the lines satisfying the equations x+2y=3 and 3y+ax=2 are perpendicular to each other. find the value of 'a'. please answer and explain how you got, please!!
slope of first line = -1/2
slope of 2nd line = -a/3
to be perpendicular, they must be negative reciprocals of each other, or, when multiplied we should get -1
(-1/2)(-a/3) = -1
a/6 = -1
a = -6
To find the value of 'a' such that the lines are perpendicular to each other, we can use the slope-intercept form of the equations.
The given equations are:
1) x + 2y = 3
2) 3y + ax = 2
To determine the slope of a line in the slope-intercept form (y = mx + c), where 'm' is the slope, we need to rearrange the equations in this form.
1) x + 2y = 3
Rearrange as: 2y = -x + 3
Divide by 2: y = -1/2x + 3/2
2) 3y + ax = 2
Rearrange as: 3y = -ax + 2
Divide by 3: y = -ax/3 + 2/3
The slopes of the lines are the coefficients of 'x' in each equation. So, the slopes are:
1) Slope of line 1: -1/2
2) Slope of line 2: -a/3
For two lines to be perpendicular, the product of their slopes should be -1.
Therefore, we can set up the equation:
(-1/2) * (-a/3) = -1
a/6 = -1
Multiply both sides by 6:
a = -6
Hence, the value of 'a' that satisfies the condition of perpendicular lines is -6.