Enter in standard form the equation of the line passing through the given point and having the given slope.
B(2, - 5), m = -2/3
y + 5 = (-2/3)(x-2)
times 3
3y + 15 = -2x + 4
2x + 3y = -11 or 2x + 3y + 11 = 0
(some texts or authors labe
ax + by + c = 0 as the 'general' form
and ax + by = k as the 'standard' form
You will have to decide which of my versions to use.
To find the equation of a line passing through a given point and having a given slope, you can use the point-slope form of a linear equation.
The point-slope form is given by:
y - y1 = m(x - x1)
Where (x1, y1) represents the given point, and m represents the slope.
In this case, the given point is B(2, -5), and the slope is m = -2/3.
Substituting these values into the point-slope form, we get:
y - (-5) = (-2/3)(x - 2)
Simplifying, we have:
y + 5 = (-2/3)(x - 2)
Now, to convert the equation into standard form, we need to eliminate any fractions and move all the terms to one side.
Multiplying both sides of the equation by 3 to eliminate the fraction, we get:
3(y + 5) = -2(x - 2)
Expanding both sides, we have:
3y + 15 = -2x + 4
Next, let's move the terms involving x to the left side by adding 2x to both sides:
2x + 3y + 15 = 4
To make the equation in standard form, we need the coefficients of x, y, and the constant term to be integers with a positive coefficient for x. So, we can multiply both sides of the equation by -1:
-2x - 3y - 15 = -4
Finally, rearranging the terms, we obtain the equation in standard form:
2x + 3y = -11
Therefore, the equation of the line passing through the point B(2, -5) with a slope of -2/3 is 2x + 3y = -11 in standard form.