P and Q are two points on the line x - y + 1 = 0 and are at distance 5 from the origin. Find the area of the triangle OPQ.

P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.

Step-by-step explanation:

P and Q are the intersection points of

x-y+1 = 0 and the circle x^2 + y^2 = 25

sub y = x+1 into the circle

x^2 + (x+1)^2 = 25

x^2 + x^2 + 2x + 1 - 25 = 0

x^2 + x - 12 = 0

(x+4)(x-3) = 0

x = 3 or x = -4

y = 4 or y = -3

so P(3,4) and Q(-4,3) are our two points

Height of triangle.

h = |0 - 0 + 1|/√2 = 1/√2

PQ = √( (-7)^2 + 1^2) = √50 = 5√2

area POQ = (1/2)(1/√2)(5√2) = 5/2 square units

hope this helped

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

translate k to n and reflect across the line containing jk.