Figure OPQR is a square. Point O is the origin, and point Q has coordinates (2,2). What are the coordinates for T so that the area of triangle PQT equals the area of square OPQR? (Picture not drawn to scale)

(A) (−6,0) (B) (−4,0) (C) (−2,0) (D) (2,0) (E) (4,0)

C is the answer, based on coordinates T = -2,0 they both distribute better as QPT = RA triangle base 10cm and point of origin is central to the base at point O. Length PO = Length TO

Therefore, Coordinates Q (2,2) can prove T = (-2.0) as xy is 2 units from point of origin for both sides, each.

Step-by-step explanation:

Proof same length/area.

Lets just say area is 25 and all lengths of 5 each 5x5 -25

Then for triangle QPT to have same area it would need to have one side 5 and then we work a second right angle at OQ = diagonal =  √ 25+ √ 25 =  √ 50 =  √ QP +  √ PO =  √ OQ Diagonal  =  √ 50 = 7.07106781187

OQ = 7.07

Cos (45)degree x 7.07106781187  = 5

We just need to redo as a long rectangle to find the longer diagonal using cos also.

We make a new letter above the T called S

SRQ = 15 sq +5sq =  √225 + √25

= SRQsq √250 = 15.81cm

We can check area OPQR = 5x5 = 25

We can now check area  QPT 7.5 x 5= 37.5

We can see its not to scale so length PT must be 10 to 1/2 and be 10 value to share same area. Area QPT = 1/2 10 x 5 = 25 and now match QPT.

coordinates 2, 2 = Q

coordinates -8, -3 = T = as we found -10 in run and -5 in rise

But as Q = 2,2 meaning 2 in run from origin zero this means equal side from origin would be found at T.

Therefore T = -2, 0

We do not need to count the length we only need to match Q and recognize Q's value to the origin =x= 2

T therefore = x=-2

All options show y value 0, which means the scale of the square = 2cm