88317
The area of the triangle formed by the coordinate axes and the line \(4 x+5 y=20\) is (in square unit)
1 5
2 10
3 15
4 20
Explanation:
(B) : Given the line \(4 x+5 y=20\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(\therefore 4 \mathrm{x}+0=20\) \(x=5\) At \(\mathrm{y}-\operatorname{axis} \mathrm{x}=0\) \(\mathrm{y}=\frac{20}{5}=4\) \(\therefore\) Area of the triangle \(=\frac{1}{2} \times 4 \times 5=10\) Square units
Co-Ordinate system
88350
The centroid of the triangle formed by the lines \(x+y=1\) and \(2 y^{2}-x y-6 x^{2}=0\) is
1 \((0,0)\)
2 \(\left(\frac{5}{9}, \frac{11}{9}\right)\)
3 \(\left(\frac{-5}{9}, \frac{11}{9}\right)\)
4 \(\left(\frac{5}{9}, \frac{-11}{9}\right)\)
Explanation:
(C) : We have, \(2 y^{2}-x y-6 x^{2}=0\) \(2 y^{2}-4 x y+3 x y-6 x^{2}=0\) \(2 y(y-2 x)+3 x(y-2 x)=0\) \((y-2 x)(2 y+3 x)=0\) \(\therefore\) Equations of sides of triangle are \(\mathrm{x}+\mathrm{y}=1\), \(y-2 x=0,2 y+3 x=0\) On solving the above equations, we get the vertices of triangle as \(\left(\frac{1}{3}, \frac{2}{3}\right),(0,0)\) and \((-2,3)\), \(\therefore\) Centroid of triangle \(=\left(\frac{\frac{1}{3}+0-2}{3}, \frac{\frac{2}{3}+0+3}{3}\right)\) \(=\left(\frac{-5}{9}, \frac{11}{9}\right)\)
TS EAMCET-2020-10.09.2020
Co-Ordinate system
88270
Orthocentre of the triangle formed by the lines \(x+y=1\) and \(x y=0\) is
1 \((0,0)\)
2 \((0,1)\)
3 \((1,0)\)
4 \(-1,1)\)
Explanation:
(A) : Given lines are \(x+y=1\) and \(x y=0\) \(x y=0\) represents line \(x=0\) and \(y=0\). Triangle formed by lines \(\mathrm{x}+\mathrm{y}=1, \mathrm{x}=0\) And \(\mathrm{y}=0\) is \(\triangle \mathrm{ABC}\). \(\because \triangle \mathrm{ABC}\) is right angled triangle at \(\angle \mathrm{B}\). Orthocentre of \(\triangle A B C\) is at \(B(0,0)\).
COMEDK-2011
Co-Ordinate system
88286
If the lines \(x+3 y-9=0,4 x+b y-2=0\) and \(2 x-y-4=0\) are concurrent, then \(b\) equal to
1 -6
2 5
3 1
4 -5
Explanation:
(D): Given, lines \(x+3 y-9=0,4 x+\) by \(-2=0\) and \(2 \mathrm{x}-\mathrm{y}-4=0\) are concurrent \(\therefore\) The determinant is zero for concurrent line - i.e. \(\quad\left|\begin{array}{ccc}1 3 -9 \\ 4 \mathrm{~b} -2 \\ 2 -1 -4\end{array}\right|=0\) \(\Rightarrow 1(-4 b-2)-3(-16+4)-9(-4-2 b)=0\) \(\Rightarrow-4 \mathrm{~b}-2+36+36+18 \mathrm{~b}=0\) \(14 b=-70\) \(b=-5\)
BCECE-2016
Co-Ordinate system
88262
\(\mathrm{O}(0,0), \mathrm{A}(1,2), \mathrm{B}(3,4)\) are the vertices of \(\triangle O A B\). The joint equation of the altitude and median drawn from \(O\) is
1 \(x^{2}+7 x y-y^{2}=0\)
2 \(x^{2}+7 x y+y^{2}=0\)
3 \(3 x^{2}-x y-2 y^{2}=0\)
4 \(3 x^{2}+x y-2 y^{2}=0\)
Explanation:
(D) : \(D\) is midpoint of \(A B \Rightarrow D \equiv(2,3)\) Hence equation of \(\mathrm{OD}\) \(\frac{y-0}{3-0}=\frac{x-0}{2-0}\) \(2 y=3 x \Rightarrow 3 x-2 y=0 \tag{i}\) Slope of \(A B=\frac{4-2}{3-1}=1 \Rightarrow\) slope of OD is -1 . Equation of OP, \((y-0)=-(x-0)\) \(x+y=0 \tag{ii}\) Required joint equation is \((3 x-2 y)(x+y)=0\) \(3 x^{2}+x y-2 y^{2}=0\)
88317
The area of the triangle formed by the coordinate axes and the line \(4 x+5 y=20\) is (in square unit)
1 5
2 10
3 15
4 20
Explanation:
(B) : Given the line \(4 x+5 y=20\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(\therefore 4 \mathrm{x}+0=20\) \(x=5\) At \(\mathrm{y}-\operatorname{axis} \mathrm{x}=0\) \(\mathrm{y}=\frac{20}{5}=4\) \(\therefore\) Area of the triangle \(=\frac{1}{2} \times 4 \times 5=10\) Square units
Co-Ordinate system
88350
The centroid of the triangle formed by the lines \(x+y=1\) and \(2 y^{2}-x y-6 x^{2}=0\) is
1 \((0,0)\)
2 \(\left(\frac{5}{9}, \frac{11}{9}\right)\)
3 \(\left(\frac{-5}{9}, \frac{11}{9}\right)\)
4 \(\left(\frac{5}{9}, \frac{-11}{9}\right)\)
Explanation:
(C) : We have, \(2 y^{2}-x y-6 x^{2}=0\) \(2 y^{2}-4 x y+3 x y-6 x^{2}=0\) \(2 y(y-2 x)+3 x(y-2 x)=0\) \((y-2 x)(2 y+3 x)=0\) \(\therefore\) Equations of sides of triangle are \(\mathrm{x}+\mathrm{y}=1\), \(y-2 x=0,2 y+3 x=0\) On solving the above equations, we get the vertices of triangle as \(\left(\frac{1}{3}, \frac{2}{3}\right),(0,0)\) and \((-2,3)\), \(\therefore\) Centroid of triangle \(=\left(\frac{\frac{1}{3}+0-2}{3}, \frac{\frac{2}{3}+0+3}{3}\right)\) \(=\left(\frac{-5}{9}, \frac{11}{9}\right)\)
TS EAMCET-2020-10.09.2020
Co-Ordinate system
88270
Orthocentre of the triangle formed by the lines \(x+y=1\) and \(x y=0\) is
1 \((0,0)\)
2 \((0,1)\)
3 \((1,0)\)
4 \(-1,1)\)
Explanation:
(A) : Given lines are \(x+y=1\) and \(x y=0\) \(x y=0\) represents line \(x=0\) and \(y=0\). Triangle formed by lines \(\mathrm{x}+\mathrm{y}=1, \mathrm{x}=0\) And \(\mathrm{y}=0\) is \(\triangle \mathrm{ABC}\). \(\because \triangle \mathrm{ABC}\) is right angled triangle at \(\angle \mathrm{B}\). Orthocentre of \(\triangle A B C\) is at \(B(0,0)\).
COMEDK-2011
Co-Ordinate system
88286
If the lines \(x+3 y-9=0,4 x+b y-2=0\) and \(2 x-y-4=0\) are concurrent, then \(b\) equal to
1 -6
2 5
3 1
4 -5
Explanation:
(D): Given, lines \(x+3 y-9=0,4 x+\) by \(-2=0\) and \(2 \mathrm{x}-\mathrm{y}-4=0\) are concurrent \(\therefore\) The determinant is zero for concurrent line - i.e. \(\quad\left|\begin{array}{ccc}1 3 -9 \\ 4 \mathrm{~b} -2 \\ 2 -1 -4\end{array}\right|=0\) \(\Rightarrow 1(-4 b-2)-3(-16+4)-9(-4-2 b)=0\) \(\Rightarrow-4 \mathrm{~b}-2+36+36+18 \mathrm{~b}=0\) \(14 b=-70\) \(b=-5\)
BCECE-2016
Co-Ordinate system
88262
\(\mathrm{O}(0,0), \mathrm{A}(1,2), \mathrm{B}(3,4)\) are the vertices of \(\triangle O A B\). The joint equation of the altitude and median drawn from \(O\) is
1 \(x^{2}+7 x y-y^{2}=0\)
2 \(x^{2}+7 x y+y^{2}=0\)
3 \(3 x^{2}-x y-2 y^{2}=0\)
4 \(3 x^{2}+x y-2 y^{2}=0\)
Explanation:
(D) : \(D\) is midpoint of \(A B \Rightarrow D \equiv(2,3)\) Hence equation of \(\mathrm{OD}\) \(\frac{y-0}{3-0}=\frac{x-0}{2-0}\) \(2 y=3 x \Rightarrow 3 x-2 y=0 \tag{i}\) Slope of \(A B=\frac{4-2}{3-1}=1 \Rightarrow\) slope of OD is -1 . Equation of OP, \((y-0)=-(x-0)\) \(x+y=0 \tag{ii}\) Required joint equation is \((3 x-2 y)(x+y)=0\) \(3 x^{2}+x y-2 y^{2}=0\)
88317
The area of the triangle formed by the coordinate axes and the line \(4 x+5 y=20\) is (in square unit)
1 5
2 10
3 15
4 20
Explanation:
(B) : Given the line \(4 x+5 y=20\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(\therefore 4 \mathrm{x}+0=20\) \(x=5\) At \(\mathrm{y}-\operatorname{axis} \mathrm{x}=0\) \(\mathrm{y}=\frac{20}{5}=4\) \(\therefore\) Area of the triangle \(=\frac{1}{2} \times 4 \times 5=10\) Square units
Co-Ordinate system
88350
The centroid of the triangle formed by the lines \(x+y=1\) and \(2 y^{2}-x y-6 x^{2}=0\) is
1 \((0,0)\)
2 \(\left(\frac{5}{9}, \frac{11}{9}\right)\)
3 \(\left(\frac{-5}{9}, \frac{11}{9}\right)\)
4 \(\left(\frac{5}{9}, \frac{-11}{9}\right)\)
Explanation:
(C) : We have, \(2 y^{2}-x y-6 x^{2}=0\) \(2 y^{2}-4 x y+3 x y-6 x^{2}=0\) \(2 y(y-2 x)+3 x(y-2 x)=0\) \((y-2 x)(2 y+3 x)=0\) \(\therefore\) Equations of sides of triangle are \(\mathrm{x}+\mathrm{y}=1\), \(y-2 x=0,2 y+3 x=0\) On solving the above equations, we get the vertices of triangle as \(\left(\frac{1}{3}, \frac{2}{3}\right),(0,0)\) and \((-2,3)\), \(\therefore\) Centroid of triangle \(=\left(\frac{\frac{1}{3}+0-2}{3}, \frac{\frac{2}{3}+0+3}{3}\right)\) \(=\left(\frac{-5}{9}, \frac{11}{9}\right)\)
TS EAMCET-2020-10.09.2020
Co-Ordinate system
88270
Orthocentre of the triangle formed by the lines \(x+y=1\) and \(x y=0\) is
1 \((0,0)\)
2 \((0,1)\)
3 \((1,0)\)
4 \(-1,1)\)
Explanation:
(A) : Given lines are \(x+y=1\) and \(x y=0\) \(x y=0\) represents line \(x=0\) and \(y=0\). Triangle formed by lines \(\mathrm{x}+\mathrm{y}=1, \mathrm{x}=0\) And \(\mathrm{y}=0\) is \(\triangle \mathrm{ABC}\). \(\because \triangle \mathrm{ABC}\) is right angled triangle at \(\angle \mathrm{B}\). Orthocentre of \(\triangle A B C\) is at \(B(0,0)\).
COMEDK-2011
Co-Ordinate system
88286
If the lines \(x+3 y-9=0,4 x+b y-2=0\) and \(2 x-y-4=0\) are concurrent, then \(b\) equal to
1 -6
2 5
3 1
4 -5
Explanation:
(D): Given, lines \(x+3 y-9=0,4 x+\) by \(-2=0\) and \(2 \mathrm{x}-\mathrm{y}-4=0\) are concurrent \(\therefore\) The determinant is zero for concurrent line - i.e. \(\quad\left|\begin{array}{ccc}1 3 -9 \\ 4 \mathrm{~b} -2 \\ 2 -1 -4\end{array}\right|=0\) \(\Rightarrow 1(-4 b-2)-3(-16+4)-9(-4-2 b)=0\) \(\Rightarrow-4 \mathrm{~b}-2+36+36+18 \mathrm{~b}=0\) \(14 b=-70\) \(b=-5\)
BCECE-2016
Co-Ordinate system
88262
\(\mathrm{O}(0,0), \mathrm{A}(1,2), \mathrm{B}(3,4)\) are the vertices of \(\triangle O A B\). The joint equation of the altitude and median drawn from \(O\) is
1 \(x^{2}+7 x y-y^{2}=0\)
2 \(x^{2}+7 x y+y^{2}=0\)
3 \(3 x^{2}-x y-2 y^{2}=0\)
4 \(3 x^{2}+x y-2 y^{2}=0\)
Explanation:
(D) : \(D\) is midpoint of \(A B \Rightarrow D \equiv(2,3)\) Hence equation of \(\mathrm{OD}\) \(\frac{y-0}{3-0}=\frac{x-0}{2-0}\) \(2 y=3 x \Rightarrow 3 x-2 y=0 \tag{i}\) Slope of \(A B=\frac{4-2}{3-1}=1 \Rightarrow\) slope of OD is -1 . Equation of OP, \((y-0)=-(x-0)\) \(x+y=0 \tag{ii}\) Required joint equation is \((3 x-2 y)(x+y)=0\) \(3 x^{2}+x y-2 y^{2}=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Co-Ordinate system
88317
The area of the triangle formed by the coordinate axes and the line \(4 x+5 y=20\) is (in square unit)
1 5
2 10
3 15
4 20
Explanation:
(B) : Given the line \(4 x+5 y=20\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(\therefore 4 \mathrm{x}+0=20\) \(x=5\) At \(\mathrm{y}-\operatorname{axis} \mathrm{x}=0\) \(\mathrm{y}=\frac{20}{5}=4\) \(\therefore\) Area of the triangle \(=\frac{1}{2} \times 4 \times 5=10\) Square units
Co-Ordinate system
88350
The centroid of the triangle formed by the lines \(x+y=1\) and \(2 y^{2}-x y-6 x^{2}=0\) is
1 \((0,0)\)
2 \(\left(\frac{5}{9}, \frac{11}{9}\right)\)
3 \(\left(\frac{-5}{9}, \frac{11}{9}\right)\)
4 \(\left(\frac{5}{9}, \frac{-11}{9}\right)\)
Explanation:
(C) : We have, \(2 y^{2}-x y-6 x^{2}=0\) \(2 y^{2}-4 x y+3 x y-6 x^{2}=0\) \(2 y(y-2 x)+3 x(y-2 x)=0\) \((y-2 x)(2 y+3 x)=0\) \(\therefore\) Equations of sides of triangle are \(\mathrm{x}+\mathrm{y}=1\), \(y-2 x=0,2 y+3 x=0\) On solving the above equations, we get the vertices of triangle as \(\left(\frac{1}{3}, \frac{2}{3}\right),(0,0)\) and \((-2,3)\), \(\therefore\) Centroid of triangle \(=\left(\frac{\frac{1}{3}+0-2}{3}, \frac{\frac{2}{3}+0+3}{3}\right)\) \(=\left(\frac{-5}{9}, \frac{11}{9}\right)\)
TS EAMCET-2020-10.09.2020
Co-Ordinate system
88270
Orthocentre of the triangle formed by the lines \(x+y=1\) and \(x y=0\) is
1 \((0,0)\)
2 \((0,1)\)
3 \((1,0)\)
4 \(-1,1)\)
Explanation:
(A) : Given lines are \(x+y=1\) and \(x y=0\) \(x y=0\) represents line \(x=0\) and \(y=0\). Triangle formed by lines \(\mathrm{x}+\mathrm{y}=1, \mathrm{x}=0\) And \(\mathrm{y}=0\) is \(\triangle \mathrm{ABC}\). \(\because \triangle \mathrm{ABC}\) is right angled triangle at \(\angle \mathrm{B}\). Orthocentre of \(\triangle A B C\) is at \(B(0,0)\).
COMEDK-2011
Co-Ordinate system
88286
If the lines \(x+3 y-9=0,4 x+b y-2=0\) and \(2 x-y-4=0\) are concurrent, then \(b\) equal to
1 -6
2 5
3 1
4 -5
Explanation:
(D): Given, lines \(x+3 y-9=0,4 x+\) by \(-2=0\) and \(2 \mathrm{x}-\mathrm{y}-4=0\) are concurrent \(\therefore\) The determinant is zero for concurrent line - i.e. \(\quad\left|\begin{array}{ccc}1 3 -9 \\ 4 \mathrm{~b} -2 \\ 2 -1 -4\end{array}\right|=0\) \(\Rightarrow 1(-4 b-2)-3(-16+4)-9(-4-2 b)=0\) \(\Rightarrow-4 \mathrm{~b}-2+36+36+18 \mathrm{~b}=0\) \(14 b=-70\) \(b=-5\)
BCECE-2016
Co-Ordinate system
88262
\(\mathrm{O}(0,0), \mathrm{A}(1,2), \mathrm{B}(3,4)\) are the vertices of \(\triangle O A B\). The joint equation of the altitude and median drawn from \(O\) is
1 \(x^{2}+7 x y-y^{2}=0\)
2 \(x^{2}+7 x y+y^{2}=0\)
3 \(3 x^{2}-x y-2 y^{2}=0\)
4 \(3 x^{2}+x y-2 y^{2}=0\)
Explanation:
(D) : \(D\) is midpoint of \(A B \Rightarrow D \equiv(2,3)\) Hence equation of \(\mathrm{OD}\) \(\frac{y-0}{3-0}=\frac{x-0}{2-0}\) \(2 y=3 x \Rightarrow 3 x-2 y=0 \tag{i}\) Slope of \(A B=\frac{4-2}{3-1}=1 \Rightarrow\) slope of OD is -1 . Equation of OP, \((y-0)=-(x-0)\) \(x+y=0 \tag{ii}\) Required joint equation is \((3 x-2 y)(x+y)=0\) \(3 x^{2}+x y-2 y^{2}=0\)
88317
The area of the triangle formed by the coordinate axes and the line \(4 x+5 y=20\) is (in square unit)
1 5
2 10
3 15
4 20
Explanation:
(B) : Given the line \(4 x+5 y=20\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(\therefore 4 \mathrm{x}+0=20\) \(x=5\) At \(\mathrm{y}-\operatorname{axis} \mathrm{x}=0\) \(\mathrm{y}=\frac{20}{5}=4\) \(\therefore\) Area of the triangle \(=\frac{1}{2} \times 4 \times 5=10\) Square units
Co-Ordinate system
88350
The centroid of the triangle formed by the lines \(x+y=1\) and \(2 y^{2}-x y-6 x^{2}=0\) is
1 \((0,0)\)
2 \(\left(\frac{5}{9}, \frac{11}{9}\right)\)
3 \(\left(\frac{-5}{9}, \frac{11}{9}\right)\)
4 \(\left(\frac{5}{9}, \frac{-11}{9}\right)\)
Explanation:
(C) : We have, \(2 y^{2}-x y-6 x^{2}=0\) \(2 y^{2}-4 x y+3 x y-6 x^{2}=0\) \(2 y(y-2 x)+3 x(y-2 x)=0\) \((y-2 x)(2 y+3 x)=0\) \(\therefore\) Equations of sides of triangle are \(\mathrm{x}+\mathrm{y}=1\), \(y-2 x=0,2 y+3 x=0\) On solving the above equations, we get the vertices of triangle as \(\left(\frac{1}{3}, \frac{2}{3}\right),(0,0)\) and \((-2,3)\), \(\therefore\) Centroid of triangle \(=\left(\frac{\frac{1}{3}+0-2}{3}, \frac{\frac{2}{3}+0+3}{3}\right)\) \(=\left(\frac{-5}{9}, \frac{11}{9}\right)\)
TS EAMCET-2020-10.09.2020
Co-Ordinate system
88270
Orthocentre of the triangle formed by the lines \(x+y=1\) and \(x y=0\) is
1 \((0,0)\)
2 \((0,1)\)
3 \((1,0)\)
4 \(-1,1)\)
Explanation:
(A) : Given lines are \(x+y=1\) and \(x y=0\) \(x y=0\) represents line \(x=0\) and \(y=0\). Triangle formed by lines \(\mathrm{x}+\mathrm{y}=1, \mathrm{x}=0\) And \(\mathrm{y}=0\) is \(\triangle \mathrm{ABC}\). \(\because \triangle \mathrm{ABC}\) is right angled triangle at \(\angle \mathrm{B}\). Orthocentre of \(\triangle A B C\) is at \(B(0,0)\).
COMEDK-2011
Co-Ordinate system
88286
If the lines \(x+3 y-9=0,4 x+b y-2=0\) and \(2 x-y-4=0\) are concurrent, then \(b\) equal to
1 -6
2 5
3 1
4 -5
Explanation:
(D): Given, lines \(x+3 y-9=0,4 x+\) by \(-2=0\) and \(2 \mathrm{x}-\mathrm{y}-4=0\) are concurrent \(\therefore\) The determinant is zero for concurrent line - i.e. \(\quad\left|\begin{array}{ccc}1 3 -9 \\ 4 \mathrm{~b} -2 \\ 2 -1 -4\end{array}\right|=0\) \(\Rightarrow 1(-4 b-2)-3(-16+4)-9(-4-2 b)=0\) \(\Rightarrow-4 \mathrm{~b}-2+36+36+18 \mathrm{~b}=0\) \(14 b=-70\) \(b=-5\)
BCECE-2016
Co-Ordinate system
88262
\(\mathrm{O}(0,0), \mathrm{A}(1,2), \mathrm{B}(3,4)\) are the vertices of \(\triangle O A B\). The joint equation of the altitude and median drawn from \(O\) is
1 \(x^{2}+7 x y-y^{2}=0\)
2 \(x^{2}+7 x y+y^{2}=0\)
3 \(3 x^{2}-x y-2 y^{2}=0\)
4 \(3 x^{2}+x y-2 y^{2}=0\)
Explanation:
(D) : \(D\) is midpoint of \(A B \Rightarrow D \equiv(2,3)\) Hence equation of \(\mathrm{OD}\) \(\frac{y-0}{3-0}=\frac{x-0}{2-0}\) \(2 y=3 x \Rightarrow 3 x-2 y=0 \tag{i}\) Slope of \(A B=\frac{4-2}{3-1}=1 \Rightarrow\) slope of OD is -1 . Equation of OP, \((y-0)=-(x-0)\) \(x+y=0 \tag{ii}\) Required joint equation is \((3 x-2 y)(x+y)=0\) \(3 x^{2}+x y-2 y^{2}=0\)
