show that the point (7,10),(-2,5)and (8-4) are the vertices of an isosceles right triangle
Answers
Answer:
This is your answer hope it's help u
Step-by-step explanation:
Given :-
The point (7,10),(-2,5)and (8-4)
To find :-
Show that the point (7,10),(-2,5)and (8-4) are the vertices of an isosceles right triangle ?
Solution :-
Given points are (7,10),(-2,5)and (8-4)
Using Concept :-
To show that the given points are the vertices of an Isosceles right angled triangle , then the satisfies the following conditions.
i) Any lengths of the two sides formed by the given points are equal.
ii) The sum of squares of any two sides is equal to the square of the third side.
Let A = (7,10)
Let B = (-2,5)
Let C = (8,-4)
Length of AB :-
Let (x1, y1) = (7,10) => x1 = 7 and y1 = 10
Let (x2, y2) = (-2,5) = x2 = -2 and y2 = 5
We know that
The distance between two points (x1, y1) and
( x2, y2) is √[(x2-x1)²+(y2-y1)²] sq.units
Distance between A and B
=>AB = √[(-2-7)²+(5-10)²]
=> AB = √[(-9)²+(-5)²]
=> AB = √(81+25)
AB = √ 106 units
Length of BC :-
Let (x1, y1) = (-2,5) => x1 = -2 and y1 = 5
Let (x2, y2) = (8,-4) = x2 = 8 and y2 = -4
We know that
The distance between two points (x1, y1) and
( x2, y2) is √[(x2-x1)²+(y2-y1)²] sq.units
Distance between B and C
=> BC = √[(8-(-2))²+(-4-5)²]
=> BC = √[(8+2)²+(-9)²]
=> BC = √[(10)²+(-9)²]
=> BC =√(100+81)
BC = √181 units
Length of AC :-
Let (x1, y1) = (7,10) => x1 = 7 and y1 = 10
Let (x2, y2) = (8,-4) = x2 = 8 and y2 = -4
We know that
The distance between two points (x1, y1) and
( x2, y2) is √[(x2-x1)²+(y2-y1)²] sq.units
Distance between A and C = AC
=> √[(8-7)²+(-4-10)²]
=> √[(1)²+(-14)²]
=> √(1+196)
=> √197 units
AC = √ 197 units
From (1),(2)&(3)
AB ≠ BC ≠ CA
They are not the vertices of an Isosceles triangle
Now,
AB = √106
=> AB² = [√106]² = 106
BC =√181
=>BC² = (√181)² = 181
and
AC = √197
=> AC² = (√197)² = 197
Sum of the squares of any two sides
=> AB²+BC²
=> 106+181
=> 287 ≠ 197
=> AB² + BC² ≠ AC²
So they are not the vertices of the right angled triangle.
Therefore, Given points are not the vertices of the Isosceles right angled triangle.
Answer :-
The points (7,10),(-2,5)and (8-4) are not the vertices of an isosceles right angled triangle.
Used Concept :-
To show that the given points are the vertices of an Isosceles right angled triangle , then the satisfies the following conditions.
i) Any lengths of the two sides formed by the given points are equal.
ii) The sum of squares of any two sides is equal to the square of the third side.
Used formulae:-
Distance Formula :-
The distance between two points (x1, y1) and
( x2, y2) is √[(x2-x1)²+(y2-y1)²] sq.units
Isosceles triangle :-
If the lengths of any two sides are equal then the triangle is an Isosceles triangle.
In ∆ ABC , AB = BC then ABC is an Isosceles triangle.
Pythagorous Theorem :-
In a right angled triangle The square of the hypotenuse is equal to the sum of the squares of any two sides.
In ∆ ABC, AB² + BC² = AC² then A,B,C forms a right angled triangle.
Note :-
If the points (7,10) ,(-2,5) and (3,-4) forms an Isosceles right angled triangle.