Given that `P(3,1),Q(6. 5),` and `R(x , y)` are three points such that the angle `P Q R` is a right angle and the area of ` R Q P` is 7,
find the value of `4x-3y+5`

Explanation

Clearly, R lies on the circle with P and Q as end points of diameter.

Now, `PQ=sqrt((6-3)^2+(5-1)^2)=5`
`therfore` Radius, `r=2.5`
Now, area of triangle `PQR=(1/2)RM.PQ=7`
`rArrRm=(14)/(5)=2.8`,
which is not possible as RM cannot be more than radius Hence, no such triangle is possible.