The equation of the hypotenuse of an isosceles right angled triangle is $$x + 3y = 3.$$

The right angle is at the vertex $C(−2, 0)$.

(a) Find the two other vertices of the triangle.

(b) Find the equation of the circumscribed circle of the triangle.


Let $\Delta ABC$ is a given triangle.

Hence, the altitude to $AB$ is $$\frac{|1\cdot(-2)+3\cdot0-3|}{\sqrt{1^2+3^2}}=\sqrt{\frac{5}{2}},$$ which gives $AC=BC=\sqrt5$.

Now, for finding $A$ and $B$ we need to solve a system: $(x+2)^2+y^2=5$ and $x=3-3y$, which gives $y^2-3y+2=0$ and $(0,1)$ and $(-3,2)$.

Thus, the middle point of $AB$ it's $\left(-\frac{3}{2},\frac{3}{2}\right)$ and the equation of the circle is $$\left(x+\frac{3}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{5}{2}$$ Done!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.