# If A(1;-2) B(3;t) C(9;-3) determine t for the angle BAC=90 degrees

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### 3 Answers

We have the three points A(1,-2) B(3, t) and C(9,-3).

We have to find t for the angle BAC to be equal to 90 degree.

Now in the triangle ABC if , angle BAC is equal to 90 degree, according to the Pythagorean Theorem, we have

BC^2 = AB^2 + AC^2

BC^2 = (3 - 9)^2 + (t + 3)^2 = 36 + (t + 3)^2

AB^2 = (1 - 3)^2 + (-2 - t)^2 = 4 + (-2 - t)^2

AC^2 = ( 1- 9)^2 + (-2 +3)^2 = 64 + 1 = 65

So 36 + (t + 3)^2 = 4 + (-2 - t)^2 + 65

=> t^2 + 9 +6t + 36 = 4 + 4 + t^2 + 4t + 65

=> 2t = 28

=> t = 28/2

=> t = 14

**Therefore t = 14.**

If A(1,-2), B(3,t) and C(9,-3) determine t for the angle BAC=90

degrees.

We determine the slope of AB and AC :

Slope of AB = m1= (yB-YA)/(xB-xA) = (t-(2))/((3-1) = (t+2)/2

Slope of AC= m2 = (yC-yA)/(xC-xA) = (-3-(-2))/(9-1) = -1/8.

Since Angle BAC = 90 degrees +m1*m2 = -1.

=> {(t+2)/2}{-1/8) = -1.

=> -(t+2) = -16.

=> t+2 = 16.

=> t= 16-2 = 14.

Therefore t = 14.

If the measure of the angle BAC = 90 degrees, then the triangle ABC is a right angle triangle and BC is the hypotenuse.

We'll apply the Pythagorean theorem in the triangle ABC:

BC^2 = AB^2 + AC^2

BC^2 = (xC - xB)^2 + (yC - yB)^2

BC^2 = (9 - 3)^2 + (-3 - t)^2

BC^2 = 36 + 9 + 6t + t^2

BC^2 = 45 + 6t + t^2 (1)

AC^2 = (xC - xA)^2 + (yC - yA)^2

AC^2 = (9-1)^2 + (-3+2)^2

AC^2 = 64 + 1

AC^2 = 65 (2)

AB^2 = (xB - xA)^2 + (yB - yA)^2

AB^2 = (3 - 1)^2 + (t+2)^2

AB^2 = 4 + t^2 + 4t + 4

AB^2 = t^2 + 4t + 8 (3)

We'll substitute (1) , (2) , (3) in the formula of Pythagorean Theorem:

BC^2 = AB^2 + AC^2

45 + 6t + t^2 = t^2 + 4t + 8 + 65

We'll combine and eliminate like terms:

45 + 6t - (4t + 8 + 65) = 0

2t - 28 = 0

2t = 28

**t = 14**