88729
A line passes through \((2,2)\) and is perpendicular to the line \(3 x+y=3\). Its \(y-\) itercept is
1 \(2 / 3\)
2 1
3 \(1 / 3\)
4 \(4 / 3\)
Explanation:
(D) : Given, equation of line is,
\(3 x+y=3\)
\(y=-3 x+3\)
The slope of the line \((m)=-3\)
And, the slope of perpendicular line is \((m)=\frac{1}{3}\)
Now, the equation of line passing through \((2,2)\) with slope \(\mathrm{m}=\frac{1}{3}\) is
\(y-2=\left(\frac{1}{3}\right)(x-2) \Rightarrow 3 y-6=x-2\)
\(3 y=x+4\) and \(x-3 y=-4\) or \(\frac{x}{-4}+\frac{y}{(4 / 3)}=1\)
Comparing the above equation with \(\frac{x}{a}+\frac{y}{b}=1\)
So, \(y\)-intercept of the line is \(\frac{4}{3}\)
Karnataka CET-2015
Straight Line
88728
The equation of a line passing through \(\left(\operatorname{acos}^{3} \theta\right.\), \(\left.\operatorname{asin}^{3} \theta\right)\) and perpendicular to the line \(x \sec \theta+y\) \(\operatorname{cosec} \theta=\mathbf{a}\) is
(A) :Given, \(\mathrm{x} \sec \theta+\mathrm{y} \operatorname{cosec} \theta=\mathrm{a}\)
Slope \(=-\tan \theta=\frac{\sec \theta}{\operatorname{cosec} \theta}\)
Now, slope of line which is perpendicular to equation (i) is,
\(\frac{-1}{-\tan \theta}=\cot \theta\)
Equation of line passing through \(\left(a \cos ^{3} \theta, a \sin ^{3} \theta\right)\) and having slope \(\cot \theta\) is given by
\(\left(\mathrm{y}-\mathrm{a} \sin ^{3} \theta\right)=\cot \theta\left(\mathrm{x}-\mathrm{a} \cos ^{3} \theta\right)\) \(y \sin \theta-a \sin ^{4} \theta=x \cos \theta-a \cos ^{4} \theta\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{4} \theta-\sin ^{4} \theta\right)\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)
\(\mathrm{x} \cos \theta-\mathrm{y} \sin \theta-\mathrm{a} \cos 2 \theta=0\)
SRM JEE-2013
Straight Line
88730
If the point \(\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) lies on the curve traced by the mid-points of the line segments of the lines \(x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)\) between the co-ordinates axes, then \(\alpha\) is equal to
(A) : Given the equation of line is,
\(\mathrm{x} \sec \alpha+\mathrm{y} \tan \alpha=\mathrm{p}\)
\(\frac{\mathrm{x}}{\mathrm{p} \cos \alpha}+\frac{\mathrm{y}}{\mathrm{p} \cot \alpha}=1\)
\(\therefore A=(p \cos \alpha, 0)\) and \(B=(0, p \cot \alpha)\)
If \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is the mid-point of \(\mathrm{AB}\), then
\(2 x=p \cos \alpha\) and
\(2 y=p \cot \alpha\)
\(\therefore \sec \alpha=\frac{p}{2 x}\) and \(\Rightarrow \tan \alpha=\frac{p}{2 y}\)
Since, \(\sec ^{2} \alpha-\tan ^{2} \alpha=1\)
So, \(\frac{p^{2}}{4 x^{2}}-\frac{p^{2}}{4 y^{2}}=1 \Rightarrow \frac{p^{2}}{4 x^{2}}=1+\frac{p^{2}}{4 y^{2}}\)
COMEDK-2020
Straight Line
88732
Find the equation of a straight line passing through \((-5,6)\) and cutting off equal intercepts on the co-ordinate axes.
88729
A line passes through \((2,2)\) and is perpendicular to the line \(3 x+y=3\). Its \(y-\) itercept is
1 \(2 / 3\)
2 1
3 \(1 / 3\)
4 \(4 / 3\)
Explanation:
(D) : Given, equation of line is,
\(3 x+y=3\)
\(y=-3 x+3\)
The slope of the line \((m)=-3\)
And, the slope of perpendicular line is \((m)=\frac{1}{3}\)
Now, the equation of line passing through \((2,2)\) with slope \(\mathrm{m}=\frac{1}{3}\) is
\(y-2=\left(\frac{1}{3}\right)(x-2) \Rightarrow 3 y-6=x-2\)
\(3 y=x+4\) and \(x-3 y=-4\) or \(\frac{x}{-4}+\frac{y}{(4 / 3)}=1\)
Comparing the above equation with \(\frac{x}{a}+\frac{y}{b}=1\)
So, \(y\)-intercept of the line is \(\frac{4}{3}\)
Karnataka CET-2015
Straight Line
88728
The equation of a line passing through \(\left(\operatorname{acos}^{3} \theta\right.\), \(\left.\operatorname{asin}^{3} \theta\right)\) and perpendicular to the line \(x \sec \theta+y\) \(\operatorname{cosec} \theta=\mathbf{a}\) is
(A) :Given, \(\mathrm{x} \sec \theta+\mathrm{y} \operatorname{cosec} \theta=\mathrm{a}\)
Slope \(=-\tan \theta=\frac{\sec \theta}{\operatorname{cosec} \theta}\)
Now, slope of line which is perpendicular to equation (i) is,
\(\frac{-1}{-\tan \theta}=\cot \theta\)
Equation of line passing through \(\left(a \cos ^{3} \theta, a \sin ^{3} \theta\right)\) and having slope \(\cot \theta\) is given by
\(\left(\mathrm{y}-\mathrm{a} \sin ^{3} \theta\right)=\cot \theta\left(\mathrm{x}-\mathrm{a} \cos ^{3} \theta\right)\) \(y \sin \theta-a \sin ^{4} \theta=x \cos \theta-a \cos ^{4} \theta\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{4} \theta-\sin ^{4} \theta\right)\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)
\(\mathrm{x} \cos \theta-\mathrm{y} \sin \theta-\mathrm{a} \cos 2 \theta=0\)
SRM JEE-2013
Straight Line
88730
If the point \(\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) lies on the curve traced by the mid-points of the line segments of the lines \(x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)\) between the co-ordinates axes, then \(\alpha\) is equal to
(A) : Given the equation of line is,
\(\mathrm{x} \sec \alpha+\mathrm{y} \tan \alpha=\mathrm{p}\)
\(\frac{\mathrm{x}}{\mathrm{p} \cos \alpha}+\frac{\mathrm{y}}{\mathrm{p} \cot \alpha}=1\)
\(\therefore A=(p \cos \alpha, 0)\) and \(B=(0, p \cot \alpha)\)
If \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is the mid-point of \(\mathrm{AB}\), then
\(2 x=p \cos \alpha\) and
\(2 y=p \cot \alpha\)
\(\therefore \sec \alpha=\frac{p}{2 x}\) and \(\Rightarrow \tan \alpha=\frac{p}{2 y}\)
Since, \(\sec ^{2} \alpha-\tan ^{2} \alpha=1\)
So, \(\frac{p^{2}}{4 x^{2}}-\frac{p^{2}}{4 y^{2}}=1 \Rightarrow \frac{p^{2}}{4 x^{2}}=1+\frac{p^{2}}{4 y^{2}}\)
COMEDK-2020
Straight Line
88732
Find the equation of a straight line passing through \((-5,6)\) and cutting off equal intercepts on the co-ordinate axes.
88729
A line passes through \((2,2)\) and is perpendicular to the line \(3 x+y=3\). Its \(y-\) itercept is
1 \(2 / 3\)
2 1
3 \(1 / 3\)
4 \(4 / 3\)
Explanation:
(D) : Given, equation of line is,
\(3 x+y=3\)
\(y=-3 x+3\)
The slope of the line \((m)=-3\)
And, the slope of perpendicular line is \((m)=\frac{1}{3}\)
Now, the equation of line passing through \((2,2)\) with slope \(\mathrm{m}=\frac{1}{3}\) is
\(y-2=\left(\frac{1}{3}\right)(x-2) \Rightarrow 3 y-6=x-2\)
\(3 y=x+4\) and \(x-3 y=-4\) or \(\frac{x}{-4}+\frac{y}{(4 / 3)}=1\)
Comparing the above equation with \(\frac{x}{a}+\frac{y}{b}=1\)
So, \(y\)-intercept of the line is \(\frac{4}{3}\)
Karnataka CET-2015
Straight Line
88728
The equation of a line passing through \(\left(\operatorname{acos}^{3} \theta\right.\), \(\left.\operatorname{asin}^{3} \theta\right)\) and perpendicular to the line \(x \sec \theta+y\) \(\operatorname{cosec} \theta=\mathbf{a}\) is
(A) :Given, \(\mathrm{x} \sec \theta+\mathrm{y} \operatorname{cosec} \theta=\mathrm{a}\)
Slope \(=-\tan \theta=\frac{\sec \theta}{\operatorname{cosec} \theta}\)
Now, slope of line which is perpendicular to equation (i) is,
\(\frac{-1}{-\tan \theta}=\cot \theta\)
Equation of line passing through \(\left(a \cos ^{3} \theta, a \sin ^{3} \theta\right)\) and having slope \(\cot \theta\) is given by
\(\left(\mathrm{y}-\mathrm{a} \sin ^{3} \theta\right)=\cot \theta\left(\mathrm{x}-\mathrm{a} \cos ^{3} \theta\right)\) \(y \sin \theta-a \sin ^{4} \theta=x \cos \theta-a \cos ^{4} \theta\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{4} \theta-\sin ^{4} \theta\right)\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)
\(\mathrm{x} \cos \theta-\mathrm{y} \sin \theta-\mathrm{a} \cos 2 \theta=0\)
SRM JEE-2013
Straight Line
88730
If the point \(\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) lies on the curve traced by the mid-points of the line segments of the lines \(x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)\) between the co-ordinates axes, then \(\alpha\) is equal to
(A) : Given the equation of line is,
\(\mathrm{x} \sec \alpha+\mathrm{y} \tan \alpha=\mathrm{p}\)
\(\frac{\mathrm{x}}{\mathrm{p} \cos \alpha}+\frac{\mathrm{y}}{\mathrm{p} \cot \alpha}=1\)
\(\therefore A=(p \cos \alpha, 0)\) and \(B=(0, p \cot \alpha)\)
If \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is the mid-point of \(\mathrm{AB}\), then
\(2 x=p \cos \alpha\) and
\(2 y=p \cot \alpha\)
\(\therefore \sec \alpha=\frac{p}{2 x}\) and \(\Rightarrow \tan \alpha=\frac{p}{2 y}\)
Since, \(\sec ^{2} \alpha-\tan ^{2} \alpha=1\)
So, \(\frac{p^{2}}{4 x^{2}}-\frac{p^{2}}{4 y^{2}}=1 \Rightarrow \frac{p^{2}}{4 x^{2}}=1+\frac{p^{2}}{4 y^{2}}\)
COMEDK-2020
Straight Line
88732
Find the equation of a straight line passing through \((-5,6)\) and cutting off equal intercepts on the co-ordinate axes.
88729
A line passes through \((2,2)\) and is perpendicular to the line \(3 x+y=3\). Its \(y-\) itercept is
1 \(2 / 3\)
2 1
3 \(1 / 3\)
4 \(4 / 3\)
Explanation:
(D) : Given, equation of line is,
\(3 x+y=3\)
\(y=-3 x+3\)
The slope of the line \((m)=-3\)
And, the slope of perpendicular line is \((m)=\frac{1}{3}\)
Now, the equation of line passing through \((2,2)\) with slope \(\mathrm{m}=\frac{1}{3}\) is
\(y-2=\left(\frac{1}{3}\right)(x-2) \Rightarrow 3 y-6=x-2\)
\(3 y=x+4\) and \(x-3 y=-4\) or \(\frac{x}{-4}+\frac{y}{(4 / 3)}=1\)
Comparing the above equation with \(\frac{x}{a}+\frac{y}{b}=1\)
So, \(y\)-intercept of the line is \(\frac{4}{3}\)
Karnataka CET-2015
Straight Line
88728
The equation of a line passing through \(\left(\operatorname{acos}^{3} \theta\right.\), \(\left.\operatorname{asin}^{3} \theta\right)\) and perpendicular to the line \(x \sec \theta+y\) \(\operatorname{cosec} \theta=\mathbf{a}\) is
(A) :Given, \(\mathrm{x} \sec \theta+\mathrm{y} \operatorname{cosec} \theta=\mathrm{a}\)
Slope \(=-\tan \theta=\frac{\sec \theta}{\operatorname{cosec} \theta}\)
Now, slope of line which is perpendicular to equation (i) is,
\(\frac{-1}{-\tan \theta}=\cot \theta\)
Equation of line passing through \(\left(a \cos ^{3} \theta, a \sin ^{3} \theta\right)\) and having slope \(\cot \theta\) is given by
\(\left(\mathrm{y}-\mathrm{a} \sin ^{3} \theta\right)=\cot \theta\left(\mathrm{x}-\mathrm{a} \cos ^{3} \theta\right)\) \(y \sin \theta-a \sin ^{4} \theta=x \cos \theta-a \cos ^{4} \theta\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{4} \theta-\sin ^{4} \theta\right)\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)
\(\mathrm{x} \cos \theta-\mathrm{y} \sin \theta-\mathrm{a} \cos 2 \theta=0\)
SRM JEE-2013
Straight Line
88730
If the point \(\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) lies on the curve traced by the mid-points of the line segments of the lines \(x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)\) between the co-ordinates axes, then \(\alpha\) is equal to
(A) : Given the equation of line is,
\(\mathrm{x} \sec \alpha+\mathrm{y} \tan \alpha=\mathrm{p}\)
\(\frac{\mathrm{x}}{\mathrm{p} \cos \alpha}+\frac{\mathrm{y}}{\mathrm{p} \cot \alpha}=1\)
\(\therefore A=(p \cos \alpha, 0)\) and \(B=(0, p \cot \alpha)\)
If \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is the mid-point of \(\mathrm{AB}\), then
\(2 x=p \cos \alpha\) and
\(2 y=p \cot \alpha\)
\(\therefore \sec \alpha=\frac{p}{2 x}\) and \(\Rightarrow \tan \alpha=\frac{p}{2 y}\)
Since, \(\sec ^{2} \alpha-\tan ^{2} \alpha=1\)
So, \(\frac{p^{2}}{4 x^{2}}-\frac{p^{2}}{4 y^{2}}=1 \Rightarrow \frac{p^{2}}{4 x^{2}}=1+\frac{p^{2}}{4 y^{2}}\)
COMEDK-2020
Straight Line
88732
Find the equation of a straight line passing through \((-5,6)\) and cutting off equal intercepts on the co-ordinate axes.
88729
A line passes through \((2,2)\) and is perpendicular to the line \(3 x+y=3\). Its \(y-\) itercept is
1 \(2 / 3\)
2 1
3 \(1 / 3\)
4 \(4 / 3\)
Explanation:
(D) : Given, equation of line is,
\(3 x+y=3\)
\(y=-3 x+3\)
The slope of the line \((m)=-3\)
And, the slope of perpendicular line is \((m)=\frac{1}{3}\)
Now, the equation of line passing through \((2,2)\) with slope \(\mathrm{m}=\frac{1}{3}\) is
\(y-2=\left(\frac{1}{3}\right)(x-2) \Rightarrow 3 y-6=x-2\)
\(3 y=x+4\) and \(x-3 y=-4\) or \(\frac{x}{-4}+\frac{y}{(4 / 3)}=1\)
Comparing the above equation with \(\frac{x}{a}+\frac{y}{b}=1\)
So, \(y\)-intercept of the line is \(\frac{4}{3}\)
Karnataka CET-2015
Straight Line
88728
The equation of a line passing through \(\left(\operatorname{acos}^{3} \theta\right.\), \(\left.\operatorname{asin}^{3} \theta\right)\) and perpendicular to the line \(x \sec \theta+y\) \(\operatorname{cosec} \theta=\mathbf{a}\) is
(A) :Given, \(\mathrm{x} \sec \theta+\mathrm{y} \operatorname{cosec} \theta=\mathrm{a}\)
Slope \(=-\tan \theta=\frac{\sec \theta}{\operatorname{cosec} \theta}\)
Now, slope of line which is perpendicular to equation (i) is,
\(\frac{-1}{-\tan \theta}=\cot \theta\)
Equation of line passing through \(\left(a \cos ^{3} \theta, a \sin ^{3} \theta\right)\) and having slope \(\cot \theta\) is given by
\(\left(\mathrm{y}-\mathrm{a} \sin ^{3} \theta\right)=\cot \theta\left(\mathrm{x}-\mathrm{a} \cos ^{3} \theta\right)\) \(y \sin \theta-a \sin ^{4} \theta=x \cos \theta-a \cos ^{4} \theta\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{4} \theta-\sin ^{4} \theta\right)\)
\(x \cos \theta-y \sin \theta=a\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)
\(\mathrm{x} \cos \theta-\mathrm{y} \sin \theta-\mathrm{a} \cos 2 \theta=0\)
SRM JEE-2013
Straight Line
88730
If the point \(\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) lies on the curve traced by the mid-points of the line segments of the lines \(x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)\) between the co-ordinates axes, then \(\alpha\) is equal to
(A) : Given the equation of line is,
\(\mathrm{x} \sec \alpha+\mathrm{y} \tan \alpha=\mathrm{p}\)
\(\frac{\mathrm{x}}{\mathrm{p} \cos \alpha}+\frac{\mathrm{y}}{\mathrm{p} \cot \alpha}=1\)
\(\therefore A=(p \cos \alpha, 0)\) and \(B=(0, p \cot \alpha)\)
If \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is the mid-point of \(\mathrm{AB}\), then
\(2 x=p \cos \alpha\) and
\(2 y=p \cot \alpha\)
\(\therefore \sec \alpha=\frac{p}{2 x}\) and \(\Rightarrow \tan \alpha=\frac{p}{2 y}\)
Since, \(\sec ^{2} \alpha-\tan ^{2} \alpha=1\)
So, \(\frac{p^{2}}{4 x^{2}}-\frac{p^{2}}{4 y^{2}}=1 \Rightarrow \frac{p^{2}}{4 x^{2}}=1+\frac{p^{2}}{4 y^{2}}\)
COMEDK-2020
Straight Line
88732
Find the equation of a straight line passing through \((-5,6)\) and cutting off equal intercepts on the co-ordinate axes.
