Find the intercepts on the axes made by the straight lines :
`(i) 2x-3y+6=0`
`(ii) xcosalpha+ysinalpha=sin2alpha`

Answer

To find the intercepts on the axes made by the straight lines given in the question, we will solve each line step by step.

### (i) For the line \( 2x - 3y + 6 = 0 \):

**Step 1: Find the x-intercept.**
- To find the x-intercept, set \( y = 0 \).
- Substitute \( y = 0 \) into the equation:
\[ 2x - 3(0) + 6 = 0 \implies 2x + 6 = 0 \implies 2x = -6 \implies x = -3 \]
- Therefore, the x-intercept is \( (-3, 0) \).

**Step 2: Find the y-intercept.**
- To find the y-intercept, set \( x = 0 \).
- Substitute \( x = 0 \) into the equation:
\[ 2(0) - 3y + 6 = 0 \implies -3y + 6 = 0 \implies -3y = -6 \implies y = 2 \]
- Therefore, the y-intercept is \( (0, 2) \).

### Summary for (i):
- x-intercept: \( (-3, 0) \)
- y-intercept: \( (0, 2) \)

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### (ii) For the line \( x \cos \alpha + y \sin \alpha = \sin 2\alpha \):

**Step 1: Find the x-intercept.**
- To find the x-intercept, set \( y = 0 \).
- Substitute \( y = 0 \) into the equation:
\[ x \cos \alpha + (0) \sin \alpha = \sin 2\alpha \implies x \cos \alpha = \sin 2\alpha \]
- Therefore, solving for \( x \):
\[ x = \frac{\sin 2\alpha}{\cos \alpha} \]
- Using the identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \):
\[ x = \frac{2 \sin \alpha \cos \alpha}{\cos \alpha} = 2 \sin \alpha \]
- Thus, the x-intercept is \( (2 \sin \alpha, 0) \).

**Step 2: Find the y-intercept.**
- To find the y-intercept, set \( x = 0 \).
- Substitute \( x = 0 \) into the equation:
\[ (0) \cos \alpha + y \sin \alpha = \sin 2\alpha \implies y \sin \alpha = \sin 2\alpha \]
- Therefore, solving for \( y \):
\[ y = \frac{\sin 2\alpha}{\sin \alpha} \]
- Using the identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \):
\[ y = \frac{2 \sin \alpha \cos \alpha}{\sin \alpha} = 2 \cos \alpha \]
- Thus, the y-intercept is \( (0, 2 \cos \alpha) \).

### Summary for (ii):
- x-intercept: \( (2 \sin \alpha, 0) \)
- y-intercept: \( (0, 2 \cos \alpha) \)

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