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Solution


Q.

Two vertices of a triangle ABC are A(3, –1) and B(–2, 3) and its orthocentre is P(1, 1). If the coordinates of the point C are (α,β) and the centre of the circle circumscribing the triangle PAB is (h, k), then the value of (α+β)+2(h+k) equals          [2024]
1 15  
2 5  
3 51  
4 81  

Ans.

(2)

Slope of AB = -1 - 33 + 2 = -45

Slope of PE = 54

Equation of PC is given y - 1 = 54 (x - 1)

⇒   4y – 4 = 5x – 5

⇒   5x – 4y – 1 = 0         ... (i)

Now slope of AP = -1 - 13 - 1 = -1

⇒ Slope of BC = 1

Equation of BC is given by

y – 3 = 1(x + 2)

⇒   x – y + 5 = 0         ... (ii)

Solving (i) and (ii), we get

β = y = 26             [∵  C is point of intersection of PC and BC]

and α = x = 21

Centre of circle circumseribing ∆PAB is point of intersection of perpendicular bisector of AP, AB and AC.

Equation of perpendicular bisector of AP

(y - (-1 + 1)2) = 1(x - (3 + 1)2)

⇒  y – 0 = x – 2

⇒  y = x – 2         ... (iii)

Equation of perpendicular bisector of AB

y - (3 - 12) = 54 (x - (-2 + 32))

⇒   y - 1 = 54 (x - 12)        ... (iv)

On solving (iii) and (iv), we get

x = h = -192  and  y = k= -232

⇒   2(h + k) = -42

So, (α+β) + 2(h + k) = 47 - 42 = 5.