An equilateral triangle has each of its sides of length 6,, cm . If (x₁, y₁) ; (x₂, y₂) ,, a

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9.Straight Line

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An equilateral triangle has each of its sides of length $6\,\, cm$ . If $(x_1, y_1) ; (x_2, y_2) \,\, and \,\, (x_3, y_3)$ are its vertices then the value of the determinant,${{\left| {\,\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}\,} \right|}^2}$ is equal to :

$192$

$243$

$486$

$972$

Solution

The area of a triangle having $\left(x_{1}, y_{1}\right) ;\left(x_{2}, y_{2}\right) \;and\;\left(x_{3}, y_{3}\right)$ as its vertices is, $\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=A($ let $)$

Now, $\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|^{2}=4 A^{2}$

The given triangle is equilateral and each side is of $6 \mathrm{cm}$. Then area $(A)=\frac{\sqrt{3}}{4} \times 6^{2}$

Then $4 A^{2}=4 \times \frac{3}{4^{2}} \times 36 \times 36=3 \times 9 \times 36=972$

Standard 11

Mathematics

If the coordinates of the vertices of the triangle $ABC$ be $(-1, 6)$, $(-3, -9)$, and $(5, -8)$ respectively, then the equation of the median through $C$ is

easy

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$- axis , $y$ axis and the $AB$ at $C,\,D$ and $E$ respectively. If $O$ is the origin of coordinates, then the area of $OCEB$ is

hard

(IIT-1976)

The vertex of a right angle of a right angled triangle lies on the straight line $2x + y – 10 = 0$ and the two other vertices, at points $(2, -3)$ and $(4, 1)$ then the area of triangle in sq. units is

normal

A straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y -$ axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\triangle OAB$ is $\frac{98}{3} \sqrt{3}$, then $a ^2- b ^2$ is equal to:

hard

(JEE MAIN-2023)

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $AB$ is

hard

Solution

Similar Questions

If the coordinates of the vertices of the triangle $ABC$ be $(-1, 6)$, $(-3, -9)$, and $(5, -8)$ respectively, then the equation of the median through $C$ is

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$- axis , $y$ axis and the $AB$ at $C,\,D$ and $E$ respectively. If $O$ is the origin of coordinates, then the area of $OCEB$ is

The vertex of a right angle of a right angled triangle lies on the straight line $2x + y – 10 = 0$ and the two other vertices, at points $(2, -3)$ and $(4, 1)$ then the area of triangle in sq. units is

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha } + \frac{y}{\beta } = 1$and $\frac{x}{\beta } + \frac{y}{\alpha } = 1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid point of $AB$ is

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