Find b such that the points A(2,b),B(5,5)and C(-6,0) are the vertices of right angled  with angle < BAC = 90 degree.
2 answers:
The <BAC angle isn't a perfect right angle because the image was a photo
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Answer:
- b = -3 or b = 8
Step-by-step explanation:
We know that the perpendicular lines have negative-reciprocal slopes.
In other words, their product is - 1.
<u>Find the slopes of AB and AC:</u>
- m(AB) = (5 - b)/(5 - 2) = (5 - b) / 3
- m(AC) = (0 - b)/(- 6 - 2) = - b / - 8 = b/8
<u>Find the the product of slopes and solve for b:</u>
- (5 - b)/3 * b/8 = - 1
- b(b - 5)/24 = 1
- b(b - 5) = 24
- b² - 5b - 24 = 0
- b² - 8b + 3b - 24 = 0
- b(b - 8) + 3(b - 8) = 0
- (b + 3)(b - 8) = 0
- b = - 3 or b = 8
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The rest of the question is the attached figure.
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Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17
C.
A = A0e^-0.000103k