Question

The points A=7(4,7),B(P,3)and C(7,3) are the vertices of a triangle,right angled at B.Find the value of P.

Answer 1

Given,

∆ABC is a rt. angled ∆, rt. angled at point

B.

Coordinates of point A = (4 , 7)

Coordinates of point B = (p , 3)

Coordinates of point C = (7 , 3)

Using distance formula, we get

➡ AB = √(p – 4)² + (3 – 7)²

➡ AB = √(p² + 16 – 8p + 16)

➡ AB = √(p² – 8p + 32) units

Now,

➡ BC = √(p – 7)² + (3 – 3)²

➡ BC = √(p² + 49 – 14p) units

And,

➡ AC = √(4 – 7)² + (7 – 3)²

➡ AC = √9 + 16

➡ AC = √25

➡ AC = 5 units.

Now,

Using Pythagoras theorem, we get

➡ AC² = AB² + BC²

➡ 5² = p² – 8p + 32 + p² + 49 – 14p

➡ 25 = 2p² – 22p + 81

➡ 2p² – 22p + 56 = 0

➡ p² – 11p + 28 = 0

Now, using quadratic formula, we get

➡ p = 11 ± √(121 – 112)/2

➡ p = 11 ± √(9)/2

➡ p = 11 ± 3/2

Here,

➡ p = 11 + 3/2

➡ p = 14/2

➡ p = 7

OR

➡ p = 11 – 3/2

➡ p = 8/2

➡ p = 4

So, we get p = 7 or 4.