Let LRM is the perpendicular bisector of PQ at R .
[∵ Slope of line passing through (x1,y1) and (x2,y2) is
y2−y1
x2−x1
] R=(
1+k
2
,
4+3
2
) R=(
k+1
2
,
7
2
) [∵ mid-point of line with end points (x1,y1),(x2,y2)=(
x1+x2
2
,
y1+y2
2
) ] Also slope of PQ=
3−4
k−1
mPQ=
−1
k−1
∵PQ and LM are perpendicular to each other. ∴mLM=
−1
mPQ
=
−1
−
1
k−1
mLM=k−1 ∴ Equation of LM which has slope (k−1) has Y -intercept −4 ⇒ Equation of LM : y=(k−1)x−4 Since, Eq. (i) passes through R. Hence, point R will satisfy Eq. (i) y=(k−1)x−4 ⇒72=(k−1)(