An isosceles right angled triangle whose sides are $1, 1, \sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the coordinates axes. The locus of its centroid is ${(3 x - y)}^{2} + {(x - 3 y)}^{2} = \frac{p}{9}$ . Then p =

Open in App

Solution

AS per the question
Hypotenuse cuts the cordinate axes
Let the required cordinates are (a,0) and (0,b)

and the Centroid be G(h,k)
$\to h = \frac{a}{3}, k = \frac{b}{3}$
$\to a = 3 h, b = 3 k$

Also the length of hypotenuse = $\sqrt{2}$
$\to a^{2} + b^{2} = 2$
$\to 9 h^{2} + 9 k^{2} = 2$
$h^{2} + k^{2} = \frac{2}{9}$

On comparing ith the given equation we get p=2

Suggest Corrections

Similar questions

An isosceles right angled triangle is inscribed in the circle $x^{2} + y^{2} = r^{2} .$ If the coordinates of an end of the hypotenuse are (a, b), the coordinates of the vertex are

Q. In the given figure vertices of

△ A B C

lie on y =

f (x) = a x^{2} + b x + c

. (vertex lies on y axis). The

△ A B C

is right angled isosceles triangle whose hypotenuse

A C = 8 u n i t s

, then

Q. Equation of one of the sides of an isosceles right angled triangle whose hypotenuse is

3 x + 4 y = 4

and the opposite vertex of the hypotenuse is

(2, 2)

, will be ?

Q. The equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of hypotenuse is (2, 2) is

Q. Find the equation of one of the sides of an isosceles right-angled triangle whose hypotenuse is given by

3 x + 4 y = 4

and the opposite vertex of the hypotenuse is

(2, 2) .

Join BYJU'S Learning Program

An isosceles right angled triangle whose sides are 1,1,√2 lies entirely in the first quadrant with the ends of the hypotenuse on the coordinates axes. The locus of its centroid is (3x−y)2+(x−3y)2=p9. Then p =

AS per the questionHypotenuse cuts the cordinate axesLet the required cordinates are (a,0) and (0,b)and the Centroid be G(h,k)→h=a3,k=b3→a=3h,b=3kAlso the length of hypotenuse =√2 →a2+b2=2 →9h2+9k2=2 h2+k2=29On comparing ith the given equation we get p=2

An isosceles right angled triangle whose sides are $1, 1, \sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the coordinates axes. The locus of its centroid is ${(3 x - y)}^{2} + {(x - 3 y)}^{2} = \frac{p}{9}$ . Then p =