Question

Which equation represents a line which is perpendicular to the line 7, x, minus, 8, y, equals, 167x−8y=16? Answer Multiple Choice Answers y, equals, start fraction, 8, divided by, 7, end fraction, x, minus, 1y= 7 8 ​ x−1 y, equals, start fraction, 7, divided by, 8, end fraction, x, plus, 8y= 8 7 ​ x+8 y, equals, minus, start fraction, 8, divided by, 7, end fraction, x, minus, 7y=− 7 8 ​ x−7 y, equals, minus, start fraction, 7, divided by, 8, end fraction, x, plus, 4y=− 8 7 ​ x+4

Answer 1

The equation representing a line perpendicular to 7x−8y=16 is y=− 8/7 x+4.

To find the equation of a line perpendicular to 7x−8y=16, we need to consider the slope of the given line. The given line can be rewritten as 8y=7x−16, and when solving for y, we get y= 7/8x−2. The slope of this line is 7/8.

The negative reciprocal of 7/8 is −8/7, which is the slope of a line perpendicular to the given line. Therefore, the equation of the perpendicular line has the form y=− 8/7x+b, where b is the y-intercept.

Comparing this with the provided options, the correct equation is y=− 8/7x+4, where b=4. This line is perpendicular to the original line 7x−8y=16, and the negative reciprocal slopes ensure the perpendicular relationship.