Question

points A(3,0), B(a,-2)and C(4,-1) are vertices of triangle ABC right angled at vertex A. Find value of a

Answer 1

$\textbf{Given:}$

$\text{ABC is a right angled triangle with vertices}$

$\text{A(3,0), B(a,-2) and C(4,-1)}$

$\textbf{To find:}$

$\text{Value of a}$

$\textbf{Solution:}$

$\text{First we find, lengths of the sides AB, BC and AC}$

$AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$AB=\sqrt{(3-a)^2+(0+2)^2}$

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

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

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

$AC=\sqrt{(3-4)^2+(0+1)^2}$

$AC=\sqrt{1+1}$

$AC=\sqrt{2}$

$\text{Since ABC is a right angled triangled at A, by pythagoras theorem}$

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

$(a-4)^2+1=(3-a)^2+4+2$

$a^2+16-8a+1=9+a^2-6a+6$

$17-8a=-6a+15$

$-8a+6a=15-17$

$-2a=-2$

$\implies\boxed{\bf\,a=1}$

$\therefore\textbf{The value of a is 1}$

Find more:

In triangle ABC angle BAC is 90 degree seg BL and seg CM are medians of triangle ABC. Then prove that 4( BL²+CM²)= 5 BC²

https://brainly.in/question/8096479

In triangle LMN, LM = 8cm, MN = 6cm and LMN = 90°

X and Y are the midpoints of MN and LN respectively.

Find YXN and YN.​

https://brainly.in/question/15654414

Answer 2

Answer:

hope it helps

please mark me brainliest

Step-by-step explanation:

Given:

\text{ABC is a right angled triangle with vertices}ABC is a right angled triangle with vertices

\text{A(3,0), B(a,-2) and C(4,-1)}A(3,0), B(a,-2) and C(4,-1)

\textbf{To find:}To find:

\text{Value of a}Value of a

\textbf{Solution:}Solution:

\text{First we find, lengths of the sides AB, BC and AC}First we find, lengths of the sides AB, BC and AC

AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}AB=

(x

1

−x

2

)

2

+(y

1

−y

2

)

2

AB=\sqrt{(3-a)^2+(0+2)^2}AB=

(3−a)

2

+(0+2)

2

AB=\sqrt{(3-a)^2+4}AB=

(3−a)

2

+4

BC=\sqrt{(a-4)^2+(-2+1)^2}BC=

(a−4)

2

+(−2+1)

2

BC=\sqrt{(a-4)^2+1}BC=

(a−4)

2

+1

AC=\sqrt{(3-4)^2+(0+1)^2}AC=

(3−4)

2

+(0+1)

2

AC=\sqrt{1+1}AC=

1+1

AC=\sqrt{2}AC=

2

\text{Since ABC is a right angled triangled at A, by pythagoras theorem}Since ABC is a right angled triangled at A, by pythagoras theorem

BC^2=AB^2+AC^2BC

2

=AB

2

+AC

2

(a-4)^2+1=(3-a)^2+4+2(a−4)

2

+1=(3−a)

2

+4+2

a^2+16-8a+1=9+a^2-6a+6a

2

+16−8a+1=9+a

2

−6a+6

17-8a=-6a+1517−8a=−6a+15

-8a+6a=15-17−8a+6a=15−17

-2a=-2−2a=−2

\implies\boxed{\bf\,a=1}⟹

a=1

\therefore\textbf{The value of a is 1}∴The value of a is 1

Find more:

In triangle ABC angle BAC is 90 degree seg BL and seg CM are medians of triangle ABC. Then prove that 4( BL²+CM²)= 5 BC²

https://brainly.in/question/8096479

In triangle LMN, LM = 8cm, MN = 6cm and LMN = 90°

X and Y are the midpoints of MN and LN respectively.

Find YXN and YN.

https://brainly.in/question/15654414