88636
A straight line through a fixed point \((2,3)\) intersects the coordinate axes at distinct points \(P\) and \(Q\). If \(O\) is the origin and the rectangle OPRQ is completed, then the locus of \(R\) is :
1 \(2 x+3 y=x y\)
2 \(3 x+2 y=x y\)
3 \(3 x+2 y=6 x y\)
4 \(3 x+2 y=6\)
Explanation:
Equation of PQ is : \(\frac{x}{h}+\frac{y}{k} = 1\) …..(i)
Since, (1) passes through the fixed point \((2,3)\)
Then, \(\frac{2}{\mathrm{~h}}+\frac{3}{\mathrm{k}}=1\)
Then, the locus of \(\mathrm{R}\) is \(\frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}=1\)
or \(3 x+2 y=x y\).
BITSAT-2018
Straight Line
88637
If \((a,-2 a), a>0\) is the midpoint of a line segment intercepted between the co-ordinate axis, then the equation of the line is
1 \(2 x-y=4 a\)
2 \(x-2 y=5 a\)
3 \(2 x-y+4 a=0\)
4 \(x-2 y+4 a=0\)
Explanation:
(A) : Given, point \((a,-2 a)\) is the midpoint of the intercepts.
let \((A, 0)\) and \((0, B)\) are \(x\) and \(y\) intercept
Then, the equation of line is,
\(\frac{\mathrm{x}}{\mathrm{A}}+\frac{\mathrm{y}}{\mathrm{B}}=1\)
Since, \((a,-2 a)\) is a mid-point then,
\(\quad \begin{aligned} \mathrm{a} =\frac{\mathrm{A}}{2} \text { and }-2 \mathrm{a}=\frac{\mathrm{B}}{2} \\ \mathrm{~A} =2 \mathrm{a} \text { and } \mathrm{B}=-4 \mathrm{a}\end{aligned}\)
Hence required equation of line is,
\(\frac{\mathrm{x}}{2 \mathrm{a}}-\frac{\mathrm{y}}{4 \mathrm{a}}=1 \quad 2 \mathrm{x}-\mathrm{y}=4 \mathrm{a}\)
MHT CET-2020
Straight Line
88638
The equations of the lines which make intercepts on the axes whose sum is 8 and product is 15 are
(B) :
Let ' \(a\) ' and ' \(b\) ' be the intercepts made by line.
We have \(a+b=8\) and \(\mathrm{ab}=15\)
Solving, we get \((a, b)=(3,5)\) or \((5,3)\)
Let equation of line be \(\frac{x}{a}+\frac{y}{b}=1\)
When \(\mathrm{a}=3, \mathrm{~b}=5\) from (1), we get
\(\frac{x}{3}+\frac{y}{5}=1 \quad \Rightarrow \quad 5 x+3 y-15=0\)
When \(a=5, b=3\) from (1), we get
\(\frac{x}{5}+\frac{y}{3}=1 \quad \Rightarrow \quad 3 x+5 y-15=0\)
Hence, the equation of line is \(3 x+5 y-15=0\) and \(5 x+\) \(3 y-15=0\)
MHT CET-2020
Straight Line
88639
The equation of a line passing through the point \((7,-4)\) and perpendicular to the line passing through the points \((2,3)\) and \((1,-2)\) is
1 \(x-2 y-15=0\)
2 \(x+2 y+1=0\)
3 \(x-5 y-13=0\)
4 \(x+5 y+13=0\)
Explanation:
(D) : Slope between two given point \((2,3)\) and (1,2) is,
\(\mathrm{m}=\frac{-2-3}{1-2}=5\)
Perpendicular slope is \(\mathrm{m}_{1}=-\frac{1}{5}\)
The equation of line with this slope is,
\(y=-\frac{1}{5} x+c \tag{i}\)
Now equation of required line who passing through the point \((7,-4)\) is
\((y+4)=-\frac{1}{5}(x-7) \Rightarrow 5 y+20=-x+7\)
\(x+5 y+13=0\)
MHT CET-2020
Straight Line
88640
The \(x\)-intercept of the line which passes through points \(A(-1,2)\) and \(B(2,3)\) is
1 -7
2 -3
3 7
4 3
Explanation:
(A) :
Equation of line \(\mathrm{AB}\) is
\(\frac{y-2}{2-3}=\frac{x-(-1)}{-1-2}\)
\(\Rightarrow \frac{y-2}{-1}=\frac{x+1}{-3}\)
\(\Rightarrow-3(y-2)=-1(x+1)\)
\(-3 y+6=-x-1 \Rightarrow x-3 y+7=0\)
\(\therefore x \text { intercept }=\frac{-7}{1}=-7\)
88636
A straight line through a fixed point \((2,3)\) intersects the coordinate axes at distinct points \(P\) and \(Q\). If \(O\) is the origin and the rectangle OPRQ is completed, then the locus of \(R\) is :
1 \(2 x+3 y=x y\)
2 \(3 x+2 y=x y\)
3 \(3 x+2 y=6 x y\)
4 \(3 x+2 y=6\)
Explanation:
Equation of PQ is : \(\frac{x}{h}+\frac{y}{k} = 1\) …..(i)
Since, (1) passes through the fixed point \((2,3)\)
Then, \(\frac{2}{\mathrm{~h}}+\frac{3}{\mathrm{k}}=1\)
Then, the locus of \(\mathrm{R}\) is \(\frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}=1\)
or \(3 x+2 y=x y\).
BITSAT-2018
Straight Line
88637
If \((a,-2 a), a>0\) is the midpoint of a line segment intercepted between the co-ordinate axis, then the equation of the line is
1 \(2 x-y=4 a\)
2 \(x-2 y=5 a\)
3 \(2 x-y+4 a=0\)
4 \(x-2 y+4 a=0\)
Explanation:
(A) : Given, point \((a,-2 a)\) is the midpoint of the intercepts.
let \((A, 0)\) and \((0, B)\) are \(x\) and \(y\) intercept
Then, the equation of line is,
\(\frac{\mathrm{x}}{\mathrm{A}}+\frac{\mathrm{y}}{\mathrm{B}}=1\)
Since, \((a,-2 a)\) is a mid-point then,
\(\quad \begin{aligned} \mathrm{a} =\frac{\mathrm{A}}{2} \text { and }-2 \mathrm{a}=\frac{\mathrm{B}}{2} \\ \mathrm{~A} =2 \mathrm{a} \text { and } \mathrm{B}=-4 \mathrm{a}\end{aligned}\)
Hence required equation of line is,
\(\frac{\mathrm{x}}{2 \mathrm{a}}-\frac{\mathrm{y}}{4 \mathrm{a}}=1 \quad 2 \mathrm{x}-\mathrm{y}=4 \mathrm{a}\)
MHT CET-2020
Straight Line
88638
The equations of the lines which make intercepts on the axes whose sum is 8 and product is 15 are
(B) :
Let ' \(a\) ' and ' \(b\) ' be the intercepts made by line.
We have \(a+b=8\) and \(\mathrm{ab}=15\)
Solving, we get \((a, b)=(3,5)\) or \((5,3)\)
Let equation of line be \(\frac{x}{a}+\frac{y}{b}=1\)
When \(\mathrm{a}=3, \mathrm{~b}=5\) from (1), we get
\(\frac{x}{3}+\frac{y}{5}=1 \quad \Rightarrow \quad 5 x+3 y-15=0\)
When \(a=5, b=3\) from (1), we get
\(\frac{x}{5}+\frac{y}{3}=1 \quad \Rightarrow \quad 3 x+5 y-15=0\)
Hence, the equation of line is \(3 x+5 y-15=0\) and \(5 x+\) \(3 y-15=0\)
MHT CET-2020
Straight Line
88639
The equation of a line passing through the point \((7,-4)\) and perpendicular to the line passing through the points \((2,3)\) and \((1,-2)\) is
1 \(x-2 y-15=0\)
2 \(x+2 y+1=0\)
3 \(x-5 y-13=0\)
4 \(x+5 y+13=0\)
Explanation:
(D) : Slope between two given point \((2,3)\) and (1,2) is,
\(\mathrm{m}=\frac{-2-3}{1-2}=5\)
Perpendicular slope is \(\mathrm{m}_{1}=-\frac{1}{5}\)
The equation of line with this slope is,
\(y=-\frac{1}{5} x+c \tag{i}\)
Now equation of required line who passing through the point \((7,-4)\) is
\((y+4)=-\frac{1}{5}(x-7) \Rightarrow 5 y+20=-x+7\)
\(x+5 y+13=0\)
MHT CET-2020
Straight Line
88640
The \(x\)-intercept of the line which passes through points \(A(-1,2)\) and \(B(2,3)\) is
1 -7
2 -3
3 7
4 3
Explanation:
(A) :
Equation of line \(\mathrm{AB}\) is
\(\frac{y-2}{2-3}=\frac{x-(-1)}{-1-2}\)
\(\Rightarrow \frac{y-2}{-1}=\frac{x+1}{-3}\)
\(\Rightarrow-3(y-2)=-1(x+1)\)
\(-3 y+6=-x-1 \Rightarrow x-3 y+7=0\)
\(\therefore x \text { intercept }=\frac{-7}{1}=-7\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Straight Line
88636
A straight line through a fixed point \((2,3)\) intersects the coordinate axes at distinct points \(P\) and \(Q\). If \(O\) is the origin and the rectangle OPRQ is completed, then the locus of \(R\) is :
1 \(2 x+3 y=x y\)
2 \(3 x+2 y=x y\)
3 \(3 x+2 y=6 x y\)
4 \(3 x+2 y=6\)
Explanation:
Equation of PQ is : \(\frac{x}{h}+\frac{y}{k} = 1\) …..(i)
Since, (1) passes through the fixed point \((2,3)\)
Then, \(\frac{2}{\mathrm{~h}}+\frac{3}{\mathrm{k}}=1\)
Then, the locus of \(\mathrm{R}\) is \(\frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}=1\)
or \(3 x+2 y=x y\).
BITSAT-2018
Straight Line
88637
If \((a,-2 a), a>0\) is the midpoint of a line segment intercepted between the co-ordinate axis, then the equation of the line is
1 \(2 x-y=4 a\)
2 \(x-2 y=5 a\)
3 \(2 x-y+4 a=0\)
4 \(x-2 y+4 a=0\)
Explanation:
(A) : Given, point \((a,-2 a)\) is the midpoint of the intercepts.
let \((A, 0)\) and \((0, B)\) are \(x\) and \(y\) intercept
Then, the equation of line is,
\(\frac{\mathrm{x}}{\mathrm{A}}+\frac{\mathrm{y}}{\mathrm{B}}=1\)
Since, \((a,-2 a)\) is a mid-point then,
\(\quad \begin{aligned} \mathrm{a} =\frac{\mathrm{A}}{2} \text { and }-2 \mathrm{a}=\frac{\mathrm{B}}{2} \\ \mathrm{~A} =2 \mathrm{a} \text { and } \mathrm{B}=-4 \mathrm{a}\end{aligned}\)
Hence required equation of line is,
\(\frac{\mathrm{x}}{2 \mathrm{a}}-\frac{\mathrm{y}}{4 \mathrm{a}}=1 \quad 2 \mathrm{x}-\mathrm{y}=4 \mathrm{a}\)
MHT CET-2020
Straight Line
88638
The equations of the lines which make intercepts on the axes whose sum is 8 and product is 15 are
(B) :
Let ' \(a\) ' and ' \(b\) ' be the intercepts made by line.
We have \(a+b=8\) and \(\mathrm{ab}=15\)
Solving, we get \((a, b)=(3,5)\) or \((5,3)\)
Let equation of line be \(\frac{x}{a}+\frac{y}{b}=1\)
When \(\mathrm{a}=3, \mathrm{~b}=5\) from (1), we get
\(\frac{x}{3}+\frac{y}{5}=1 \quad \Rightarrow \quad 5 x+3 y-15=0\)
When \(a=5, b=3\) from (1), we get
\(\frac{x}{5}+\frac{y}{3}=1 \quad \Rightarrow \quad 3 x+5 y-15=0\)
Hence, the equation of line is \(3 x+5 y-15=0\) and \(5 x+\) \(3 y-15=0\)
MHT CET-2020
Straight Line
88639
The equation of a line passing through the point \((7,-4)\) and perpendicular to the line passing through the points \((2,3)\) and \((1,-2)\) is
1 \(x-2 y-15=0\)
2 \(x+2 y+1=0\)
3 \(x-5 y-13=0\)
4 \(x+5 y+13=0\)
Explanation:
(D) : Slope between two given point \((2,3)\) and (1,2) is,
\(\mathrm{m}=\frac{-2-3}{1-2}=5\)
Perpendicular slope is \(\mathrm{m}_{1}=-\frac{1}{5}\)
The equation of line with this slope is,
\(y=-\frac{1}{5} x+c \tag{i}\)
Now equation of required line who passing through the point \((7,-4)\) is
\((y+4)=-\frac{1}{5}(x-7) \Rightarrow 5 y+20=-x+7\)
\(x+5 y+13=0\)
MHT CET-2020
Straight Line
88640
The \(x\)-intercept of the line which passes through points \(A(-1,2)\) and \(B(2,3)\) is
1 -7
2 -3
3 7
4 3
Explanation:
(A) :
Equation of line \(\mathrm{AB}\) is
\(\frac{y-2}{2-3}=\frac{x-(-1)}{-1-2}\)
\(\Rightarrow \frac{y-2}{-1}=\frac{x+1}{-3}\)
\(\Rightarrow-3(y-2)=-1(x+1)\)
\(-3 y+6=-x-1 \Rightarrow x-3 y+7=0\)
\(\therefore x \text { intercept }=\frac{-7}{1}=-7\)
88636
A straight line through a fixed point \((2,3)\) intersects the coordinate axes at distinct points \(P\) and \(Q\). If \(O\) is the origin and the rectangle OPRQ is completed, then the locus of \(R\) is :
1 \(2 x+3 y=x y\)
2 \(3 x+2 y=x y\)
3 \(3 x+2 y=6 x y\)
4 \(3 x+2 y=6\)
Explanation:
Equation of PQ is : \(\frac{x}{h}+\frac{y}{k} = 1\) …..(i)
Since, (1) passes through the fixed point \((2,3)\)
Then, \(\frac{2}{\mathrm{~h}}+\frac{3}{\mathrm{k}}=1\)
Then, the locus of \(\mathrm{R}\) is \(\frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}=1\)
or \(3 x+2 y=x y\).
BITSAT-2018
Straight Line
88637
If \((a,-2 a), a>0\) is the midpoint of a line segment intercepted between the co-ordinate axis, then the equation of the line is
1 \(2 x-y=4 a\)
2 \(x-2 y=5 a\)
3 \(2 x-y+4 a=0\)
4 \(x-2 y+4 a=0\)
Explanation:
(A) : Given, point \((a,-2 a)\) is the midpoint of the intercepts.
let \((A, 0)\) and \((0, B)\) are \(x\) and \(y\) intercept
Then, the equation of line is,
\(\frac{\mathrm{x}}{\mathrm{A}}+\frac{\mathrm{y}}{\mathrm{B}}=1\)
Since, \((a,-2 a)\) is a mid-point then,
\(\quad \begin{aligned} \mathrm{a} =\frac{\mathrm{A}}{2} \text { and }-2 \mathrm{a}=\frac{\mathrm{B}}{2} \\ \mathrm{~A} =2 \mathrm{a} \text { and } \mathrm{B}=-4 \mathrm{a}\end{aligned}\)
Hence required equation of line is,
\(\frac{\mathrm{x}}{2 \mathrm{a}}-\frac{\mathrm{y}}{4 \mathrm{a}}=1 \quad 2 \mathrm{x}-\mathrm{y}=4 \mathrm{a}\)
MHT CET-2020
Straight Line
88638
The equations of the lines which make intercepts on the axes whose sum is 8 and product is 15 are
(B) :
Let ' \(a\) ' and ' \(b\) ' be the intercepts made by line.
We have \(a+b=8\) and \(\mathrm{ab}=15\)
Solving, we get \((a, b)=(3,5)\) or \((5,3)\)
Let equation of line be \(\frac{x}{a}+\frac{y}{b}=1\)
When \(\mathrm{a}=3, \mathrm{~b}=5\) from (1), we get
\(\frac{x}{3}+\frac{y}{5}=1 \quad \Rightarrow \quad 5 x+3 y-15=0\)
When \(a=5, b=3\) from (1), we get
\(\frac{x}{5}+\frac{y}{3}=1 \quad \Rightarrow \quad 3 x+5 y-15=0\)
Hence, the equation of line is \(3 x+5 y-15=0\) and \(5 x+\) \(3 y-15=0\)
MHT CET-2020
Straight Line
88639
The equation of a line passing through the point \((7,-4)\) and perpendicular to the line passing through the points \((2,3)\) and \((1,-2)\) is
1 \(x-2 y-15=0\)
2 \(x+2 y+1=0\)
3 \(x-5 y-13=0\)
4 \(x+5 y+13=0\)
Explanation:
(D) : Slope between two given point \((2,3)\) and (1,2) is,
\(\mathrm{m}=\frac{-2-3}{1-2}=5\)
Perpendicular slope is \(\mathrm{m}_{1}=-\frac{1}{5}\)
The equation of line with this slope is,
\(y=-\frac{1}{5} x+c \tag{i}\)
Now equation of required line who passing through the point \((7,-4)\) is
\((y+4)=-\frac{1}{5}(x-7) \Rightarrow 5 y+20=-x+7\)
\(x+5 y+13=0\)
MHT CET-2020
Straight Line
88640
The \(x\)-intercept of the line which passes through points \(A(-1,2)\) and \(B(2,3)\) is
1 -7
2 -3
3 7
4 3
Explanation:
(A) :
Equation of line \(\mathrm{AB}\) is
\(\frac{y-2}{2-3}=\frac{x-(-1)}{-1-2}\)
\(\Rightarrow \frac{y-2}{-1}=\frac{x+1}{-3}\)
\(\Rightarrow-3(y-2)=-1(x+1)\)
\(-3 y+6=-x-1 \Rightarrow x-3 y+7=0\)
\(\therefore x \text { intercept }=\frac{-7}{1}=-7\)
88636
A straight line through a fixed point \((2,3)\) intersects the coordinate axes at distinct points \(P\) and \(Q\). If \(O\) is the origin and the rectangle OPRQ is completed, then the locus of \(R\) is :
1 \(2 x+3 y=x y\)
2 \(3 x+2 y=x y\)
3 \(3 x+2 y=6 x y\)
4 \(3 x+2 y=6\)
Explanation:
Equation of PQ is : \(\frac{x}{h}+\frac{y}{k} = 1\) …..(i)
Since, (1) passes through the fixed point \((2,3)\)
Then, \(\frac{2}{\mathrm{~h}}+\frac{3}{\mathrm{k}}=1\)
Then, the locus of \(\mathrm{R}\) is \(\frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}=1\)
or \(3 x+2 y=x y\).
BITSAT-2018
Straight Line
88637
If \((a,-2 a), a>0\) is the midpoint of a line segment intercepted between the co-ordinate axis, then the equation of the line is
1 \(2 x-y=4 a\)
2 \(x-2 y=5 a\)
3 \(2 x-y+4 a=0\)
4 \(x-2 y+4 a=0\)
Explanation:
(A) : Given, point \((a,-2 a)\) is the midpoint of the intercepts.
let \((A, 0)\) and \((0, B)\) are \(x\) and \(y\) intercept
Then, the equation of line is,
\(\frac{\mathrm{x}}{\mathrm{A}}+\frac{\mathrm{y}}{\mathrm{B}}=1\)
Since, \((a,-2 a)\) is a mid-point then,
\(\quad \begin{aligned} \mathrm{a} =\frac{\mathrm{A}}{2} \text { and }-2 \mathrm{a}=\frac{\mathrm{B}}{2} \\ \mathrm{~A} =2 \mathrm{a} \text { and } \mathrm{B}=-4 \mathrm{a}\end{aligned}\)
Hence required equation of line is,
\(\frac{\mathrm{x}}{2 \mathrm{a}}-\frac{\mathrm{y}}{4 \mathrm{a}}=1 \quad 2 \mathrm{x}-\mathrm{y}=4 \mathrm{a}\)
MHT CET-2020
Straight Line
88638
The equations of the lines which make intercepts on the axes whose sum is 8 and product is 15 are
(B) :
Let ' \(a\) ' and ' \(b\) ' be the intercepts made by line.
We have \(a+b=8\) and \(\mathrm{ab}=15\)
Solving, we get \((a, b)=(3,5)\) or \((5,3)\)
Let equation of line be \(\frac{x}{a}+\frac{y}{b}=1\)
When \(\mathrm{a}=3, \mathrm{~b}=5\) from (1), we get
\(\frac{x}{3}+\frac{y}{5}=1 \quad \Rightarrow \quad 5 x+3 y-15=0\)
When \(a=5, b=3\) from (1), we get
\(\frac{x}{5}+\frac{y}{3}=1 \quad \Rightarrow \quad 3 x+5 y-15=0\)
Hence, the equation of line is \(3 x+5 y-15=0\) and \(5 x+\) \(3 y-15=0\)
MHT CET-2020
Straight Line
88639
The equation of a line passing through the point \((7,-4)\) and perpendicular to the line passing through the points \((2,3)\) and \((1,-2)\) is
1 \(x-2 y-15=0\)
2 \(x+2 y+1=0\)
3 \(x-5 y-13=0\)
4 \(x+5 y+13=0\)
Explanation:
(D) : Slope between two given point \((2,3)\) and (1,2) is,
\(\mathrm{m}=\frac{-2-3}{1-2}=5\)
Perpendicular slope is \(\mathrm{m}_{1}=-\frac{1}{5}\)
The equation of line with this slope is,
\(y=-\frac{1}{5} x+c \tag{i}\)
Now equation of required line who passing through the point \((7,-4)\) is
\((y+4)=-\frac{1}{5}(x-7) \Rightarrow 5 y+20=-x+7\)
\(x+5 y+13=0\)
MHT CET-2020
Straight Line
88640
The \(x\)-intercept of the line which passes through points \(A(-1,2)\) and \(B(2,3)\) is
1 -7
2 -3
3 7
4 3
Explanation:
(A) :
Equation of line \(\mathrm{AB}\) is
\(\frac{y-2}{2-3}=\frac{x-(-1)}{-1-2}\)
\(\Rightarrow \frac{y-2}{-1}=\frac{x+1}{-3}\)
\(\Rightarrow-3(y-2)=-1(x+1)\)
\(-3 y+6=-x-1 \Rightarrow x-3 y+7=0\)
\(\therefore x \text { intercept }=\frac{-7}{1}=-7\)
