Question

Helppppppp. Find k if the lines 3x-5y=9 & 2x+ky=11 are a)parallel and b)perpendicular

Answer 1

For lines to be parallel, they have to have the same slopes. For the equation given as Ax + By + C = 0, the slope is -A / B. Equating the slopes of the given equations of lines,
                                      - (3 / -5) = - (2 / k)
The value of k is equal to -10/3. 

Answer 2

Answer:

a) $k=-\frac{10}{3}$

b) $k=\frac{6}{5}$

Step-by-step explanation:

If two lines $y = m_1x+c_1$ and $y=m_2x+c_2$ ( where, $m_1$ and $m_2$ are their slope respectively ) are,

Parallel, if,

$m_1=m_2$

Perpendicular, if,

$m_1\times m_2 = -1$

Here, the lines are,

3x - 5y = 9 ⇒ 5y = 3x - 9 ⇒ $y=\frac{3}{5}x-\frac{9}{5}$

2x + ky = 11 ⇒ ky = -2x + 11 ⇒ $y=-\frac{2}{k}x + 11$

a) If they are parallel,

$\frac{3}{5}=-\frac{2}{k}\implies 3k=-10\implies k=-\frac{10}{3}$

b) If they are perpendicular,

$\frac{3}{5}\times -\frac{2}{k}=-1\implies -\frac{6}{5k}=-1\implies k=\frac{6}{5}$