VIT - VITEEE 2025
ApplyNational level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Area of Triangle is considered one the most difficult concept.
55 Questions around this concept.
Let a line $y=m x(m>0)$ intersect the parabola, $y^2=x_{\text {at a point } \mathrm{P} \text {, other than the }}$ origin. Let the tangent to it at P meet the x-axis at the point Q . If area $(\triangle O P Q)=4$ sq.units, then m is equal to $\qquad$
Drawn from the origin are two mutually perpendicular lines forming an isosceles triangle together with a straight line , then the area of this triangle is:
Let three vectors $\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, $\vec{b}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ from a triangle such that $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$ and the area of the triangle is $5 \sqrt{6}$. if $\alpha$ is a positive real number, then $|\overrightarrow{\mathrm{c}}|^2$ is :
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The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $(0, 0)$, $(0, 41)$ and $(41, 0)$ is :
Let $A=(4,0), B=(0,12)$ be two points in the plane. The locus of a point $C$ such that the area of triangle $A B C 18 \mathrm{sq}$. units are
Let the parabola $y=x^2+\mathrm{px}-3$, meet the coordinate axes at the points $\mathrm{P}, \mathrm{Q}$ and R . If the circle C with centre at $(-1,-1)$ passes through the points $P, Q$ and $R$, then the area of $\triangle P Q R$ is :
Area of Triangle
If vertices of a triangle ABC given as A (x1, y1), B (x2, y2) and C(x3, y3), then area of ΔABC is
\begin{equation}
\left|\frac{1}{2}\right| \begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\left|\left|=\frac{1}{2}\right| \mathrm{x}_1\left(\mathrm{y}_2-\mathrm{y}_3\right)+\mathrm{x}_2\left(\mathrm{y}_3-\mathrm{y}_1\right)+\mathrm{x}_3\left(\mathrm{y}_1-\mathrm{y}_2\right)\right|
\end{equation}
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