Suppose, in triangle △ABC,x−y+5=0,x+2y=0 are respectively the equations of the perpendicular bisectors of the sides AB and AC. If A is (1,−2), the equation of the line joining B and C is
Let the perpendicular bisectors x−y+5=0 and x+2y=0 of the sides AB and AC intersect at D and E, respectively.
Let (x1,y1) and (x2,y2) be coordinates of B and C. Thus, the coordinates of D=(‌
x1+1
2
,‌
y1−2
2
) and the coordinates of E=(‌
x2+1
2
,‌
y2−2
2
) The point D lies on the line x−y+5=0 x1−y1+13=0........(i) The point E lies on the line x+2y=0 x2+2y2−3=0.......(ii) Since, side AB perpendicular to line x−y+5=0 ‌⇒‌‌1×(‌
y1+2
x1−1
)=−1 ‌⇒‌‌x1+y1+1=0.......(iii) Similarly, side AC is perpendicular to the line ‌x+2y=0 ⇒2x2−y2=4......(iv) Solving Eqs: (i) and (iii), we get x1=−7, and y1=6 Solving Eqs. (ii) and (iv), we get x2=‌
11
5
,y2=‌
2
5
Thus, coordinate of C are (‌
11
5
,‌
2
5
) and coordinate of B are (−7,6). Thus, equation of the line BC is ‌(y−6)=‌