A and B are constants such that the graphs of the lines 3x - 4y = 7 and 8x + Ay = B are perpendicular and intersect at (5,2). What is A+B?
4 y = 3 x - 7 ----> y = (3/4) x - 7/4
slope = (3/4)
so the slope of the second one = (-4/3)
A y = - 8 x + B
y = -8/A x + B/A
so
(-4/3) = -8/A
looks like A = 6
goes through (5,2)
2 = (-4/3)(5) + B/6
12 = -40 + B
B = 52
6+52 = 58
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You are welcome.
To determine the values of A and B, we need to use the information provided about the lines being perpendicular and intersecting at the point (5,2).
First, let's find the slope of the line 3x - 4y = 7. We can rearrange this equation to the slope-intercept form (y = mx + c), where m represents the slope and c is the y-intercept:
3x - 4y = 7
-4y = -3x + 7
y = (3/4)x - 7/4
From this equation, we can see that the slope of the line is 3/4.
If two lines are perpendicular, the product of their slopes is -1. Therefore, the slope of the second line 8x + Ay = B must be the negative reciprocal of the slope of the first line.
The negative reciprocal of 3/4 is -4/3. So, we have:
-4/3 = -A/8
Solving for A, we get A = -32/3.
Now, substituting the coordinates (5,2) into the equation 8x + Ay = B, we can find B:
8(5) + (-32/3)(2) = B
40 - 64/3 = B
(120 - 64)/3 = B
56/3 = B
Therefore, A = -32/3 and B = 56/3.
Finally, to find A+B, we can add the values:
A + B = (-32/3) + (56/3)
A + B = 24/3
A + B = 8
Therefore, A + B = 8.