A straight line 2x+3y=6, passes through the point (-1,2). Find the equation of the line.
Answer Details
To find the equation of a straight line, we need to know either two points that the line passes through or a point and the slope of the line. In this case, we are given one point that the line passes through, (-1,2), and the equation of the line, 2x+3y=6.
We can use the point-slope form of the equation of a line to find the equation of the line passing through the given point. The point-slope form is:
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope of the line.
To find the slope of the given line, we can rearrange the equation 2x+3y=6 into the slope-intercept form:
y = (-2/3)x + 2
where the slope is -2/3.
Substituting the values of the given point and slope into the point-slope form, we get:
y - 2 = (-2/3)(x - (-1))
Simplifying this equation, we get:
y - 2 = (-2/3)x - (2/3)
Multiplying both sides by 3 to eliminate the fraction, we get:
3y - 6 = -2x - 2
Adding 2x to both sides, we get:
2x + 3y = 4
Therefore, the equation of the line passing through the given point is 2x+3y=4.
Hence, the correct answer is option (D) 2x+3y=4.