Question

7. Show that the points (1, 7), (2, 4) and (5, 5)
are the vertices of an isosceles right angled
triangle.​

Answer 1

$\Large{\underline{\underline{\mathfrak{\red{\bf{Solution}}}}}}$

$\Large{\underline{\mathfrak{\orange{\bf{Given}}}}}$

• Vertices of isosceles triangle (1,7) , (2,4) , (5,5)

$\Large{\underline{\mathfrak{\orange{\bf{Prove}}}}}$

• Given point be are vertices of isosceles triangle

$\Large{\underline{\underline{\mathfrak{\red{\bf{Explanation}}}}}}$

Let,

ABC be a isosceles triangle.

Where, Vertices are A(1,7) , B(2,4) , C(5,5)

we prove here these vertices are isosceles triangle .

For this if length of any two side of triangle is equal , then we can say that our triangle is isosceles .

Distance Formula

$\small\boxed{\tt{\green{\:Distance\:=\:\sqrt{(x-x1)^2+(y-y1)^2}}}}$

Where,

• (x,y) & (x1, y1) are any point.

Now, First calculate AB

$\mapsto{\:AB\:=\:\sqrt{(1-2)^2+(7-4)^2}} \\ \\ \mapsto{\:AB\:=\sqrt{(-1)^2+(3)^2}} \\ \\ \mapsto{\:AB\:=\sqrt{1+9}} \\ \\ \mapsto{\:AB\:=\sqrt{10}}$

Now, Calculate BC

$\mapsto{\:BC\:=\:\sqrt{(2-5)^2+(4-5)^2}} \\ \\ \mapsto{\:BC\:=\:\sqrt{(-3)^2+(-1)^2}} \\ \\ \mapsto{\:BC\:=\:\sqrt{9+1}} \\ \\ \mapsto{\:BC\:=\:\sqrt{10}}$

Now, Calculate CA

$\mapsto{\:CA\:=\:\sqrt{(5-1)^2+(5-7)^2}} \\ \\ \mapsto{\:CA\:=\:\sqrt{(4)^2+(-2)^2}} \\ \\ \mapsto{\:CA\:=\:\sqrt{16+4}} \\ \\ \mapsto{\:CA\:=\:\sqrt{20}}$

We can see here,

Distance between two side of triangle are equal .

• AB = BC ≠ CA

$\Large{\underline{\underline{\mathfrak{\red{\bf{Hence}}}}}}$

• We can say that this vertices of an isosceles triangle .

________________

Answer 2

Answer:

Step-by-step explanation: