Find an equation of the line that passes through the point (2, 3) and is perpendicular to the line 4x – 3y = 10

To find the equation of the line that passes through the point (2, 3) and is perpendicular to the given line, we first need to determine the slope of the given line.

The equation of the line is given in standard form:

4x – 3y = 10

To find the slope, we can rewrite this equation in slope-intercept form (y = mx + b), where m is the slope. We will rearrange the equation:

3y = 4x - 10

Next, divide everything by 3:

y = (4/3)x - (10/3)

From this, we can see that the slope (m) of the given line is 4/3.

The slope of a line that is perpendicular to another is the negative reciprocal of the original slope. Therefore, the slope of the line we want to find is:

-3/4

Now, we have a slope of -3/4 and a point that the line passes through, which is (2, 3). We can use the point-slope form of a line’s equation, which is:

y - y₁ = m(x - x₁)

Here, (x₁, y₁) is the point (2, 3) and m is the slope -3/4. Substituting these values in gives us:

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

Now, we’ll simplify this equation:

y - 3 = -3/4x + 3/2

y = -3/4x + 3/2 + 3

y = -3/4x + 3/2 + 6/2

y = -3/4x + 9/2

Thus, the equation of the line that passes through the point (2, 3) and is perpendicular to the line 4x – 3y = 10 is:

y = -3/4x + 9/2