88273
A straight line meets the coordinates axes at \(\mathbf{A}\) and \(B\), so that the centroid of the triangle \(O A B\) is \((1,2)\) Then the equation of the line \(A B\) is
1 \(x+y=6\)
2 \(2 x+y=6\)
3 \(x+2 y=6\)
4 \(3 x+y=6\)
Explanation:
(B) : Since, straight line meets the coordinate axes at \(\mathrm{A}\) and \(\mathrm{B}\), so equation of line in intercept form is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\). \(\mathrm{G}\left(\frac{0+\mathrm{a}+0}{3}, \frac{0+0+\mathrm{b}}{3}\right)=(1,2)\) \(\Rightarrow \frac{\mathrm{a}}{3}=1 \Rightarrow \mathrm{a}=3, \frac{\mathrm{b}}{3}=2 \Rightarrow \mathrm{b}=6\) Hence, required equation of line is \(\frac{x}{3}+\frac{y}{6}=1 \Rightarrow 2 x+y=6\)
COMEDK-2015
Co-Ordinate system
88274
The medians \(A D\) and \(B E\) of a triangle with vertices \(A(0, b), B(0,0)\) and \(C(a, 0)\) are perpendicular to each other, if
1 \(a=\frac{b}{2}\)
2 \(\mathrm{b}=\frac{\mathrm{a}}{2}\)
3 \(\mathrm{ab}=1\)
4 \(a= \pm \sqrt{2} b\)
Explanation:
(D) : We have, BE and AD are the medians SO, E and \(\mathrm{D}\) are the mid points of \(\mathrm{AC}\) and \(\mathrm{BC}\) respectively. \(\therefore\) Coordinates of \(\mathrm{E}=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) And coordinates of \(\mathrm{D}=\left(\frac{\mathrm{a}}{2}, 0\right)\) Now, slope of median BE, \(\mathrm{m}_{1}=\frac{\mathrm{b}}{\mathrm{a}}\) Also, slope of median AD, \(\mathrm{m}_{2}=\frac{-2 \mathrm{~b}}{\mathrm{a}}\) Now, \(\mathrm{m}_{1} \& \mathrm{~m}_{2}\) are perpendicular if \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}} \times \frac{-2 \mathrm{~b}}{\mathrm{a}}=-1 \Rightarrow 2 \mathrm{~b}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{a}= \pm \sqrt{2} \mathrm{~b}\)
COMEDK-2015
Co-Ordinate system
88276
A straight line through the point \(A(3,4)\) is such that its intercept between the axes is bisected at \(A\), its equation is
1 \(3 x-4 y+7=0\)
2 \(4 x+3 y=24\)
3 \(3 x+4 y=25\)
4 \(x+y=7\)
Explanation:
(B) : A is mid point of line PQ. \(\therefore 3=\frac{a+0}{2} \Rightarrow \mathrm{a}=6\) and \(4=\frac{0+\mathrm{b}}{2} \Rightarrow \mathrm{b}=8\) Thus, equation of line is \(\frac{x}{6}+\frac{y}{8}=1\) \(\Rightarrow 4 \mathrm{x}+3 \mathrm{y}=24\)
VITEEE-2012
Co-Ordinate system
88277
If a plane meets the coordinate axes at \(A, B\) and C such that the centroid of the triangle is \((1,2\), 4), then the equation of the plane is
1 \(x+2 y+4 z=12\)
2 \(4 x+2 y+z=12\)
3 \(x+2 y+4 z=3\)
4 \(4 x+2 y+z=3\)
Explanation:
(B) : Let the equation of the plane is \(\frac{\mathrm{x}}{\alpha}+\frac{\mathrm{y}}{\beta}+\frac{\mathrm{z}}{\gamma}=1\) Then, \(\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)\) and \(\mathrm{C}(0,0, \gamma)\) are the points on the co-ordinate axes, The centroid of the triangle is \((1,2,4)\). \(\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3\) \(\frac{\beta}{3}=2 \Rightarrow \beta=6\) and \(\quad \frac{\gamma}{3}=4 \Rightarrow \gamma=12\) \(\therefore \quad\) The equation of the plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \Rightarrow 4 x+2 y+z=12\)
VITEEE-2014
Co-Ordinate system
88278
The normals at three points \(P, Q\) and \(R\) of the parabola \(y^{2}=4 a x\) meet at \((h, k)\). The centroid of the \(\triangle \mathrm{PQR}\) lies on
1 \(x=0\)
2 \(y=0\)
3 \(x=-a\)
4 \(y=a\)
Explanation:
(B) : We know that, the sum of ordinates of feet of normals draw from a point the parabola, \(y^{2}=4 a x\) is always zero. Now, as normals at three points \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\) of points \(\mathrm{P}\), \(Q\) and \(R\) of parabola \(y^{2}=4 a x\) meet at \((h, k)\). \(\Rightarrow\) The normals from \((\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}^{2}=4 \mathrm{ax}\) meet the parabola at \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\). \(\Rightarrow \mathrm{y}\)-coordinate \(\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}\) of these points and \(\mathrm{R}\) will be zero. \(\Rightarrow y\)-coordinate of the centroid of \(\triangle \mathrm{PQR} \text { i.e., }\) \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}=\frac{0}{3}=0\) \(\therefore \quad\) Centroid lies on \(\mathrm{y}=0\)
88273
A straight line meets the coordinates axes at \(\mathbf{A}\) and \(B\), so that the centroid of the triangle \(O A B\) is \((1,2)\) Then the equation of the line \(A B\) is
1 \(x+y=6\)
2 \(2 x+y=6\)
3 \(x+2 y=6\)
4 \(3 x+y=6\)
Explanation:
(B) : Since, straight line meets the coordinate axes at \(\mathrm{A}\) and \(\mathrm{B}\), so equation of line in intercept form is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\). \(\mathrm{G}\left(\frac{0+\mathrm{a}+0}{3}, \frac{0+0+\mathrm{b}}{3}\right)=(1,2)\) \(\Rightarrow \frac{\mathrm{a}}{3}=1 \Rightarrow \mathrm{a}=3, \frac{\mathrm{b}}{3}=2 \Rightarrow \mathrm{b}=6\) Hence, required equation of line is \(\frac{x}{3}+\frac{y}{6}=1 \Rightarrow 2 x+y=6\)
COMEDK-2015
Co-Ordinate system
88274
The medians \(A D\) and \(B E\) of a triangle with vertices \(A(0, b), B(0,0)\) and \(C(a, 0)\) are perpendicular to each other, if
1 \(a=\frac{b}{2}\)
2 \(\mathrm{b}=\frac{\mathrm{a}}{2}\)
3 \(\mathrm{ab}=1\)
4 \(a= \pm \sqrt{2} b\)
Explanation:
(D) : We have, BE and AD are the medians SO, E and \(\mathrm{D}\) are the mid points of \(\mathrm{AC}\) and \(\mathrm{BC}\) respectively. \(\therefore\) Coordinates of \(\mathrm{E}=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) And coordinates of \(\mathrm{D}=\left(\frac{\mathrm{a}}{2}, 0\right)\) Now, slope of median BE, \(\mathrm{m}_{1}=\frac{\mathrm{b}}{\mathrm{a}}\) Also, slope of median AD, \(\mathrm{m}_{2}=\frac{-2 \mathrm{~b}}{\mathrm{a}}\) Now, \(\mathrm{m}_{1} \& \mathrm{~m}_{2}\) are perpendicular if \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}} \times \frac{-2 \mathrm{~b}}{\mathrm{a}}=-1 \Rightarrow 2 \mathrm{~b}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{a}= \pm \sqrt{2} \mathrm{~b}\)
COMEDK-2015
Co-Ordinate system
88276
A straight line through the point \(A(3,4)\) is such that its intercept between the axes is bisected at \(A\), its equation is
1 \(3 x-4 y+7=0\)
2 \(4 x+3 y=24\)
3 \(3 x+4 y=25\)
4 \(x+y=7\)
Explanation:
(B) : A is mid point of line PQ. \(\therefore 3=\frac{a+0}{2} \Rightarrow \mathrm{a}=6\) and \(4=\frac{0+\mathrm{b}}{2} \Rightarrow \mathrm{b}=8\) Thus, equation of line is \(\frac{x}{6}+\frac{y}{8}=1\) \(\Rightarrow 4 \mathrm{x}+3 \mathrm{y}=24\)
VITEEE-2012
Co-Ordinate system
88277
If a plane meets the coordinate axes at \(A, B\) and C such that the centroid of the triangle is \((1,2\), 4), then the equation of the plane is
1 \(x+2 y+4 z=12\)
2 \(4 x+2 y+z=12\)
3 \(x+2 y+4 z=3\)
4 \(4 x+2 y+z=3\)
Explanation:
(B) : Let the equation of the plane is \(\frac{\mathrm{x}}{\alpha}+\frac{\mathrm{y}}{\beta}+\frac{\mathrm{z}}{\gamma}=1\) Then, \(\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)\) and \(\mathrm{C}(0,0, \gamma)\) are the points on the co-ordinate axes, The centroid of the triangle is \((1,2,4)\). \(\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3\) \(\frac{\beta}{3}=2 \Rightarrow \beta=6\) and \(\quad \frac{\gamma}{3}=4 \Rightarrow \gamma=12\) \(\therefore \quad\) The equation of the plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \Rightarrow 4 x+2 y+z=12\)
VITEEE-2014
Co-Ordinate system
88278
The normals at three points \(P, Q\) and \(R\) of the parabola \(y^{2}=4 a x\) meet at \((h, k)\). The centroid of the \(\triangle \mathrm{PQR}\) lies on
1 \(x=0\)
2 \(y=0\)
3 \(x=-a\)
4 \(y=a\)
Explanation:
(B) : We know that, the sum of ordinates of feet of normals draw from a point the parabola, \(y^{2}=4 a x\) is always zero. Now, as normals at three points \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\) of points \(\mathrm{P}\), \(Q\) and \(R\) of parabola \(y^{2}=4 a x\) meet at \((h, k)\). \(\Rightarrow\) The normals from \((\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}^{2}=4 \mathrm{ax}\) meet the parabola at \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\). \(\Rightarrow \mathrm{y}\)-coordinate \(\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}\) of these points and \(\mathrm{R}\) will be zero. \(\Rightarrow y\)-coordinate of the centroid of \(\triangle \mathrm{PQR} \text { i.e., }\) \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}=\frac{0}{3}=0\) \(\therefore \quad\) Centroid lies on \(\mathrm{y}=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Co-Ordinate system
88273
A straight line meets the coordinates axes at \(\mathbf{A}\) and \(B\), so that the centroid of the triangle \(O A B\) is \((1,2)\) Then the equation of the line \(A B\) is
1 \(x+y=6\)
2 \(2 x+y=6\)
3 \(x+2 y=6\)
4 \(3 x+y=6\)
Explanation:
(B) : Since, straight line meets the coordinate axes at \(\mathrm{A}\) and \(\mathrm{B}\), so equation of line in intercept form is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\). \(\mathrm{G}\left(\frac{0+\mathrm{a}+0}{3}, \frac{0+0+\mathrm{b}}{3}\right)=(1,2)\) \(\Rightarrow \frac{\mathrm{a}}{3}=1 \Rightarrow \mathrm{a}=3, \frac{\mathrm{b}}{3}=2 \Rightarrow \mathrm{b}=6\) Hence, required equation of line is \(\frac{x}{3}+\frac{y}{6}=1 \Rightarrow 2 x+y=6\)
COMEDK-2015
Co-Ordinate system
88274
The medians \(A D\) and \(B E\) of a triangle with vertices \(A(0, b), B(0,0)\) and \(C(a, 0)\) are perpendicular to each other, if
1 \(a=\frac{b}{2}\)
2 \(\mathrm{b}=\frac{\mathrm{a}}{2}\)
3 \(\mathrm{ab}=1\)
4 \(a= \pm \sqrt{2} b\)
Explanation:
(D) : We have, BE and AD are the medians SO, E and \(\mathrm{D}\) are the mid points of \(\mathrm{AC}\) and \(\mathrm{BC}\) respectively. \(\therefore\) Coordinates of \(\mathrm{E}=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) And coordinates of \(\mathrm{D}=\left(\frac{\mathrm{a}}{2}, 0\right)\) Now, slope of median BE, \(\mathrm{m}_{1}=\frac{\mathrm{b}}{\mathrm{a}}\) Also, slope of median AD, \(\mathrm{m}_{2}=\frac{-2 \mathrm{~b}}{\mathrm{a}}\) Now, \(\mathrm{m}_{1} \& \mathrm{~m}_{2}\) are perpendicular if \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}} \times \frac{-2 \mathrm{~b}}{\mathrm{a}}=-1 \Rightarrow 2 \mathrm{~b}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{a}= \pm \sqrt{2} \mathrm{~b}\)
COMEDK-2015
Co-Ordinate system
88276
A straight line through the point \(A(3,4)\) is such that its intercept between the axes is bisected at \(A\), its equation is
1 \(3 x-4 y+7=0\)
2 \(4 x+3 y=24\)
3 \(3 x+4 y=25\)
4 \(x+y=7\)
Explanation:
(B) : A is mid point of line PQ. \(\therefore 3=\frac{a+0}{2} \Rightarrow \mathrm{a}=6\) and \(4=\frac{0+\mathrm{b}}{2} \Rightarrow \mathrm{b}=8\) Thus, equation of line is \(\frac{x}{6}+\frac{y}{8}=1\) \(\Rightarrow 4 \mathrm{x}+3 \mathrm{y}=24\)
VITEEE-2012
Co-Ordinate system
88277
If a plane meets the coordinate axes at \(A, B\) and C such that the centroid of the triangle is \((1,2\), 4), then the equation of the plane is
1 \(x+2 y+4 z=12\)
2 \(4 x+2 y+z=12\)
3 \(x+2 y+4 z=3\)
4 \(4 x+2 y+z=3\)
Explanation:
(B) : Let the equation of the plane is \(\frac{\mathrm{x}}{\alpha}+\frac{\mathrm{y}}{\beta}+\frac{\mathrm{z}}{\gamma}=1\) Then, \(\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)\) and \(\mathrm{C}(0,0, \gamma)\) are the points on the co-ordinate axes, The centroid of the triangle is \((1,2,4)\). \(\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3\) \(\frac{\beta}{3}=2 \Rightarrow \beta=6\) and \(\quad \frac{\gamma}{3}=4 \Rightarrow \gamma=12\) \(\therefore \quad\) The equation of the plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \Rightarrow 4 x+2 y+z=12\)
VITEEE-2014
Co-Ordinate system
88278
The normals at three points \(P, Q\) and \(R\) of the parabola \(y^{2}=4 a x\) meet at \((h, k)\). The centroid of the \(\triangle \mathrm{PQR}\) lies on
1 \(x=0\)
2 \(y=0\)
3 \(x=-a\)
4 \(y=a\)
Explanation:
(B) : We know that, the sum of ordinates of feet of normals draw from a point the parabola, \(y^{2}=4 a x\) is always zero. Now, as normals at three points \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\) of points \(\mathrm{P}\), \(Q\) and \(R\) of parabola \(y^{2}=4 a x\) meet at \((h, k)\). \(\Rightarrow\) The normals from \((\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}^{2}=4 \mathrm{ax}\) meet the parabola at \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\). \(\Rightarrow \mathrm{y}\)-coordinate \(\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}\) of these points and \(\mathrm{R}\) will be zero. \(\Rightarrow y\)-coordinate of the centroid of \(\triangle \mathrm{PQR} \text { i.e., }\) \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}=\frac{0}{3}=0\) \(\therefore \quad\) Centroid lies on \(\mathrm{y}=0\)
88273
A straight line meets the coordinates axes at \(\mathbf{A}\) and \(B\), so that the centroid of the triangle \(O A B\) is \((1,2)\) Then the equation of the line \(A B\) is
1 \(x+y=6\)
2 \(2 x+y=6\)
3 \(x+2 y=6\)
4 \(3 x+y=6\)
Explanation:
(B) : Since, straight line meets the coordinate axes at \(\mathrm{A}\) and \(\mathrm{B}\), so equation of line in intercept form is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\). \(\mathrm{G}\left(\frac{0+\mathrm{a}+0}{3}, \frac{0+0+\mathrm{b}}{3}\right)=(1,2)\) \(\Rightarrow \frac{\mathrm{a}}{3}=1 \Rightarrow \mathrm{a}=3, \frac{\mathrm{b}}{3}=2 \Rightarrow \mathrm{b}=6\) Hence, required equation of line is \(\frac{x}{3}+\frac{y}{6}=1 \Rightarrow 2 x+y=6\)
COMEDK-2015
Co-Ordinate system
88274
The medians \(A D\) and \(B E\) of a triangle with vertices \(A(0, b), B(0,0)\) and \(C(a, 0)\) are perpendicular to each other, if
1 \(a=\frac{b}{2}\)
2 \(\mathrm{b}=\frac{\mathrm{a}}{2}\)
3 \(\mathrm{ab}=1\)
4 \(a= \pm \sqrt{2} b\)
Explanation:
(D) : We have, BE and AD are the medians SO, E and \(\mathrm{D}\) are the mid points of \(\mathrm{AC}\) and \(\mathrm{BC}\) respectively. \(\therefore\) Coordinates of \(\mathrm{E}=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) And coordinates of \(\mathrm{D}=\left(\frac{\mathrm{a}}{2}, 0\right)\) Now, slope of median BE, \(\mathrm{m}_{1}=\frac{\mathrm{b}}{\mathrm{a}}\) Also, slope of median AD, \(\mathrm{m}_{2}=\frac{-2 \mathrm{~b}}{\mathrm{a}}\) Now, \(\mathrm{m}_{1} \& \mathrm{~m}_{2}\) are perpendicular if \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}} \times \frac{-2 \mathrm{~b}}{\mathrm{a}}=-1 \Rightarrow 2 \mathrm{~b}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{a}= \pm \sqrt{2} \mathrm{~b}\)
COMEDK-2015
Co-Ordinate system
88276
A straight line through the point \(A(3,4)\) is such that its intercept between the axes is bisected at \(A\), its equation is
1 \(3 x-4 y+7=0\)
2 \(4 x+3 y=24\)
3 \(3 x+4 y=25\)
4 \(x+y=7\)
Explanation:
(B) : A is mid point of line PQ. \(\therefore 3=\frac{a+0}{2} \Rightarrow \mathrm{a}=6\) and \(4=\frac{0+\mathrm{b}}{2} \Rightarrow \mathrm{b}=8\) Thus, equation of line is \(\frac{x}{6}+\frac{y}{8}=1\) \(\Rightarrow 4 \mathrm{x}+3 \mathrm{y}=24\)
VITEEE-2012
Co-Ordinate system
88277
If a plane meets the coordinate axes at \(A, B\) and C such that the centroid of the triangle is \((1,2\), 4), then the equation of the plane is
1 \(x+2 y+4 z=12\)
2 \(4 x+2 y+z=12\)
3 \(x+2 y+4 z=3\)
4 \(4 x+2 y+z=3\)
Explanation:
(B) : Let the equation of the plane is \(\frac{\mathrm{x}}{\alpha}+\frac{\mathrm{y}}{\beta}+\frac{\mathrm{z}}{\gamma}=1\) Then, \(\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)\) and \(\mathrm{C}(0,0, \gamma)\) are the points on the co-ordinate axes, The centroid of the triangle is \((1,2,4)\). \(\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3\) \(\frac{\beta}{3}=2 \Rightarrow \beta=6\) and \(\quad \frac{\gamma}{3}=4 \Rightarrow \gamma=12\) \(\therefore \quad\) The equation of the plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \Rightarrow 4 x+2 y+z=12\)
VITEEE-2014
Co-Ordinate system
88278
The normals at three points \(P, Q\) and \(R\) of the parabola \(y^{2}=4 a x\) meet at \((h, k)\). The centroid of the \(\triangle \mathrm{PQR}\) lies on
1 \(x=0\)
2 \(y=0\)
3 \(x=-a\)
4 \(y=a\)
Explanation:
(B) : We know that, the sum of ordinates of feet of normals draw from a point the parabola, \(y^{2}=4 a x\) is always zero. Now, as normals at three points \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\) of points \(\mathrm{P}\), \(Q\) and \(R\) of parabola \(y^{2}=4 a x\) meet at \((h, k)\). \(\Rightarrow\) The normals from \((\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}^{2}=4 \mathrm{ax}\) meet the parabola at \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\). \(\Rightarrow \mathrm{y}\)-coordinate \(\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}\) of these points and \(\mathrm{R}\) will be zero. \(\Rightarrow y\)-coordinate of the centroid of \(\triangle \mathrm{PQR} \text { i.e., }\) \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}=\frac{0}{3}=0\) \(\therefore \quad\) Centroid lies on \(\mathrm{y}=0\)
88273
A straight line meets the coordinates axes at \(\mathbf{A}\) and \(B\), so that the centroid of the triangle \(O A B\) is \((1,2)\) Then the equation of the line \(A B\) is
1 \(x+y=6\)
2 \(2 x+y=6\)
3 \(x+2 y=6\)
4 \(3 x+y=6\)
Explanation:
(B) : Since, straight line meets the coordinate axes at \(\mathrm{A}\) and \(\mathrm{B}\), so equation of line in intercept form is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\). \(\mathrm{G}\left(\frac{0+\mathrm{a}+0}{3}, \frac{0+0+\mathrm{b}}{3}\right)=(1,2)\) \(\Rightarrow \frac{\mathrm{a}}{3}=1 \Rightarrow \mathrm{a}=3, \frac{\mathrm{b}}{3}=2 \Rightarrow \mathrm{b}=6\) Hence, required equation of line is \(\frac{x}{3}+\frac{y}{6}=1 \Rightarrow 2 x+y=6\)
COMEDK-2015
Co-Ordinate system
88274
The medians \(A D\) and \(B E\) of a triangle with vertices \(A(0, b), B(0,0)\) and \(C(a, 0)\) are perpendicular to each other, if
1 \(a=\frac{b}{2}\)
2 \(\mathrm{b}=\frac{\mathrm{a}}{2}\)
3 \(\mathrm{ab}=1\)
4 \(a= \pm \sqrt{2} b\)
Explanation:
(D) : We have, BE and AD are the medians SO, E and \(\mathrm{D}\) are the mid points of \(\mathrm{AC}\) and \(\mathrm{BC}\) respectively. \(\therefore\) Coordinates of \(\mathrm{E}=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) And coordinates of \(\mathrm{D}=\left(\frac{\mathrm{a}}{2}, 0\right)\) Now, slope of median BE, \(\mathrm{m}_{1}=\frac{\mathrm{b}}{\mathrm{a}}\) Also, slope of median AD, \(\mathrm{m}_{2}=\frac{-2 \mathrm{~b}}{\mathrm{a}}\) Now, \(\mathrm{m}_{1} \& \mathrm{~m}_{2}\) are perpendicular if \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}} \times \frac{-2 \mathrm{~b}}{\mathrm{a}}=-1 \Rightarrow 2 \mathrm{~b}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{a}= \pm \sqrt{2} \mathrm{~b}\)
COMEDK-2015
Co-Ordinate system
88276
A straight line through the point \(A(3,4)\) is such that its intercept between the axes is bisected at \(A\), its equation is
1 \(3 x-4 y+7=0\)
2 \(4 x+3 y=24\)
3 \(3 x+4 y=25\)
4 \(x+y=7\)
Explanation:
(B) : A is mid point of line PQ. \(\therefore 3=\frac{a+0}{2} \Rightarrow \mathrm{a}=6\) and \(4=\frac{0+\mathrm{b}}{2} \Rightarrow \mathrm{b}=8\) Thus, equation of line is \(\frac{x}{6}+\frac{y}{8}=1\) \(\Rightarrow 4 \mathrm{x}+3 \mathrm{y}=24\)
VITEEE-2012
Co-Ordinate system
88277
If a plane meets the coordinate axes at \(A, B\) and C such that the centroid of the triangle is \((1,2\), 4), then the equation of the plane is
1 \(x+2 y+4 z=12\)
2 \(4 x+2 y+z=12\)
3 \(x+2 y+4 z=3\)
4 \(4 x+2 y+z=3\)
Explanation:
(B) : Let the equation of the plane is \(\frac{\mathrm{x}}{\alpha}+\frac{\mathrm{y}}{\beta}+\frac{\mathrm{z}}{\gamma}=1\) Then, \(\mathrm{A}(\alpha, 0,0), \mathrm{B}(0, \beta, 0)\) and \(\mathrm{C}(0,0, \gamma)\) are the points on the co-ordinate axes, The centroid of the triangle is \((1,2,4)\). \(\therefore \frac{\alpha}{3}=1 \Rightarrow \alpha=3\) \(\frac{\beta}{3}=2 \Rightarrow \beta=6\) and \(\quad \frac{\gamma}{3}=4 \Rightarrow \gamma=12\) \(\therefore \quad\) The equation of the plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1 \Rightarrow 4 x+2 y+z=12\)
VITEEE-2014
Co-Ordinate system
88278
The normals at three points \(P, Q\) and \(R\) of the parabola \(y^{2}=4 a x\) meet at \((h, k)\). The centroid of the \(\triangle \mathrm{PQR}\) lies on
1 \(x=0\)
2 \(y=0\)
3 \(x=-a\)
4 \(y=a\)
Explanation:
(B) : We know that, the sum of ordinates of feet of normals draw from a point the parabola, \(y^{2}=4 a x\) is always zero. Now, as normals at three points \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\) of points \(\mathrm{P}\), \(Q\) and \(R\) of parabola \(y^{2}=4 a x\) meet at \((h, k)\). \(\Rightarrow\) The normals from \((\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}^{2}=4 \mathrm{ax}\) meet the parabola at \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\). \(\Rightarrow \mathrm{y}\)-coordinate \(\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}\) of these points and \(\mathrm{R}\) will be zero. \(\Rightarrow y\)-coordinate of the centroid of \(\triangle \mathrm{PQR} \text { i.e., }\) \(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}=\frac{0}{3}=0\) \(\therefore \quad\) Centroid lies on \(\mathrm{y}=0\)
