88738
The nearest point on the line \(3 x+4 y=12\) from the origin is
1 \(\left(\frac{36}{25}, \frac{48}{25}\right)\)
2 \(\left(3, \frac{3}{4}\right)\)
3 \(\left(2, \frac{3}{2}\right)\)
4 None of these
Explanation:
(A) : If ' \(D\) ' be the foot of altitude, drawn from origin to the given line, then ' \(\mathrm{D}\) ' is the required point. Let \(\angle \mathrm{OBA}=\theta\)
\(\Rightarrow \tan \theta=4 / 3\)
\(\Rightarrow \angle \mathrm{DOA}=\theta\)
we have
\(\mathrm{OD}=12 / 5\)
If \(D\) is \((h, k)\) then \(h=O D \cos \theta, k=O D \sin \theta\)
\(\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25\).
BITSAT-2012
Straight Line
88739
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 \mathrm{x}+72=5 \mathrm{y}-10\)
or \(12 \mathrm{x}-5 \mathrm{y}+72+10=0 \quad \Rightarrow 12 \mathrm{x}-5 \mathrm{y}+82=0\)
The perpendicular distance from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\)
the line \(a x+b y+c=0\) is \(\frac{\left(a x_{1}+b y_{1}+c\right)}{\sqrt{a^{2}+b^{2}}}\).
The point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is
\(=\frac{|-12-5+82|}{\sqrt{12^{2}+(-5)^{2}}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)
BITSAT-2011
Straight Line
88737
The distance between the lines represented by the equation \(9 x^{2}+24 x y+16 y^{2}-12 x+16 y-12\) \(=\mathbf{0}\) is
88740
Point \((2,4)\) is translated through a distance \(3 \sqrt{2}\) units measured parallel to the line \(y-x=1\), in the direction of decreasing ordinates, to reach at \(Q\). if \(R\) is the image of \(Q\) with respect to the line \(y-x=1\), then coordinates of \(\mathbf{R}\) are
1 \((0,0)\)
2 \((-1,1)\)
3 \((6,6)\)
4 \((5,7)\)
Explanation:
(A) : The equation of a line through \(\mathrm{P}(2,4)\) and parallel to \(y-x=1\) is
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}\)
The coordinates of \(\mathrm{Q}\) are given by
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}=3 \sqrt{2} \Rightarrow x=-1, y=1\)
Thus, the coordinates of \(\mathrm{Q}\) are \((-1,1) . \mathrm{R}\) is the image of \(\mathrm{Q}\) with respect to the line \(\mathrm{y}-\mathrm{x}=1\)
Therefore, coordinates of \(\mathrm{R}\) are given by
\(\frac{x+1}{-1}=\frac{y-1}{1}=\frac{-2(1-1)}{(-1)^{2}+(1)^{2}} \Rightarrow \frac{x+1}{-1}=\frac{y-1}{1}=-1\)
\(\Rightarrow \mathrm{x}=0\) and \(\mathrm{y}=0\)
Hence, the coordinates of \(\mathrm{R}\) are \((0,0)\).
BITSAT-2015
Straight Line
88742
The length of the perpendicular from the point \(P(a, b)\) to the line \(\frac{x}{a}+\frac{y}{b}=1\) is
1 \(\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\) units
2 \(\left|\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
3 \(\left|\frac{\sqrt{a^{2}+b^{2}}}{a b}\right|\) units
4 \(\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
Explanation:
(B) : Given, equation of line
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
We have line \(b x+a y-a b=0\)
Length of perpendicular from \(P(a, b)\) on the given line is
\(d=\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\)
88738
The nearest point on the line \(3 x+4 y=12\) from the origin is
1 \(\left(\frac{36}{25}, \frac{48}{25}\right)\)
2 \(\left(3, \frac{3}{4}\right)\)
3 \(\left(2, \frac{3}{2}\right)\)
4 None of these
Explanation:
(A) : If ' \(D\) ' be the foot of altitude, drawn from origin to the given line, then ' \(\mathrm{D}\) ' is the required point. Let \(\angle \mathrm{OBA}=\theta\)
\(\Rightarrow \tan \theta=4 / 3\)
\(\Rightarrow \angle \mathrm{DOA}=\theta\)
we have
\(\mathrm{OD}=12 / 5\)
If \(D\) is \((h, k)\) then \(h=O D \cos \theta, k=O D \sin \theta\)
\(\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25\).
BITSAT-2012
Straight Line
88739
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 \mathrm{x}+72=5 \mathrm{y}-10\)
or \(12 \mathrm{x}-5 \mathrm{y}+72+10=0 \quad \Rightarrow 12 \mathrm{x}-5 \mathrm{y}+82=0\)
The perpendicular distance from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\)
the line \(a x+b y+c=0\) is \(\frac{\left(a x_{1}+b y_{1}+c\right)}{\sqrt{a^{2}+b^{2}}}\).
The point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is
\(=\frac{|-12-5+82|}{\sqrt{12^{2}+(-5)^{2}}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)
BITSAT-2011
Straight Line
88737
The distance between the lines represented by the equation \(9 x^{2}+24 x y+16 y^{2}-12 x+16 y-12\) \(=\mathbf{0}\) is
88740
Point \((2,4)\) is translated through a distance \(3 \sqrt{2}\) units measured parallel to the line \(y-x=1\), in the direction of decreasing ordinates, to reach at \(Q\). if \(R\) is the image of \(Q\) with respect to the line \(y-x=1\), then coordinates of \(\mathbf{R}\) are
1 \((0,0)\)
2 \((-1,1)\)
3 \((6,6)\)
4 \((5,7)\)
Explanation:
(A) : The equation of a line through \(\mathrm{P}(2,4)\) and parallel to \(y-x=1\) is
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}\)
The coordinates of \(\mathrm{Q}\) are given by
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}=3 \sqrt{2} \Rightarrow x=-1, y=1\)
Thus, the coordinates of \(\mathrm{Q}\) are \((-1,1) . \mathrm{R}\) is the image of \(\mathrm{Q}\) with respect to the line \(\mathrm{y}-\mathrm{x}=1\)
Therefore, coordinates of \(\mathrm{R}\) are given by
\(\frac{x+1}{-1}=\frac{y-1}{1}=\frac{-2(1-1)}{(-1)^{2}+(1)^{2}} \Rightarrow \frac{x+1}{-1}=\frac{y-1}{1}=-1\)
\(\Rightarrow \mathrm{x}=0\) and \(\mathrm{y}=0\)
Hence, the coordinates of \(\mathrm{R}\) are \((0,0)\).
BITSAT-2015
Straight Line
88742
The length of the perpendicular from the point \(P(a, b)\) to the line \(\frac{x}{a}+\frac{y}{b}=1\) is
1 \(\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\) units
2 \(\left|\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
3 \(\left|\frac{\sqrt{a^{2}+b^{2}}}{a b}\right|\) units
4 \(\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
Explanation:
(B) : Given, equation of line
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
We have line \(b x+a y-a b=0\)
Length of perpendicular from \(P(a, b)\) on the given line is
\(d=\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\)
88738
The nearest point on the line \(3 x+4 y=12\) from the origin is
1 \(\left(\frac{36}{25}, \frac{48}{25}\right)\)
2 \(\left(3, \frac{3}{4}\right)\)
3 \(\left(2, \frac{3}{2}\right)\)
4 None of these
Explanation:
(A) : If ' \(D\) ' be the foot of altitude, drawn from origin to the given line, then ' \(\mathrm{D}\) ' is the required point. Let \(\angle \mathrm{OBA}=\theta\)
\(\Rightarrow \tan \theta=4 / 3\)
\(\Rightarrow \angle \mathrm{DOA}=\theta\)
we have
\(\mathrm{OD}=12 / 5\)
If \(D\) is \((h, k)\) then \(h=O D \cos \theta, k=O D \sin \theta\)
\(\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25\).
BITSAT-2012
Straight Line
88739
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 \mathrm{x}+72=5 \mathrm{y}-10\)
or \(12 \mathrm{x}-5 \mathrm{y}+72+10=0 \quad \Rightarrow 12 \mathrm{x}-5 \mathrm{y}+82=0\)
The perpendicular distance from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\)
the line \(a x+b y+c=0\) is \(\frac{\left(a x_{1}+b y_{1}+c\right)}{\sqrt{a^{2}+b^{2}}}\).
The point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is
\(=\frac{|-12-5+82|}{\sqrt{12^{2}+(-5)^{2}}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)
BITSAT-2011
Straight Line
88737
The distance between the lines represented by the equation \(9 x^{2}+24 x y+16 y^{2}-12 x+16 y-12\) \(=\mathbf{0}\) is
88740
Point \((2,4)\) is translated through a distance \(3 \sqrt{2}\) units measured parallel to the line \(y-x=1\), in the direction of decreasing ordinates, to reach at \(Q\). if \(R\) is the image of \(Q\) with respect to the line \(y-x=1\), then coordinates of \(\mathbf{R}\) are
1 \((0,0)\)
2 \((-1,1)\)
3 \((6,6)\)
4 \((5,7)\)
Explanation:
(A) : The equation of a line through \(\mathrm{P}(2,4)\) and parallel to \(y-x=1\) is
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}\)
The coordinates of \(\mathrm{Q}\) are given by
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}=3 \sqrt{2} \Rightarrow x=-1, y=1\)
Thus, the coordinates of \(\mathrm{Q}\) are \((-1,1) . \mathrm{R}\) is the image of \(\mathrm{Q}\) with respect to the line \(\mathrm{y}-\mathrm{x}=1\)
Therefore, coordinates of \(\mathrm{R}\) are given by
\(\frac{x+1}{-1}=\frac{y-1}{1}=\frac{-2(1-1)}{(-1)^{2}+(1)^{2}} \Rightarrow \frac{x+1}{-1}=\frac{y-1}{1}=-1\)
\(\Rightarrow \mathrm{x}=0\) and \(\mathrm{y}=0\)
Hence, the coordinates of \(\mathrm{R}\) are \((0,0)\).
BITSAT-2015
Straight Line
88742
The length of the perpendicular from the point \(P(a, b)\) to the line \(\frac{x}{a}+\frac{y}{b}=1\) is
1 \(\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\) units
2 \(\left|\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
3 \(\left|\frac{\sqrt{a^{2}+b^{2}}}{a b}\right|\) units
4 \(\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
Explanation:
(B) : Given, equation of line
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
We have line \(b x+a y-a b=0\)
Length of perpendicular from \(P(a, b)\) on the given line is
\(d=\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Straight Line
88738
The nearest point on the line \(3 x+4 y=12\) from the origin is
1 \(\left(\frac{36}{25}, \frac{48}{25}\right)\)
2 \(\left(3, \frac{3}{4}\right)\)
3 \(\left(2, \frac{3}{2}\right)\)
4 None of these
Explanation:
(A) : If ' \(D\) ' be the foot of altitude, drawn from origin to the given line, then ' \(\mathrm{D}\) ' is the required point. Let \(\angle \mathrm{OBA}=\theta\)
\(\Rightarrow \tan \theta=4 / 3\)
\(\Rightarrow \angle \mathrm{DOA}=\theta\)
we have
\(\mathrm{OD}=12 / 5\)
If \(D\) is \((h, k)\) then \(h=O D \cos \theta, k=O D \sin \theta\)
\(\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25\).
BITSAT-2012
Straight Line
88739
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 \mathrm{x}+72=5 \mathrm{y}-10\)
or \(12 \mathrm{x}-5 \mathrm{y}+72+10=0 \quad \Rightarrow 12 \mathrm{x}-5 \mathrm{y}+82=0\)
The perpendicular distance from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\)
the line \(a x+b y+c=0\) is \(\frac{\left(a x_{1}+b y_{1}+c\right)}{\sqrt{a^{2}+b^{2}}}\).
The point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is
\(=\frac{|-12-5+82|}{\sqrt{12^{2}+(-5)^{2}}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)
BITSAT-2011
Straight Line
88737
The distance between the lines represented by the equation \(9 x^{2}+24 x y+16 y^{2}-12 x+16 y-12\) \(=\mathbf{0}\) is
88740
Point \((2,4)\) is translated through a distance \(3 \sqrt{2}\) units measured parallel to the line \(y-x=1\), in the direction of decreasing ordinates, to reach at \(Q\). if \(R\) is the image of \(Q\) with respect to the line \(y-x=1\), then coordinates of \(\mathbf{R}\) are
1 \((0,0)\)
2 \((-1,1)\)
3 \((6,6)\)
4 \((5,7)\)
Explanation:
(A) : The equation of a line through \(\mathrm{P}(2,4)\) and parallel to \(y-x=1\) is
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}\)
The coordinates of \(\mathrm{Q}\) are given by
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}=3 \sqrt{2} \Rightarrow x=-1, y=1\)
Thus, the coordinates of \(\mathrm{Q}\) are \((-1,1) . \mathrm{R}\) is the image of \(\mathrm{Q}\) with respect to the line \(\mathrm{y}-\mathrm{x}=1\)
Therefore, coordinates of \(\mathrm{R}\) are given by
\(\frac{x+1}{-1}=\frac{y-1}{1}=\frac{-2(1-1)}{(-1)^{2}+(1)^{2}} \Rightarrow \frac{x+1}{-1}=\frac{y-1}{1}=-1\)
\(\Rightarrow \mathrm{x}=0\) and \(\mathrm{y}=0\)
Hence, the coordinates of \(\mathrm{R}\) are \((0,0)\).
BITSAT-2015
Straight Line
88742
The length of the perpendicular from the point \(P(a, b)\) to the line \(\frac{x}{a}+\frac{y}{b}=1\) is
1 \(\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\) units
2 \(\left|\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
3 \(\left|\frac{\sqrt{a^{2}+b^{2}}}{a b}\right|\) units
4 \(\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
Explanation:
(B) : Given, equation of line
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
We have line \(b x+a y-a b=0\)
Length of perpendicular from \(P(a, b)\) on the given line is
\(d=\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\)
88738
The nearest point on the line \(3 x+4 y=12\) from the origin is
1 \(\left(\frac{36}{25}, \frac{48}{25}\right)\)
2 \(\left(3, \frac{3}{4}\right)\)
3 \(\left(2, \frac{3}{2}\right)\)
4 None of these
Explanation:
(A) : If ' \(D\) ' be the foot of altitude, drawn from origin to the given line, then ' \(\mathrm{D}\) ' is the required point. Let \(\angle \mathrm{OBA}=\theta\)
\(\Rightarrow \tan \theta=4 / 3\)
\(\Rightarrow \angle \mathrm{DOA}=\theta\)
we have
\(\mathrm{OD}=12 / 5\)
If \(D\) is \((h, k)\) then \(h=O D \cos \theta, k=O D \sin \theta\)
\(\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25\).
BITSAT-2012
Straight Line
88739
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 \mathrm{x}+72=5 \mathrm{y}-10\)
or \(12 \mathrm{x}-5 \mathrm{y}+72+10=0 \quad \Rightarrow 12 \mathrm{x}-5 \mathrm{y}+82=0\)
The perpendicular distance from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\)
the line \(a x+b y+c=0\) is \(\frac{\left(a x_{1}+b y_{1}+c\right)}{\sqrt{a^{2}+b^{2}}}\).
The point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is
\(=\frac{|-12-5+82|}{\sqrt{12^{2}+(-5)^{2}}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)
BITSAT-2011
Straight Line
88737
The distance between the lines represented by the equation \(9 x^{2}+24 x y+16 y^{2}-12 x+16 y-12\) \(=\mathbf{0}\) is
88740
Point \((2,4)\) is translated through a distance \(3 \sqrt{2}\) units measured parallel to the line \(y-x=1\), in the direction of decreasing ordinates, to reach at \(Q\). if \(R\) is the image of \(Q\) with respect to the line \(y-x=1\), then coordinates of \(\mathbf{R}\) are
1 \((0,0)\)
2 \((-1,1)\)
3 \((6,6)\)
4 \((5,7)\)
Explanation:
(A) : The equation of a line through \(\mathrm{P}(2,4)\) and parallel to \(y-x=1\) is
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}\)
The coordinates of \(\mathrm{Q}\) are given by
\(\frac{x-2}{\cos \frac{\pi}{4}}=\frac{y-4}{\sin \frac{\pi}{4}}=3 \sqrt{2} \Rightarrow x=-1, y=1\)
Thus, the coordinates of \(\mathrm{Q}\) are \((-1,1) . \mathrm{R}\) is the image of \(\mathrm{Q}\) with respect to the line \(\mathrm{y}-\mathrm{x}=1\)
Therefore, coordinates of \(\mathrm{R}\) are given by
\(\frac{x+1}{-1}=\frac{y-1}{1}=\frac{-2(1-1)}{(-1)^{2}+(1)^{2}} \Rightarrow \frac{x+1}{-1}=\frac{y-1}{1}=-1\)
\(\Rightarrow \mathrm{x}=0\) and \(\mathrm{y}=0\)
Hence, the coordinates of \(\mathrm{R}\) are \((0,0)\).
BITSAT-2015
Straight Line
88742
The length of the perpendicular from the point \(P(a, b)\) to the line \(\frac{x}{a}+\frac{y}{b}=1\) is
1 \(\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\) units
2 \(\left|\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
3 \(\left|\frac{\sqrt{a^{2}+b^{2}}}{a b}\right|\) units
4 \(\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
Explanation:
(B) : Given, equation of line
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
We have line \(b x+a y-a b=0\)
Length of perpendicular from \(P(a, b)\) on the given line is
\(d=\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\)
