If area of triangle is 35 mathrm{sq} mathrm{units} with vertices (2,-6),(5,4) and (mathrm{k}, 4) . T

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3 and 4 .Determinants and Matrices

medium

If area of triangle is $35$ $\mathrm{sq}$ $\mathrm{units}$ with vertices $(2,-6),(5,4)$ and $(\mathrm{k}, 4) .$ Then $\mathrm{k}$ is

$12$

$-2$

$12,-2$

$-12,-2$

Solution

The area of the triangle with vertices $(2,-6),(5,4)$ and $(k, 4)$ is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1\end{array}\right|$

$=\frac{1}{2}[2(4-4)+6(5-k)+1(20-4 k)]$

$=\frac{1}{2}[30-6 k+20-4 k]$

$=\frac{1}{2}[50-10 k]$

$=25-5 k$

It is given that the area of the triangle id $\pm 35$.

Therefore, we have:

$\Rightarrow 25-5 k=\pm 35$

$\Rightarrow 5(5-k)=\pm 35$

$\Rightarrow 5-k=\pm 7$

When $5-k=-7, k=5+7=12$

When $5-k=-7, k=5-7=-2$

Hence, $k=12,-2$

The correct answer is $C$.

Standard 12

Mathematics

If $A\, = \,\left[ \begin{gathered}
1\ \ \ \,1\ \ \ \,2\ \ \ \hfill \\
0\ \ \ \,2\ \ \ \,1\ \ \ \hfill \\
1\ \ \ \,0\ \ \ \,2\ \ \ \hfill \\
\end{gathered} \right]$ and $A^3 = (aA-I) (bA-I)$,where $a, b$ are integers and $I$ is a $3 × 3$ unit matrix then value of $(a + b)$ is equal to

normal

If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to

hard

(JEE MAIN-2020)

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|$ is

easy

If ${\Delta _1} = \left| {\,\begin{array}{{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then

hard

Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ – 3i}&1\\4&{3i}&{ – 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then

easy

(IIT-1998)

If area of triangle is $35$ $\mathrm{sq}$ $\mathrm{units}$ with vertices $(2,-6),(5,4)$ and $(\mathrm{k}, 4) .$ Then $\mathrm{k}$ is

Solution

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