Question

If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.


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Answer 1

Answer:-

Given:-

A(4 ,3) , B (- 1 , y) and C (3, 4) are the vertices of a right angled triangle.

And, A is the right angle. So, the opposite side (BC) would be the Hypotenuse.

We know that,

In a right angled triangle,

(Base)² + (Perpendicular)² = (Hypotenuse)²

Let the Base be AC and Perpendicular be AB.

We know that,

Distance between two points (x₁ , y₁) & (x₂ , y₂) is:

$\red{ \sf \: \sqrt{(x_2 - x_1) ^{2} + ( {y_2 - y_1)}^{2} } }$

So,

For Base (AC);

Let

• x₁ = 4
• y₁ = 3
• x₂ = 3
• y₂ = 4

Hence,

$\sf \: AC \: = \sqrt{ {(3 - 4)}^{2} + ( {4 - 3)}^{2} } \\ \\ \\ \implies \sf \: AC \: = \sqrt{( - 1) ^{2} + {1}^{2} } \\ \\ \\ \implies \sf \: AC \: = \sqrt{1 + 1} \\ \\ \\ \implies \boxed{ \sf \: AC \: = \sqrt{2} }$

Similarly,

For Perpendicular (AB);

Let,

• x₁ = 4
• y₁ = 3
• x₂ = - 1
• y₂ = y.

$\implies \sf \: AB \: = \sqrt{( - 1 - 4 )^{2} + ( {y - 3)}^{2} } \\ \\ \\ \implies \sf \: AB \: = \sqrt{ {( - 5)}^{2} + {(y - 3)}^{2} } \\ \\ \\ \implies \boxed{ \sf \: AB\: = \sqrt{25 + ( {y - 3)}^{2} } }$

For Hypotenuse(BC),

Let,

• x₁ = - 1
• y₁ = y
• x₂ = 3
• y₂ = 4.

$\implies \sf \: BC \: = \sqrt{(3 - ( - 1)) ^{2} + (4 - y) ^{2} } \\ \\ \\ \implies \sf \: BC \: = \sqrt{(3 + 1 )^{2} + {(4 - y)}^{2} } \\ \\ \\ \implies \sf \: BC \: = \sqrt{ {4}^{2} + (4 - y) ^{2} } \\ \\ \\ \implies \boxed{\sf \: BC \: = \sqrt{16 + ( {4 - y)}^{2} } }$

Therefore,

$\implies \sf \: (AC) ^{2} + (AB) ^{2} = (BC)^{2} \: \\ \\ \\ \implies \sf \: {( \sqrt{2}) }^{2} + \big( \sqrt{ {25 + {(y - 3)}^{2}} } \big)^{2} = { \big( \sqrt{16 + {(4 - y)}^{2} }} \big)^{2} \\ \\ \\ \implies \sf \:2 + 25 + (y - 3) ^{2} = 16 + {(4 - y)}^{2}$

Using (a - b)² = a² + b² - 2ab we get,

$\implies \sf \: 27 + {y}^{2} + 9 - 6y = 16 + 16 + {y}^{2} - 8y \\ \\ \\ \implies \sf \:36 - 6y = 32 - 8y \\ \\ \\ \implies \sf \: - 6y + 8y = 32 - 36 \\ \\ \\ \implies \sf \:2y = - 4 \\ \\ \\ \implies \sf \: y = \frac{ - 4}{2} \\ \\ \\ \implies \boxed{ \sf \red{y = - 2 }}$

Answer 2

Answer:

Question :-

• If A (4,3), B (-1,y) and C (3,4) are the vertices of a right triangle ABC , right angled at A .Then find the value of y .

Answer :-

• The value of y=-2.

Given :-

• If A (4,3), B (-1,y) and C (3,4) are the vertices of a right triangle ABC , right angled at A.

To find :-

• The value of y .

Solution :-

• Lets first start conversation of your question

In the question given that If A (4,3),B (-1,y),C (3,4) are the vertices of a right triangle ABC , right angled at A. We should need to find the value of y .

So ,

• We know that,

${AC}^{2} + {AB}^{2} = {BC}^{2}$

• Which is ,

${base}^{2} + {perpend.}^{2} = {hyp}^{2}$

• Here we should use distance between two points formula.
• That is ,

$= \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} }$

1》

Let apply the value to perpendicular AB .

• Where x1=4,y=3,x2=-1,y2=y.

• Now applying, we get that,

• $AB = \sqrt{( - 1 - 4) {}^{2} + (y - 3) {}^{2} }$
• $AB = \sqrt{ { (- 5)}^{2} + (y - 3) {}^{2} }$
• $AB = \sqrt{25 + (y - 3) {}^{2} }$

2》

• Now for base AC.

• Where,

• x1=4,y1=3,x2=3,y2=4

• Now applying all the values for formula we get that,

$AC = \sqrt{(3 - 4) {}^{2} + (4 - 3) {}^{2} }$

• $AC = \sqrt{( { - 1)}^{2} + {1}^{2} }$
• $AC = \sqrt{2}$

Now ,

3》

• Hypothenuse BC
• Where,
• x1=-1,y1=y,x2=3,y2=4

• Now applying it for formula we get that,

$BC = \sqrt{(3 - ( - 1) {}^{2} + (4 - y) {}^{2} }$

• $BC = \sqrt{ {4}^{2} + (4 - y) {}^{2} } = \sqrt{16 + (4 - y) {}^{2} }$

Now ,

• All the values in the formula we get that,

• AC^2+AB^2=BC^2

$= ( \sqrt{2} ) {}^{2} + \sqrt{25 + (y - 3) {}^{2} } {)}^{2} = \sqrt{16 + (4 - y)^2} {}^{2}$

• $= 2 + 25 + (y - 3) {}^{2} = 16 + (4 - y) {}^{2}$

• According to (a-b)^2 formula we get that,

• $= 27 + {y}^{2} + 9 - 6y = 16 + 16 + {y}^{2} - 8y$

• $= 36 - 6y = 32 - 8y$

• $- 6y + 8y = 32 - 36 = 2y = - 4$

• $y = \frac{ - 4}{2} = - 2$

Therefore,

• The value of y is -2.

Conclusion :-

• See the given attachment for ABC right angled triangle.

Hope it helps u mate .

Thank you .