Find the real number a for which the lines −4x+ 3y = 7 and ax− 2y = 4 will be perpendicular.
3 y = 4 x + 7
y = (4/3) x + 7/3
slope is (4/3)
slope of perpendicular = -3/4
so now look at second line, has to have m = slope = -3/4
2 y = a x - 4
y = (a/2) x - 2
so
-3/4 = a/2
a = -3/2
This Question is very hard. I like for my son.
To find the real number "a" for which the given lines are perpendicular, we can use the property that the slopes of perpendicular lines are negative reciprocals of each other.
First, let's rearrange the given equations into slope-intercept form (y = mx + b), where "m" is the slope:
Line 1: -4x + 3y = 7
Rearranging, we get 3y = 4x + 7, and dividing by 3, y = (4/3)x + 7/3.
So the slope of Line 1 is m1 = 4/3.
Line 2: ax - 2y = 4
Rearranging, we get -2y = -ax + 4, and dividing by -2, y = (a/2)x - 2.
So the slope of Line 2 is m2 = a/2.
Since the slopes of perpendicular lines are negative reciprocals, we have the equation: m1 * m2 = -1.
Substituting the slopes, we get (4/3) * (a/2) = -1.
Now, let's solve for "a":
(4/3) * (a/2) = -1
(4a)/(3*2) = -1
2a/3 = -1
Multiply both sides by 3:
2a = -3
Divide both sides by 2:
a = -3/2
Therefore, the real number "a" for which the given lines are perpendicular is a = -3/2.