+1 Simplify and collect like terms: 5h2+5k2−40h−30k+125=h2+4hk+4k2−2h−4k+1
Rearranging gives: 4h2+k2−4hk−38h−26k+124=0 Finally, replace (h,k) with (x,y) : 4x2+y2−4xy−38x−26y+124=0