The equation of a line with slope $-\frac{1}{\sqrt{2}}$ and making an intercept of $2 \sqrt{2}$ units on the negative direction of the $y$-axis is:
- A$x+\sqrt{2} y+4=0$
- B$x+\sqrt{2} y+2 \sqrt{2}=0$
- C$\sqrt{2} y+x+4=0$
- D$x+\sqrt{2} y-2 \sqrt{2}=0$
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Match the items given in List-$I$ to the items given in List-$II$. List-$I$ List-$II$ $A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$ $1$. $2x - 5y + 4 = 0$ $B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes $2$. $3x + 5y = 3$ $C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is $3$. $10x - 15y + 4 = 0$ $D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is $4$. $10x - 15y = 4$ $5$. $5x - 2y - 20 = 0$
| List-$I$ | List-$II$ |
| $A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$ | $1$. $2x - 5y + 4 = 0$ |
| $B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes | $2$. $3x + 5y = 3$ |
| $C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is | $3$. $10x - 15y + 4 = 0$ |
| $D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is | $4$. $10x - 15y = 4$ |
| $5$. $5x - 2y - 20 = 0$ |
The equation of the line perpendicular to $2x - 3y + 5 = 0$ and making an intercept of $3$ with the positive $Y$-axis is
The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$ is:
DifficultTS EAMCET 2003
View SolutionIf a straight line is at a distance of $10$ units from the origin and the perpendicular drawn from the origin to it makes an angle of $\frac{\pi}{4}$ with the negative $X$-axis in the negative direction,then the equation of that line is:
The equation of a straight line passing through the points $(-5, -6)$ and $(3, 10)$ is:
Easy
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