How do you write the equation of a line with m = -3/2, passes through (4, -7) in standard from?

Question

Savannah Anderson · Accepted Answer

#3x+2y=-26#

based on the given we're going to use point-slope form

#y-y_1=m(x-x_1)#

#y--7=-3/2(x-4)#

#y+7=-3/2(x-4)#

#2(y+7)=-3(x-4)#

#2y+14=-3x-12#

#3x+2y=-12-14#

#3x+2y=-26# #rarr# standard form

#3x+2y+26=0 rarr# general form

Jameson Allen · Answer

undefined The equation of a line with slope m passing through point (x₁, y₁) is given by y - y₁ = m(x - x₁).

Substituting m = -3/2 and the point (4, -7) into the equation, we get:
y - (-7) = -3/2(x - 4)
y + 7 = -3/2(x - 4)

Expanding and rearranging terms, we get:
y + 7 = -3/2x + 6

Finally, bringing all terms to one side in standard form, we get:
3/2x + y + 1 = 0

HIX Tutor · Answer

undefined To write the equation of a line in standard form (Ax + By = C) with a slope (m) of -3/2 that passes through the point (4, -7), follow these steps:

1. **Use the point-slope form of the line equation**:
   The formula is: 
   $$
   y - y_1 = m(x - x_1)
   $$
   Here, $m = -\frac{3}{2}$, $x_1 = 4$, and $y_1 = -7$.

2. **Plug in the values**:
   $$
   y - (-7) = -\frac{3}{2}(x - 4)
   $$
   Simplify:
   $$
   y + 7 = -\frac{3}{2}(x - 4)
   $$

3. **Distribute the slope**:
   $$
   y + 7 = -\frac{3}{2}x + 6
   $$

4. **Get y by itself**:
   Subtract 7 from both sides:
   $$
   y = -\frac{3}{2}x + 6 - 7
   $$
   $$
   y = -\frac{3}{2}x - 1
   $$

5. **Convert to standard form (Ax + By = C)**:
   Move $-\frac{3}{2}x$ to the left:
   $$
   \frac{3}{2}x + y = -1
   $$

6. **Clear the fraction by multiplying by 2**:
   $$
   3x + 2y = -2
   $$

Now, the standard form of the line is:
$$
3x + 2y = -2
$$