AP EAPCET 23-Aug-2021 Shift 2 Paper

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)(

k+1

2

)−4
⇒7=k2−1−8
⇒k2−16=0
⇒k=±4