Area of triangle and collinearity of three points - class-XI
area of triangle and collinearity of three points
Questions
If $\begin{vmatrix} x _1 & y _1 & 1 \ x _2 & y _2 & 1 \ x _3 & y _3 & 1\end{vmatrix}=\begin{vmatrix} a _1 & b _1 & 1\ a _2 & b _2 & 1 \ a _3 & b _3 & 1\end{vmatrix}$, then the two triangles with vertices $(x _1, y _1), (x _2, y _2), (x _3, y _3)$ and $(a _1,b _1)$, $(a _2, b _2)$, $(a _3, b _3)$ must be congruent.
- True
- False
If the area of the triangle with vertices $(2, 5), (7, k)$ and $(3, 1)$ is $10$, then find the value of $k$.
- $-5$ or $35$
- $5$ or $-35$
- $15$ or $-5$
- $-5$ or $-25$
If $\displaystyle \left | \begin{matrix}x _{1} &y _{1} &1 \ x _{2} &y _{2} &1 \ x _{3} &y _{3} &1 \end{matrix} \right |=\left | \begin{matrix}1 &1 &1 \ b _{1} &b _{2} &b _{3} \ a _{1} &a _{2} &a _{3}\end{matrix} \right |$ then the two triangles whose vertices are $\displaystyle \left ( x _{1},y _{1} \right ), \left ( x _{2},y _{2} \right ), ( \left ( x _{3},y _{3} \right ) $ and $\displaystyle\left ( a _{1},b _{1} \right ), \left ( a _{2},b _{2} \right ), \left ( a _{13},b _{3} \right ),$ are
- congruent
- similar
- equal in area
- none of these
Let O(0, 0), P(3,4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR,PQR, OQR are of equal area. The coordinates of R are
- $\displaystyle \left ( \frac{4}{3}, 3 \right )$
- $\displaystyle \left ( 3, \frac{2}{3} \right )$
- $\displaystyle \left ( 3, \frac{4}{3} \right )$
- $\displaystyle \left ( \frac{4}{3}, \frac{2}{3} \right )$
The co-ordinates of the vertices A, B, C of a triangle are $ \displaystyle \left ( 6,3 \right ),\left ( -3,5 \right ),\left ( 4,-2 \right ) $ respectively and P is any point $ \displaystyle \left ( x,y \right ), $ then the ratio of areas of triangles PBC and ABC is
- $ \displaystyle \begin{vmatrix}x-y-2\end{vmatrix}:7 $
- $ \displaystyle \begin{vmatrix}x+y+2\end{vmatrix}:7 $
- $ \displaystyle \begin{vmatrix}x+y-2\end{vmatrix}:7 $
- None of these
if $ \displaystyle a,b,c $ as well as $ \displaystyle d,e,f $ are in G.P. with same common ratio then set of points $ \displaystyle \left ( a,d \right ),\left ( b,e \right ),\left ( c,f \right ) $ are
- collinear
- concurrent
- lies on a circle
- lie on an ellipse
The vertices of the triangle $ABC$ are $(2, 1, 1), (3, 1, 2), (-4, 0, 1)$. The area of triangle is
- $\displaystyle \frac{3\sqrt{38}}{2}$
- $\sqrt{38}$
- $\displaystyle \frac{\sqrt{38}}{2}$
- $4$
- $\displaystyle abc $
- $\displaystyle s(s-a)(s-b)(s-c)$
- $\cfrac {abc}{4} $
- $\displaystyle 4s(s-a)(s-b)(s-c)$
x _{2} & y _{2} & 1\\
x _{3} &y _{3} &1
\end{vmatrix}$.If $\displaystyle \triangle ABC$ is an equilateral triangle and $\displaystyle a = BC$ is a rational number, then $\displaystyle \triangle$ must be
- an integer
- a rational number
- an irrational number
- an imaginary number
What is the area of the triangle formed by the points $(a,c+a), (a,c)$ and $(-a,c-a)$?
- $\displaystyle- a^{2}$
- $\displaystyle \frac{1}{a^{2}}$
- $\displaystyle a^{2}+a$
- zero
What is the area of the triangle formed by the points $(a,c+a), \displaystyle \left ( a^{2},c^{2} \right )$ and $(-a, c-a)$?
- $1$
- $\displaystyle a^{2}$
- $\displaystyle \sqrt{a^{2}+c^{2}}$
- None of these
What is the area of the triangle formed by the points $(a,b+c), (b,c+a)$ and $(c,a+b)$?
- $1$
- $-1$
- $0$
- $\displaystyle \frac{1}{2}\left ( abc \right )^{2}$
The area of a triangle whose vertices are (-2,-2), (-1,-3) and (p,0) is 3 sq.units what is the value of p?
- -2
- 2
- 3
- -3
The area of a triangle, whose vertices are $(3, 2), (5, 2)$ and the point of intersection of the lines $x = a$ and $y = 5$, is $3$ square units. What is the value of $a$?
- $2$
- $3$
- $4$
- $5$
If $P=(x _{1}, y _{1}), Q=(x _{2}, y _{2})$ and $R=(x _{3}, y _{3})$ are three points of a triangle in $\mathbb{R}^{2}$. Then, area of a $\triangle PQR$ in terms of determinant of matrix $M=\begin{bmatrix} 1& 1 & 1 \ x _{1} & x _{2} & x _{3} \ y _{1} & y _{2} & y _{3}\end{bmatrix}$ is
- $-|det(M)|$
- $|det(M)|$
- $\dfrac{1}{2}|det(M)|$
- $2|det(M)|$
If $\triangle _1,\triangle _2$ be the areas of two triangles with vertices $(b,c), (c,a), (a,b)$, and $ (ac-b^2, ab-c^2),(ba-c^2, bc-a^2), (cb-a^2, ca-b^2)$, then $\ \dfrac{\triangle _1}{\triangle _2}=(a+b+c)^2$
- True
- False
If ${ \left( { x } _{ 1 }-{ { x } _{ 2 } } \right) }^{ 2 }+{ \left( { y } _{ 1 }-{ y } _{ 2 } \right) }^{ 2 }={ a }^{ 2 }$, ${ \left( x _{ 2 }-{ x } _{ 3 } \right) }^{ 2 }+{ \left( { y } _{ 2 }-{ y } _{ 3 } \right) }^{ 2 }={ b }^{ 2 }$, ${ \left( { x } _{ 3 }-{ x } _{ 1 } \right) }^{ 2 }+{ \left( { y } _{ 3 }-{ y } _{ 1 } \right) }^{ 2 }={ c }^{ 2 }$ and $k\begin{vmatrix} { x } _{ 1 } & { y } _{ 1 } & 1 \ { x } _{ 2 } & { y } _{ 2 } & 1 \ { x } _{ 3 } & { y } _{ 3 } & 1 \end{vmatrix}=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$, then the value of $k$ is
- $1$
- $2$
- $4$
- none of these
$(x _1 - x _2)^2 + (y _1 - y _2)^2 = a^2$;
$(x _2 - x _3)^2 + (y _2 - y _3)^2 = b^2$;
$(x _3 - x _1)^2 + (y _3 - y _1)^2 = c^2$;
then find $4 \begin{vmatrix}x _1 & y _1 & 1\ x _2 & y _2 & 1\ x _3 & y _3 & 1\end{vmatrix}^2 = $
- $(a+b+c) (b+c - a) (c + a - b) (a + b - c)$
- $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)$
- $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$
- $(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$