Reduce the following equation into intercept form and find its intercepts on the axes: $3y + 2 = 0$.
- A$y$-intercept = $-\frac{2}{3}$,$x$-intercept = does not exist
- B$y$-intercept = $\frac{2}{3}$,$x$-intercept = $0$
- C$y$-intercept = $-\frac{3}{2}$,$x$-intercept = does not exist
- D$y$-intercept = $0$,$x$-intercept = $-\frac{2}{3}$
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Consider the equation $y-y_1=m(x-x_1)$. If $m$ and $x_1$ are fixed and different lines are drawn for different values of $y_1$, then
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 | $4$. $10x - 15y = 4$ |
| $5$. $5x - 2y - 20 = 0$ |
Find the value of $k$ for which the line $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$ passes through the origin.
Medium
View SolutionThe straight line passing through $(-3, 6)$ and the midpoint of the line segment joining the points $(4, -5)$ and $(-2, 9)$ has an inclination of:
Find the slope of the line passing through the points $(3, -2)$ and $(-1, 4)$.
Easy
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