# The intercepts on the coordinate axes of the straight line with the given equation displaystyle{4}{x}-{3}{y}={12}.

Question
Alternate coordinate systems
The intercepts on the coordinate axes of the straight line with the given equation $$\displaystyle{4}{x}-{3}{y}={12}.$$

2021-02-26
Concept used:
If a straight line passes through the points (a, 0) (on the x-axis) and (0, b) (on the y-axis), we say that the x-intercept is a and the y-intercept is
b.
The x -intercept is found by setting $$\displaystyle{y}={0}$$ and solving for a, similarly, the y-intercept is found by setting $$\displaystyle{x}={0}$$ and solving for b.
Calculation:
For the given equation $$\displaystyle{4}{x}-{3}{y}={12}$$
$$\displaystyle{y}={0}\Rightarrow{4}{x}={12}\Rightarrow{x}-\text{intercept}={3},$$
$$\displaystyle{x}={0}\Rightarrow-{3}{y}={12}\Rightarrow{y}-\text{intercept}=-{4}$$
Conclusion:
For the given line, $$\displaystyle{x}-\text{intercept}={3}{\quad\text{and}\quad}{y}-\text{intercept}=-{4}$$

### Relevant Questions

To calculate: The intercepts on the coordinate axes of the straight line with the given equation $$\displaystyle{2}{y}-{4}={3}{x}$$

To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}$$

The volume of the largest rectangular box which lies in the first octant with three faces in the coordinate planes and its one of the vertex in the plane $$\displaystyle{x}+{2}{y}+{3}{z}={6}$$ by using Lagrange multipliers.
A golf ball lies 2.00 m directly south of the hole on a levelgreen. On the first putt, the ball travels 3.00 m along astraight-line path at an angle of 5 degrees east of north; on thesecond putt, it travels a straight-line distance of 1.20 m at anangle of 6 degrees south of west.
What would be the displacement ofa third putt that would put the ball in the hole?
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The equivalent polar equation for the given rectangular - coordinate equation.
$$\displaystyle{y}=\ -{3}$$
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle\ {x}=\ {r}{\cos{\theta}}$$
$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
b. From rectangular to polar:
$$\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}$$
$$\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}$$
Calculation:
Given: equation in rectangular-coordinate is $$\displaystyle{y}={x}$$.
Converting into equivalent polar equation -
$$\displaystyle{y}={x}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
$$\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}$$
$$\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}$$
$$\displaystyle\Rightarrow\ {\tan{\theta}}={1}$$
Thus, desired equivalent polar equation would be $$\displaystyle\theta={1}$$
$$\displaystyle{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}={0}$$
a) $$\displaystyle{y}\ge{x}^{2},{z}\ge{0},$$
b) $$\displaystyle{x}\le{y}^{2},{0}\le{z}\le{2}.$$