QBManager

Matrix and Determinant

79199 The area of triangle with vertices \((\mathrm{k}, 0),(4,0)\),
\((0,2)\)

1 8

2 0 or -8

3 0

4 0 or 8

Karnataka CET-2017

Matrix and Determinant

79200 If \(\left|\begin{array}{lll}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0\), then the area of the
triangle whose vertices are \(\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right)\)
and \(\left(\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right)\) is

1 \(\frac{1}{8} \mathrm{abc}\)

2 \(\frac{1}{8}\)

3 \(\frac{1}{4} \mathrm{abc}\)

4 \(\frac{1}{4}\)

Explanation:

(B) : It is given that,
\(=\left|\begin{array}{lll}
2 \mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ 2 \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ 2 \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \neq 0\)
\(=2\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \Rightarrow\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{4}\)
Given, The vertices of the triangle are \(\left[\frac{x_{1}}{a}, \frac{y_{1}}{a}\right]\),
\(\left[\frac{\mathrm{x}_{2}}{\mathrm{~b}}, \frac{\mathrm{y}_{2}}{\mathrm{~b}}\right] \text { and }\left[\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right]\)
Thus, the area of the triangle can be given by
\(A=\frac{1}{2}\left|\begin{array}{ccc} \frac{\mathrm{x}_{1}}{\mathrm{a}} & \frac{\mathrm{y}_{1}}{\mathrm{a}} & 1 \\ \frac{\mathrm{x}_{2}}{\mathrm{~b}} & \frac{\mathrm{y}_{2}}{\mathrm{~b}} & 1 \\ \frac{\mathrm{x}_{3}}{\mathrm{c}} & \frac{\mathrm{y}_{3}}{\mathrm{c}} & 1 \end{array}\right|\)
Taking \(\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}\) and \(\frac{1}{\mathrm{c}}\) common from \(\mathrm{R}_{1}, \mathrm{R}_{2}\) and \(\mathrm{R}_{3^{-}}\)
\(\left.\mathrm{A}=\frac{1}{2}\left|\frac{1}{\mathrm{a}} \times \frac{1}{\mathrm{~b}} \times \frac{1}{\mathrm{c}}\right| \begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array} \right\rvert\,\)
\(A=\frac{1}{2 a b c}\left|\begin{array}{lll}x_{1} & y_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array}\right| \Rightarrow \mathrm{A}=\frac{1}{2 \mathrm{abc}} \times \frac{\mathrm{abc}}{4} \Rightarrow \mathrm{A}=\frac{1}{8}\)

Karnataka CET-2015

Matrix and Determinant

79201 If the lines \(x+2 a y+a=0, x+3 b y+b=0\) and \(x+4 c y+c=0\) are concurrent, then \(a, b, c\) are in

1 \(\mathrm{AP}\)

2 GP

3 HP

4 None of these

Jamia Millia Islamia-2010

Matrix and Determinant

79202 The equation \(\left|\begin{array}{lll}\mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} \\ \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} \\ \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b}\end{array}\right|=\mathbf{0}\) where \(\mathrm{a}\),
\(b, c\) are different is satisfied by

1 \(x=0\)

2 \(\mathrm{x}=\mathrm{a}\)

3 \(x=\frac{1}{3}(a+b+c)\)

4 \(\mathrm{x}=\mathrm{a}+\mathrm{b}+\mathrm{c}\)

Jamia Millia Islamia-2010

Matrix and Determinant

79199 The area of triangle with vertices \((\mathrm{k}, 0),(4,0)\),
\((0,2)\)

1 8

2 0 or -8

3 0

4 0 or 8

Karnataka CET-2017

Matrix and Determinant

79200 If \(\left|\begin{array}{lll}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0\), then the area of the
triangle whose vertices are \(\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right)\)
and \(\left(\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right)\) is

1 \(\frac{1}{8} \mathrm{abc}\)

2 \(\frac{1}{8}\)

3 \(\frac{1}{4} \mathrm{abc}\)

4 \(\frac{1}{4}\)

Explanation:

(B) : It is given that,
\(=\left|\begin{array}{lll}
2 \mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ 2 \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ 2 \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \neq 0\)
\(=2\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \Rightarrow\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{4}\)
Given, The vertices of the triangle are \(\left[\frac{x_{1}}{a}, \frac{y_{1}}{a}\right]\),
\(\left[\frac{\mathrm{x}_{2}}{\mathrm{~b}}, \frac{\mathrm{y}_{2}}{\mathrm{~b}}\right] \text { and }\left[\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right]\)
Thus, the area of the triangle can be given by
\(A=\frac{1}{2}\left|\begin{array}{ccc} \frac{\mathrm{x}_{1}}{\mathrm{a}} & \frac{\mathrm{y}_{1}}{\mathrm{a}} & 1 \\ \frac{\mathrm{x}_{2}}{\mathrm{~b}} & \frac{\mathrm{y}_{2}}{\mathrm{~b}} & 1 \\ \frac{\mathrm{x}_{3}}{\mathrm{c}} & \frac{\mathrm{y}_{3}}{\mathrm{c}} & 1 \end{array}\right|\)
Taking \(\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}\) and \(\frac{1}{\mathrm{c}}\) common from \(\mathrm{R}_{1}, \mathrm{R}_{2}\) and \(\mathrm{R}_{3^{-}}\)
\(\left.\mathrm{A}=\frac{1}{2}\left|\frac{1}{\mathrm{a}} \times \frac{1}{\mathrm{~b}} \times \frac{1}{\mathrm{c}}\right| \begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array} \right\rvert\,\)
\(A=\frac{1}{2 a b c}\left|\begin{array}{lll}x_{1} & y_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array}\right| \Rightarrow \mathrm{A}=\frac{1}{2 \mathrm{abc}} \times \frac{\mathrm{abc}}{4} \Rightarrow \mathrm{A}=\frac{1}{8}\)

Karnataka CET-2015

Matrix and Determinant

79201 If the lines \(x+2 a y+a=0, x+3 b y+b=0\) and \(x+4 c y+c=0\) are concurrent, then \(a, b, c\) are in

1 \(\mathrm{AP}\)

2 GP

3 HP

4 None of these

Jamia Millia Islamia-2010

Matrix and Determinant

79202 The equation \(\left|\begin{array}{lll}\mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} \\ \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} \\ \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b}\end{array}\right|=\mathbf{0}\) where \(\mathrm{a}\),
\(b, c\) are different is satisfied by

1 \(x=0\)

2 \(\mathrm{x}=\mathrm{a}\)

3 \(x=\frac{1}{3}(a+b+c)\)

4 \(\mathrm{x}=\mathrm{a}+\mathrm{b}+\mathrm{c}\)

Jamia Millia Islamia-2010

Matrix and Determinant

79199 The area of triangle with vertices \((\mathrm{k}, 0),(4,0)\),
\((0,2)\)

1 8

2 0 or -8

3 0

4 0 or 8

Karnataka CET-2017

Matrix and Determinant

79200 If \(\left|\begin{array}{lll}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0\), then the area of the
triangle whose vertices are \(\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right)\)
and \(\left(\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right)\) is

1 \(\frac{1}{8} \mathrm{abc}\)

2 \(\frac{1}{8}\)

3 \(\frac{1}{4} \mathrm{abc}\)

4 \(\frac{1}{4}\)

Explanation:

(B) : It is given that,
\(=\left|\begin{array}{lll}
2 \mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ 2 \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ 2 \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \neq 0\)
\(=2\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \Rightarrow\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{4}\)
Given, The vertices of the triangle are \(\left[\frac{x_{1}}{a}, \frac{y_{1}}{a}\right]\),
\(\left[\frac{\mathrm{x}_{2}}{\mathrm{~b}}, \frac{\mathrm{y}_{2}}{\mathrm{~b}}\right] \text { and }\left[\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right]\)
Thus, the area of the triangle can be given by
\(A=\frac{1}{2}\left|\begin{array}{ccc} \frac{\mathrm{x}_{1}}{\mathrm{a}} & \frac{\mathrm{y}_{1}}{\mathrm{a}} & 1 \\ \frac{\mathrm{x}_{2}}{\mathrm{~b}} & \frac{\mathrm{y}_{2}}{\mathrm{~b}} & 1 \\ \frac{\mathrm{x}_{3}}{\mathrm{c}} & \frac{\mathrm{y}_{3}}{\mathrm{c}} & 1 \end{array}\right|\)
Taking \(\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}\) and \(\frac{1}{\mathrm{c}}\) common from \(\mathrm{R}_{1}, \mathrm{R}_{2}\) and \(\mathrm{R}_{3^{-}}\)
\(\left.\mathrm{A}=\frac{1}{2}\left|\frac{1}{\mathrm{a}} \times \frac{1}{\mathrm{~b}} \times \frac{1}{\mathrm{c}}\right| \begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array} \right\rvert\,\)
\(A=\frac{1}{2 a b c}\left|\begin{array}{lll}x_{1} & y_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array}\right| \Rightarrow \mathrm{A}=\frac{1}{2 \mathrm{abc}} \times \frac{\mathrm{abc}}{4} \Rightarrow \mathrm{A}=\frac{1}{8}\)

Karnataka CET-2015

Matrix and Determinant

79201 If the lines \(x+2 a y+a=0, x+3 b y+b=0\) and \(x+4 c y+c=0\) are concurrent, then \(a, b, c\) are in

1 \(\mathrm{AP}\)

2 GP

3 HP

4 None of these

Jamia Millia Islamia-2010

Matrix and Determinant

79202 The equation \(\left|\begin{array}{lll}\mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} \\ \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} \\ \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b}\end{array}\right|=\mathbf{0}\) where \(\mathrm{a}\),
\(b, c\) are different is satisfied by

1 \(x=0\)

2 \(\mathrm{x}=\mathrm{a}\)

3 \(x=\frac{1}{3}(a+b+c)\)

4 \(\mathrm{x}=\mathrm{a}+\mathrm{b}+\mathrm{c}\)

Jamia Millia Islamia-2010

Matrix and Determinant

79199 The area of triangle with vertices \((\mathrm{k}, 0),(4,0)\),
\((0,2)\)

1 8

2 0 or -8

3 0

4 0 or 8

Karnataka CET-2017

Matrix and Determinant

79200 If \(\left|\begin{array}{lll}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0\), then the area of the
triangle whose vertices are \(\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right)\)
and \(\left(\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right)\) is

1 \(\frac{1}{8} \mathrm{abc}\)

2 \(\frac{1}{8}\)

3 \(\frac{1}{4} \mathrm{abc}\)

4 \(\frac{1}{4}\)

Explanation:

(B) : It is given that,
\(=\left|\begin{array}{lll}
2 \mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ 2 \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ 2 \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \neq 0\)
\(=2\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{2} \Rightarrow\left|\begin{array}{lll}
\mathrm{a} & \mathrm{x}_{1} & \mathrm{y}_{1} \\ \mathrm{~b} & \mathrm{x}_{2} & \mathrm{y}_{2} \\ \mathrm{c} & \mathrm{x}_{3} & \mathrm{y}_{3} \end{array}\right|=\frac{\mathrm{abc}}{4}\)
Given, The vertices of the triangle are \(\left[\frac{x_{1}}{a}, \frac{y_{1}}{a}\right]\),
\(\left[\frac{\mathrm{x}_{2}}{\mathrm{~b}}, \frac{\mathrm{y}_{2}}{\mathrm{~b}}\right] \text { and }\left[\frac{\mathrm{x}_{3}}{\mathrm{c}}, \frac{\mathrm{y}_{3}}{\mathrm{c}}\right]\)
Thus, the area of the triangle can be given by
\(A=\frac{1}{2}\left|\begin{array}{ccc} \frac{\mathrm{x}_{1}}{\mathrm{a}} & \frac{\mathrm{y}_{1}}{\mathrm{a}} & 1 \\ \frac{\mathrm{x}_{2}}{\mathrm{~b}} & \frac{\mathrm{y}_{2}}{\mathrm{~b}} & 1 \\ \frac{\mathrm{x}_{3}}{\mathrm{c}} & \frac{\mathrm{y}_{3}}{\mathrm{c}} & 1 \end{array}\right|\)
Taking \(\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}\) and \(\frac{1}{\mathrm{c}}\) common from \(\mathrm{R}_{1}, \mathrm{R}_{2}\) and \(\mathrm{R}_{3^{-}}\)
\(\left.\mathrm{A}=\frac{1}{2}\left|\frac{1}{\mathrm{a}} \times \frac{1}{\mathrm{~b}} \times \frac{1}{\mathrm{c}}\right| \begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array} \right\rvert\,\)
\(A=\frac{1}{2 a b c}\left|\begin{array}{lll}x_{1} & y_{1} & \mathrm{a} \\ \mathrm{x}_{2} & \mathrm{y}_{2} & \mathrm{~b} \\ \mathrm{x}_{3} & \mathrm{y}_{3} & \mathrm{c}\end{array}\right| \Rightarrow \mathrm{A}=\frac{1}{2 \mathrm{abc}} \times \frac{\mathrm{abc}}{4} \Rightarrow \mathrm{A}=\frac{1}{8}\)

Karnataka CET-2015

Matrix and Determinant

79201 If the lines \(x+2 a y+a=0, x+3 b y+b=0\) and \(x+4 c y+c=0\) are concurrent, then \(a, b, c\) are in

1 \(\mathrm{AP}\)

2 GP

3 HP

4 None of these

Jamia Millia Islamia-2010

Matrix and Determinant

79202 The equation \(\left|\begin{array}{lll}\mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} \\ \mathbf{x}-\mathbf{b} & \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} \\ \mathbf{x}-\mathbf{c} & \mathbf{x}-\mathbf{a} & \mathbf{x}-\mathbf{b}\end{array}\right|=\mathbf{0}\) where \(\mathrm{a}\),
\(b, c\) are different is satisfied by

1 \(x=0\)

2 \(\mathrm{x}=\mathrm{a}\)

3 \(x=\frac{1}{3}(a+b+c)\)

4 \(\mathrm{x}=\mathrm{a}+\mathrm{b}+\mathrm{c}\)

Jamia Millia Islamia-2010