Question

in the given figure triangle pqr is an equilateral triangle with coordinates of q and r as (0,5) (0,-5) respectively.find the coordinates of vertex p.

Answer 1

The coordinates of vertex P should be (6,0).
Since you have not marked the no. on the x-axis, I have assumed it to be 6. The ordinate is definitely 0.

Answer 2

Let the coordinate of the point p be (x, y)

It is a equilateral triangle

⇒ All the lengths are equal

⇒ Distance RP= QP = RP

Find the length QR, given Q = (0, 5) and R = (0,-5):

$\text {Distance of QR} = \sqrt{(5+5)^2 + (0-0)^2}$

$\text {Distance of QR} = \sqrt{10^2$

$\text {Distance of QR} = 10$

Find the length of QP given Q = (0,5):

$\text {Distance of QP} = \sqrt{(y-5)^2 + (x-0)^2}$

$\text {Distance of QP} = \sqrt{(y-5)^2 + x^2}$

Find the length of RP given Q = (0,5):

$\text {Distance of RP} = \sqrt{(y+5)^2 + (x-0)^2}$

$\text {Distance of RP} = \sqrt{(y+5)^2 + x^2}$

Since all the 3 lengths are equal ⇒ QP = QR

√[(y - 5)² + x²] = 10

Square both sides:

(y - 5)² + x² = 100 ------------------- [ 1 ]

Since all the 3 lengths are equal ⇒ RP = QR

√[(y+5)² + x²] = 10

Square both sides:

(y + 5)² + x² = 100 ------------------- [ 2 ]

Put both equations together:

(y - 5)² + x² = 100 ------------------- [ 1 ]

(y + 5)² + x² = 100 ------------------- [ 2 ]

[1 ] - [ 2 ] :

(y - 5)²  - (y + 5)² = 0

[ (y - 5) + (y + 5) ]² [ (y - 5) - (y + 5) ]²  = 0

[ (y - 5 + y + 5 ]² [ (y - 5 - y - 5 ]²  = 0

[ 2y ]²  [-25 ]²  = 0

4y² (625) = 0

2500y² = 0

y = 0

When y = 0

(y - 5)² + x² = 100

(0 - 5)² + x² = 100

25 + x² = 100

x² = 100 - 25

x² = 75

x= √75

x = 5√3

Find the coordinate of p:

x = 5√3

y = 0

p = (5√3, 0)

Answer: p = (5√3, 0)