@Alex asked this question when I was on earlier: Let line l_1 be the graph of 5x + 8y = -9. Line l_2 is perpendicular to line l_1 and passes through the point (10,10). If line l_2 is the graph of the equation y=mx +b, then find m+b. I'm wondering like how everyone did it and is this pre-algebra, algebra 1, algebra 2, geometry, pre-calculus, or calculus. Thank you for answering!
I think algebra 1 is points and slopes
Read and study Reiny's reply.
https://www.jiskha.com/display.cgi?id=1504542568
To find the equation of line l_2, which is perpendicular to line l_1 and passes through the point (10,10), we need to determine the slope of line l_2 and its y-intercept.
First, let's find the slope of line l_1. We can rewrite the equation of line l_1 in the slope-intercept form (y = mx + b) by solving for y:
5x + 8y = -9
8y = -5x - 9
y = -5/8x - 9/8
From this equation, we can see that the slope of line l_1 is -5/8.
Since line l_2 is perpendicular to line l_1, its slope will be the negative reciprocal of the slope of line l_1. To find the slope of line l_2, we can flip the fraction and change the sign:
m_2 = -1/(-5/8) = 8/5
Now we know that the slope of line l_2 is 8/5.
To find the y-intercept (b) of line l_2, we can substitute the coordinates (10,10) into the equation y = mx + b and solve for b:
10 = (8/5)(10) + b
10 = 16 + b
b = 10 - 16
b = -6
Based on these calculations, we have determined that the slope (m) of line l_2 is 8/5 and the y-intercept (b) is -6. Therefore, m + b = 8/5 + (-6) = 8/5 - 30/5 = -22/5.
As for the grade level, this question is within the realm of algebra 1. It involves concepts such as slope, perpendicular lines, and equations of lines. It does not require knowledge of pre-calculus or calculus.