Show that the lines
x + y =0,
3x+y-4=0
and
x+3y-4=0 form
an isoceles triangle.
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Answered by
0
Answer:
x+y=0 ..... (i)
3x+y−4=0 ..... (ii)
x+3y−4=0 ....... (iii)
Solving lines (i) and (ii), we get
−x=−3x+4⟹x=2⟹y=−2
∴(i) and (ii) intersect at A=(2,−2)
Solving lines (ii) and (iii), we get
−3x+4=
3
4−x
⟹−9x+12=4−x⟹x=1⟹y=1
∴(ii) and (iii) intersect at B=(1,1)
Solving lines (i) and (iii), we get
−x=
3
4−x
⟹−3x=4−x⟹x=−2⟹y=2
∴(i) and (iii) intersect at C=(−2,2)
So, AB,BC,AC form a triangle ABC
Now, AB=
(1−2)
2
+(1+2)
2
=
10
BC=
(−2−1)
2
+(2−1)
2
=
10
AC=
(−2−2)
2
+(2+2)
2
=
32
∴AB=BC
Since, two sides of a triangle are equal then the triangle formed by A,B,C is isosceles triangle
HOPE THIS HELPS YOU!
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2
Answer:
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