Find an equation of the line passing through the point (6,5) and perpendicular to the line whose equation is 2y+3x=6
2y+3x=6 has slope -3/2
So, the line you want has slope 2/3
ow you have a point and a slope, so the new line is
y-5 = 2/3 (x-6)
To find the equation of a line perpendicular to another line, we need to know the relationship between their slopes. Two lines are perpendicular if and only if the product of their slopes is -1.
First, let's put the given equation 2y + 3x = 6 into slope-intercept form, which is in the form y = mx + b, where m is the slope and b is the y-intercept.
Starting with 2y + 3x = 6:
1. Subtract 3x from both sides to isolate the term with y:
2y = -3x + 6
2. Divide both sides by 2 to solve for y:
y = (-3/2)x + 3
From the equation y = (-3/2)x + 3, we can determine that the slope of the given line is -3/2.
Since the line we are trying to find is perpendicular to this line, the slope of our new line will be the negative reciprocal of -3/2. To find the negative reciprocal, we flip the fraction and change the sign:
The slope of the line we are trying to find is 2/3.
Now, we can use the point-slope form of a linear equation to find the equation of the line passing through the point (6, 5) with a slope of 2/3. The point-slope form is given by the equation:
y - y₁ = m(x - x₁)
where (x₁, y₁) is a point on the line, m is the slope, and (x, y) represents any other point on the line.
Using the given point (6, 5) and the slope 2/3, we substitute these values into the point-slope form equation:
y - 5 = (2/3)(x - 6)
Now, we can simplify the equation:
y - 5 = (2/3)x - 4
To get the equation in slope-intercept form, we isolate y:
y = (2/3)x + 1
Therefore, the equation of the line passing through the point (6, 5) and perpendicular to the line 2y + 3x = 6 is y = (2/3)x + 1.