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The straight lines x+y-4=0,3x+y-4=0,x+3y-4=0 form a triangle with is A isosceles B right angle C equilateral D none of these In above equation how how to determine coordinate of the vertices

The straight lines x+y-4=0,3x+y-4=0,x+3y-4=0 form a triangle with is 
A isosceles
B right angle
C equilateral
D none of these
In above equation how how to determine coordinate of the vertices

Grade:11

2 Answers

Arun
25750 Points
5 years ago
Dear student
x+y = 0 … (1) slope = -1
3x+y = 4 … (2) slope = -3/1
x+3y = 4 … (3) slope = -1/3

So no two lines are perpendicular to each other as m1xm2 is not equal to -1 for any two lines.

From (1) and (2) intersection point is

A(2,-2)

From (2) and (3) intersection point is

B(1,1)

From (3) and (1) intersection point is
C(-2,2)

AC² = [(2-(-2)] ² + [-2-2] ² = 32
BC² = [1-(-2)] ²+[1-2] ² = 10
AB²= [2-1] ² + [-2-1] ² = 10
As BC² = AB² which means BC = AB and AC is different, the triangle is a isosceles triangle
Bhuvana
15 Points
3 years ago
Write the 3 eqns
Solve 1 and 2 first....we get x and y as 2,-2  let it beA
Solve 2 and 3 now...we get x and y as 1 ,1 let it be B
Solve 3 and 1 now.....we get x and y as -2,2 let it be C
Then,
      Find distances between AB,BC,CA
We get AB = root 10
              BA = root 10
               AC= 4root 2
Since, 2 sides have equal length.....the triangle is isosceles☺️
 We get,
AB= 

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