Straight line `2x + 3y + 10 = 0` inersects coordinate axes respectively are the points

Options

A. `(- 5, 0) (0, - (10)/(3))`
B. `((10)/(3) , 0) (0, 5)`
C. `(-5, 0) , ( 0, (10)/(3))`
D. `((-10)/(3) , 0) , ( 0, ( -5)/(3))`

Answer

To find the points where the straight line \(2x + 3y + 10 = 0\) intersects the coordinate axes, we need to determine the x-intercept and y-intercept of the line.

### Step 1: Find the x-intercept
The x-intercept occurs where \(y = 0\). We substitute \(y = 0\) into the equation of the line:

\[ 2x + 3(0) + 10 = 0 \]

This simplifies to:

\[ 2x + 10 = 0 \]

Now, we solve for \(x\):

\[ 2x = -10 \]
\[ x = -5 \]

Thus, the x-intercept is the point \(Q(-5, 0)\).

### Step 2: Find the y-intercept
The y-intercept occurs where \(x = 0\). We substitute \(x = 0\) into the equation of the line:

\[ 2(0) + 3y + 10 = 0 \]

This simplifies to:

\[ 3y + 10 = 0 \]

Now, we solve for \(y\):

\[ 3y = -10 \]
\[ y = -\frac{10}{3} \]

Thus, the y-intercept is the point \(P(0, -\frac{10}{3})\).

### Final Result
The points where the line intersects the coordinate axes are:
- x-intercept: \(Q(-5, 0)\)
- y-intercept: \(P(0, -\frac{10}{3})\)