Find the intercepts made by the line $3x+4y-12=0$ on the axes:

Question

Given that $ \cot \left( \frac{A+B}{2} \right) \cdot \tan \left( \frac{A-B}{2} \right) = $, and the equation $ \frac{x}{2} + \frac{y}{3} + \frac{2}{6} - 1 = 0 $, find the area of $ \Delta ABC = 2 $.

Answer 1

The intercepts can also be found directly by setting each variable to zero in turn, without rearranging the whole equation.

To find the x-intercept, set $y=0$: $3x - 12 = 0 \Rightarrow x = 4$.
To find the y-intercept, set $x=0$: $4y - 12 = 0 \Rightarrow y = 3$.

So the line crosses the x-axis at 4 and the y-axis at 3.

The correct answer is option B, 4 and 3.

Answer 2

The intercepts can also be found directly by setting each variable to zero in turn, without rearranging the whole equation.

To find the x-intercept, set $y=0$: $3x - 12 = 0 \Rightarrow x = 4$.
To find the y-intercept, set $x=0$: $4y - 12 = 0 \Rightarrow y = 3$.

So the line crosses the x-axis at 4 and the y-axis at 3.

The correct answer is option B, 4 and 3.

Find the intercepts made by the line \(3x+4y-12=0\) on the axes:

Show Hint

The Correct Option is B

Solution and Explanation

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