the equation of lin n is 8x - 14y + 3 = 0. what is k if he graphs of n and kx - 7y + 10 = 0 are perpendicular?
To find the value of k for which the graphs of n and kx - 7y + 10 = 0 are perpendicular, we need to use the property that the slopes of perpendicular lines are negative reciprocals to each other.
Given the equation of line n as 8x - 14y + 3 = 0, we need to find its slope. To do this, we need to rearrange the equation in the form y = mx + c (where m is the slope):
8x - 14y + 3 = 0
-14y = -8x - 3
y = (8/14)x + 3/14
Comparing this equation with y = mx + c, we can see that the slope of line n is 8/14 or 4/7.
For the line kx - 7y + 10 = 0 to be perpendicular to line n, its slope should be the negative reciprocal of 4/7. The negative reciprocal of any fraction a/b is -b/a.
So, the slope of the line kx - 7y + 10 = 0 should be -7/4.
Now, comparing the equation kx - 7y + 10 = 0 with y = mx + c, we can see that the slope of the line is k/7.
Therefore, we have the equation k/7 = -7/4.
To solve for k, we cross-multiply:
k * 4 = -7 * 7
4k = -49
k = -49/4
Hence, if the graphs of n and kx - 7y + 10 = 0 are perpendicular, the value of k is -49/4.