Question

Prove that the points (-3,5), (6,-1), (10,5) are the vertices of a right angled triangle. Also, find the length of the hypotenuse and area of the triangle.

Answer 1

Answer:

Hypotenuse = 13 and Area = 39

Step-by-step explanation:

Points (-3,5),(6,-1),(10,5)

We need to find the distance between the points

Firstly,

(-3,5),(6,-1) representation (x1, y1), (x2, y2)

Distance between points

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

Distance between points (-3,5),(6,-1)

$D1 = \sqrt{ {(6 + 3)}^{2} + {( - 1 - 5)}^{2} } \\ = \sqrt{81 + 36} = \sqrt{117}$

Distance between (6,-1),(10,5)

$D2 = \sqrt{ {(10 - 6)}^{2} + { ( 5 + 1)}^{2} } \\ = \sqrt{16 + 36} \\ = \sqrt{52}$

Distance between (10,5)(-3,5)

$D3 = \sqrt{ {( - 3 - 10)}^{2} + {(5 - 5)}^{2} } \\ = \sqrt{169} \\ = 13$

We know that

Hypotenuse^2 = base^2 +height^2

We can clearly say that 13 is greater than root(52) and root(117)

So,

${13}^{2} = { \sqrt{52} }^{2} + { \sqrt{117} }^{2} \\ 169 = 52 + 117 \\ 169 = 169$

Hence they are vertices of right-angled triangle

Hypotenuse is 13

Area of right-angled triangle =

$\frac{1}{2} \times \sqrt{52} \times \sqrt{117} \\ = \frac{1}{2} \times 2 \sqrt{13} \times 3 \sqrt{13} \\ = \frac{1}{2} \times 6 \times 13 = 39$