Sure, let's solve this problem step by step.
Part (b): Lines [tex]\( ax-3y=5 \)[/tex] and [tex]\( 4x-2y=3 \)[/tex] are parallel to each other. Find the value of [tex]\( a \)[/tex].
When two lines are parallel, their slopes are equal.
The slope of a line in the form [tex]\( Ax + By = C \)[/tex] is given by [tex]\( -A/B \)[/tex].
For the line [tex]\( ax - 3y = 5 \)[/tex]:
- The slope is [tex]\( -a / (-3) = a / 3 \)[/tex].
For the line [tex]\( 4x - 2y = 3 \)[/tex]:
- The slope is [tex]\( -4 / (-2) = 2 \)[/tex].
Since the lines are parallel, their slopes are equal:
[tex]\[ \frac{a}{3} = 2 \][/tex]
To solve for [tex]\( a \)[/tex]:
[tex]\[ a = 2 \cdot 3 \][/tex]
[tex]\[ a = 6 \][/tex]
So, the value of [tex]\( a \)[/tex] is 6.
Part (c): Lines [tex]\( 5x - 4y = 13 \)[/tex] and [tex]\( ax - 5y = 21 \)[/tex] are perpendicular to each other. Find the value of [tex]\( a \)[/tex].
When two lines are perpendicular, the product of their slopes is [tex]\( -1 \)[/tex].
For the line [tex]\( 5x - 4y = 13 \)[/tex]:
- The slope is [tex]\( -5 / (-4) = 5 / 4 \)[/tex].
For the line [tex]\( ax - 5y = 21 \)[/tex]:
- The slope is [tex]\( -a / (-5) = a / 5 \)[/tex].
Since the lines are perpendicular:
[tex]\[ \left( \frac{5}{4} \right) \left( \frac{a}{5} \right) = -1 \][/tex]
Simplify:
[tex]\[ \frac{5a}{20} = -1 \][/tex]
[tex]\[ \frac{a}{4} = -1 \][/tex]
To solve for [tex]\( a \)[/tex]:
[tex]\[ a = -1 \cdot 4 \][/tex]
[tex]\[ a = -4 \][/tex]
So, the value of [tex]\( a \)[/tex] is -4.
In conclusion:
- For the lines [tex]\( ax - 3y = 5 \)[/tex] and [tex]\( 4x - 2y = 3 \)[/tex] to be parallel, [tex]\( a = 6 \)[/tex].
- For the lines [tex]\( 5x - 4y = 13 \)[/tex] and [tex]\( ax - 5y = 21 \)[/tex] to be perpendicular, [tex]\( a = -4 \)[/tex].
