To find the equation of the line that is perpendicular to the line given by the equation 4x + 5y = 10 and passes through the point (6, 3), we first need to determine the slope of the original line.
We can rewrite the equation in slope-intercept form (y = mx + b). Starting with:
4x + 5y = 10
Subtracting 4x from both sides gives:
5y = -4x + 10
Now, dividing everything by 5:
y = -\frac{4}{5}x + 2
From this, we see that the slope (m) of the original line is -\frac{4}{5}.
For the line that is perpendicular, the slope will be the negative reciprocal of -\frac{4}{5}. The negative reciprocal can be found by flipping the fraction and changing the sign:
So, the slope of the perpendicular line is \frac{5}{4}.
Now we have the slope of the line we want and a point that it passes through, (6, 3). We can use the point-slope form of the equation of a line, which is:
y – y_1 = m(x – x_1)
Substituting in our values:
y – 3 = \frac{5}{4}(x – 6)
Now we can simplify that equation. Distributing on the right side:
y – 3 = \frac{5}{4}x – \frac{30}{4}
Which simplifies to:
y – 3 = \frac{5}{4}x – 7.5
Adding 3 to both sides gives:
y = \frac{5}{4}x – 4.5
Thus, the equation of the line that passes through the point (6, 3) and is perpendicular to the line 4x + 5y = 10 is:
y = \frac{5}{4}x – 4.5