Question

What is the equation of a line perpendicular to the line through points (−3, 7) and (2, 5)?

Answer 1

The equation of a line perpendicular to the line through points (-3, 7) and (2, 5) is y = (5/2)x + 21/2.

Step-by-step explanation:

To find the equation of a line perpendicular to the line passing through points (-3, 7) and (2, 5), we first need to find the slope of the given line. The slope between two points can be calculated using the formula: m = (y2 - y1) / (x2 - x1). So, the slope of the given line is (5 - 7) / (2 - -3) = -2/5. The slope of a line perpendicular to another line is the negative reciprocal of the slope of the given line. Therefore, the slope of the perpendicular line is 5/2.

Now, we can use the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept. Since we don't have the y-intercept, we can use one of the given points to find it. Let's use point (-3, 7). So, substituting the values into the equation, we have 7 = (5/2)(-3) + b. Solving for b, we get b = 7 + (5/2)(3) = 21/2.

Therefore, the equation of the line perpendicular to the line passing through points (-3, 7) and (2, 5) is y = (5/2)x + 21/2.

Answer 2

y = 5/2x

Step-by-step explanation:

(-3, 7) and (2, 5)

m = 5-7/2+3

m = -2/5

Slope Perpendicular to this: 5/2

y = 5/2x + b

5 = 5/2(2) + b

5 = 5 + b

b = 0

y = 5/2x