Straight LineHard
Question
Let PQR be a right angled isosceles , right angled at P(2, 1) If the equation of the lineQR is 2x + y = 3 then the equation representing the pair of lines PQ and PR is
Options
A.3x2 - 3y2 + 8xy + 20x +10y + 25 = 0
B.3x2 - 3y2 + 8xy - 20x -10y + 25 = 0
C.3x2 - 3y2 + 8xy +10x +15y + 20 = 0
D.3x2 - 3y2 - 8xy - 10x - 15y - 20 = 0
Solution
Let S be the mid point of QR and given ᐃ PQR is an isosceles.
Therefore, PS ⊥ QR and S is mid point of hypotenuse,
therefore, S is equidistant fromP,Q,R.
∴ PS = QS = RS
Since, ∠ P = 90o
abd ∠Q = ∠R
But ∠P + ∠Q + ∠R = 180o
∴ 90o + ∠Q + ∠R = 180o
⇒ ∠Q = ∠R = 45o
Now, slope of QR is - 2
But QR ⏊ PS.
∴ Slope of PS is 1/2
Let m be the slope of PQ.
∴ tan
⇒
⇒ m = 3, - 1/3 Equation of PQ and PR are
y - 1 = 3(x - 2) and y - 1 = -
(x - 2)or 3(y - 1) + (x - 2) = 0
Therefore, joint equation of PQand PR is
[3(x - 2) - ( y -1)][(x - 2) + 3( y -1)] = 0
⇒ 3(x - 2)2 - 3( y -1)2 + 8(x - 2)( y -1) = 0
⇒ 3x2 - 3y2 + 8yx - 20x -10y + 25 = 0
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