the base QR of an equil. triangle PQR lies on x axis .the coordinate of the point are (-4,0) and origin is the midpoint of the base .find the coordinate of the point P and R.
pls solve
Answers
Step-by-step explanation:
We have,
PQR is an equilateral triangle such that QR lies on x−axis.
O is the midpoint of QR.
So,
OQ=QR=4 units.
⇒ Coordinates of point R are (4,0).
Now
Point P lies on the y−aLet the coordinate of point P be (0
So,
PQ=QR
⇒ √4^2 + y ^2 = √ 8^2
⇒ √16+y^2=√64
On squaring both side and we get,
16+y^2=8
on solving we get
y=±4√3
Hence, the coordinate of points P are (0,4√3) above x axis
and (0,-4√3) below x axis
HOPE IT WOULD HELP U
Answer:
We have,
PQR is an equilateral triangle such that QR lies on x−axis.
O is the midpoint of QR.
So,
OQ=QR=4 units.
⇒ Coordinates of point R are (4,0).
Now,
Point P lies on the y−axis.
Let the coordinate of point P be (0,y).
So,
PQ=QR
⇒
(0+4)
2
+(y−0)
2
=
(4+4)
2
+0
2
⇒
16+y
2
=
64
Onsquaringbothsideandweget,
⇒16+y
2
=64
⇒y
2
=64−16
⇒y
2
=48
⇒y=±4
3
Hence, the coordinate of points P are (0,4
3
) above x−axis.
And (0,−4
3
) below x−axis.
Hence, this is the answer.
solution