Question

If there is a point p in a triangle and I draw perpendiculars from point p to all sides then what can you conclude.​

Answer 1

Answer:

Let the variable line be ax + by + c = 0 …(i)

We know that the perpendicular distance from the point (x1, y1) to a line ax + by + c = 0 is given by d = |ax1 + by1 + c1|/√(a2 + b2)

Perpendicular distance between (i) and (2, 0) = |2a + 0 + c|/√(a2 + b2)

= |(2a + c)|/√(a2 + b2)

Perpendicular distance between (i) and (0, 2) = |0 + 2b + c|/√(a2 + b2)

= |(2b + c)|/√(a2 + b2)

Perpendicular distance between (i) and (1, 1) = |a + b + c|/√(a2 + b2)

= |(a + b + c)|/√(a2 + b2)

Adding the distances we get

(3a + 3b + 3c)/√(a2 + b2) = 0

=> a + b + c = 0 …(ii)

Comparing (i) and (ii), we get

x = 1, y = 1

So the point is (1, 1).

Hence option (2) is the answer.