If A (3,4), B (-2,3) and C (5,6) are the vertices of a triangle ABC, and D, E, F are mid points of sides AB, BC, CA resp., find lengths of line segments DE, EF and FD.
Answers
1
=m(x−x
1
)
y−3=m(x−2)
mx−y−2m+3=0
The above line is at a distance of
5
7
units from the point (3,2)
we know that the distance of a line ax+by+c=0 from (h,k) is given by
⇒d=
a
2
+b
2
ah+bk+c
Using the above formula we can write d=
5
7
⇒
1+m
2
3m−2−2m+3
=
5
7
⇒
1+m
2
m+1
=
5
7
squaring both sides we have ,
⇒25m
2
+25+50m=49+49m
2
⇒24m
2
−50m+24=0
⇒12m
2
−25m+12=0
⇒12m
2
−16m−9m+12=0
⇒4m(3m−4)−3(3m−4)=0
⇒(3m−4)(4m−3)=0
∴m=
3
4
or
4
3
∴ the equation of the line possible are
⇒
3
4
x−y−
3
8
−13=0 or
4
3
x−y−
4
6
+3=0
⇒4x−3y+1=0 or 3x−4y+6=0
Hence, the answer is 4x−3y+1=0 or 3x−4y+6=0.
Answer:
Answer
Let ABC be the triangle and D, E and F be the mid-point of BC, CA and AB respectively. We have to show triangle formed DEF is an equilateral triangle. We know the line segment joining the mid-points of two sides of a triangle is half of the third side.
Step-by-step explanation:
Therefore DE=
2
1
AB,EF=
2
1
BC and FD=
2
1
AC
Now, ΔABC is an equilateral triangle
⇒AB=BC=CA
⇒
2
1
AB=
2
1
BC=
2
1
CA
⇒DE=EF=FD
∴ΔDEF is an equilateral triangle.
solution