P and Q are two points on the line x-y+1=0 and at the distance of 5 units from origin. find the area of ∆OPA.

This problem is divided into two parts.

At first we have to find the coordinates of P and Q.

Assuming that you mean “the points P and Q are both distant 5 from the origin”, the locus of points which are distant 5 from the origin is a circumference whose equation is:
2+2=(5)2

We find the coordinates of P and Q as the intersection between the circumference and the line −+1=0
on which they are located.

The solution of the linear system is (3,4)
and (−4,−3)

So P(3,4) and Q(-4,-3)

The following step consists in calculating the area of a triangle with base PQ and height equal to the distance of the line from the origin, that can be calculated with the formula:

ℎ=|0−0+1|(1)2+(−1)2√

where 0
and 0

are the coordinates of the origin. Hence, the distance is equal to 2√2

and the base PQ is equal to (3−(−4))2+(4−(−3))2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=72‾√
The area of the triangle OPQ is now equal to 12∗ℎ=72

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