If the orthocentre of the triangle whose vertices are $(1,2)$, $(2,3)$ and $(3,1)$ is $(\alpha,\beta)$, then the quadratic equation whose roots are $\alpha+4\beta$ and $4\alpha+\beta$ is:
If $(a,b)$ be the orthocentre of the triangle whose vertices are $(1,2)$, $(2,3)$ and $(3,1)$, and
$I_1=\displaystyle\int_a^b x\sin(4x-x^2)\,dx,\ \ I_2=\displaystyle\int_a^b \sin(4x-x^2)\,dx,$
then $36\,\dfrac{I_1}{I_2}$ is equal to:
The equations of the sides $AB$, $BC$ and $CA$ of a triangle $ABC$ are $2x+y=0$, $x+py=39$ and $x-y=3$ respectively and $P(2,3)$ is its circumcentre. Then which of the following is NOT true?
A circle is inscribed in an equilateral triangle of side $12$. If the area and perimeter of any square inscribed in this circle are $m$ and $n$, respectively, then $m+n^{2}$ is equal to:
Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q$\left(k_{1}, k_{2}\right)$, then $k_{1}+k_{2}$ is equal to :
Let $A(\alpha,-2)$, $B(\alpha,6)$ and $C\!\left(\dfrac{\alpha}{4},-2\right)$ be vertices of $\triangle ABC$.
If $\left(5,\dfrac{\alpha}{4}\right)$ is the circumcentre of $\triangle ABC$, then which of the following is NOT correct about $\triangle ABC$?
Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $\Delta$PQR is :
Let $\triangle PQR$ be a triangle. The points $A, B,$ and $C$ are on the sides $QR, RP,$ and $PQ$ respectively such that
$\dfrac{QA}{AR}=\dfrac{RB}{BP}=\dfrac{PC}{CQ}=\dfrac{1}{2}$.
Then $\dfrac{\operatorname{Area}(\triangle PQR)}{\operatorname{Area}(\triangle ABC)}$ is equal to:
Let k be an integer such that the triangle with vertices (k,-3k), (5,k) and (-k,2) has area 28 sq. units. Then the orthocentre of this triangle is at the point :
In a $\triangle ABC$, suppose $y = x$ is the equation of the bisector of the angle $B$ and the equation of the side $AC$ is $2x - y = 2$.
If $2AB = BC$ and the points $A$ and $B$ are respectively $(4,6)$ and $(\alpha, \beta)$,
then $\alpha + 2\beta$ is equal to:
Let $P Q R$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the mid point of $\mathrm{PQ}$. The distance of the centroid of $\triangle \mathrm{PQR}$ from the point of intersection of the lines $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$
If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :
Let $ABCD$ be a tetrahedron such that the edges $AB$, $AC$ and $AD$ are mutually perpendicular.
Let the areas of the triangles $ABC$, $ACD$ and $ADB$ be $5$, $6$ and $7$ square units respectively.
Then the area (in square units) of the $\triangle BCD$ is equal to:
Two vertices of a triangle $ABC$ are $A(3,-1)$ and $B(-2,3)$, and its orthocentre is $P(1,1)$.
If the coordinates of $C$ are $(\alpha,\beta)$ and the centre of the circle circumscribing the triangle $PAB$ is $(h,k)$, then the value of $(\alpha+\beta)+2(h+k)$ equals:
Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of $\Delta$ABC, then (R + r) is equal to :
Let $A(a,b)$, $B(3,4)$ and $C(-6,-8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle.
Then, the distance of the point $P(2a+3,\ 7b+5)$ from the line
$2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ is:
The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2y + 1 = 0$ and $2x - y - 1 = 0$ and whose orthocenter is $\left( {{7 \over 3},{7 \over 3}} \right)$ is :
Let the equations of two sides of a triangle be $3x - 2y + 6 = 0$ and $4x + 5y - 20 = 0$.
If the orthocentre of this triangle is at $(1,1)$, then the equation of its third side is:
ABC is a triangle in a plane with vertices
$A(2,3,5)$, $B(-1,3,2)$ and $C(\lambda,5,\mu)$.
If the median through $A$ is equally inclined to the coordinate axes,
then the value of $(\lambda^3 + \mu^3 + 5)$ is:
If the line $3x+4y-24=0$ intersects the $x$-axis at the point $A$ and the $y$-axis at the point $B$, then the incentre of the triangle $OAB$, where $O$ is the origin, is: