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

Question

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

My Attempt:

$\triangle OPQ$ is an isosceles triangle.

We know,

$$Ar.(\triangle OPQ)=\dfrac {b}{4} \sqrt {4a^2-b^2}$$ where, $a=$equal arms and $b=$ remaining side

How do I proceed further?

P.S:The shortest possible approach is appreciated

Answer 1

The following way should be easier.

We know that $P,Q$ are both on the line $y=x+1$ and on the circle $x^2+y^2=5^2$.

Eliminating $y$ from the system gives $$x^2+(x+1)^2=5^2\iff x^2+x-12=0\iff (x+4)(x-3)=0$$

So, letting $S(0,1)$, we know that the area of $\triangle{PQR}$ is $$[\triangle{OPQ}]=[\triangle{OSP}]+[\triangle{OSQ}]=\frac{1}{2}\times 1\times (3+|-4|)=\color{red}{\frac 72}$$

Answer 2

Let the y cordinate be $a $ thus x co ordinate becomes $a-1$. Now we know $(a-1)^2+a^2=25$ as its a quadratic it will give two values of a (4,-3 specifically) that is two y coordinates . From this we can get the x coordinates. Now we know the three vertices of the triangle. Thus the side lengths. Hence the area of triangle is $\sqrt {s (s-a)(s-b)(s-c) } $. Where s is the semiperimeter. $s=\frac {a+b+c}{2} $

Answer 3

Writing the equation of the line in normal form as $(1,-1)\cdot(x,y)=-1$, we see that the perpendicular distance of the line from the origin is $$\left|{-1\over\|(1,-1)\|}\right|=\frac1{\sqrt2}.$$ This is the altitude of the triangle. The length of side $PQ$ can be found using the Pythagorean theorem: $$|PQ|=2\sqrt{|OP|^2+\left(\frac1{\sqrt2}\right)^2}=2\sqrt{5^2-\frac12}=7\sqrt2$$ and so the area of $\triangle{OPQ}$ is $\frac12\cdot7\sqrt2\cdot\frac1{\sqrt2}=\frac72.$