Question

Q14. The value of 'k'such that the line (k-2)x + (k = 3)y - 5 = 0 is
perpendicular to the line 2x - y + 7 = 0.

Answer 1

EXPLANATION.

To find value of k.

Line = ( k - 2 )x + ( k + 3 )y - 5 = 0.

→ Slope = y = Mx + c.

→ ( k + 3 )y = 5 - ( k - 2 ).

→ y = 5/k + 3 - ( k - 2/k + 3).

perpendicular to the line 2x - y + 7 = 0.

Slope of line y = 2x + 7 = 2.

Slope of perpendicular = b/a = -1/2.

$\sf \: m_{1} m_{2} \: = - 1$

→ - [ k - 2/k + 3 ] X [ 2 ] = -1.

→ - k + 2 / k + 3 X 2 = -1.

→ -2k + 4 / k + 3 = -1.

→ -2k + 4 = -k - 3.

→ -2k + k = -3 - 4.

→ K = 7.

Answer 2

(k − 2)x + (k + 3)y − 5 = 0 ....(1)

(k + 3)y = − (k − 2)x + 5

$\sf \: y = ( \frac{2 - k}{k + 3} )x + \frac{ 5}{k + 3} \\ \sf \: slope \: of \: this \: line = m \tiny{1} = \frac{2 - k}{k + 3}$

$\sf(i)2x - y + 7 = 0 \\ \sf y = 2x + 7 = 0 \\ \sf \: slope \: of \: this \: line = m {\tiny{2} } = 2$

$\sf \: line(1)is \: perpendicular \: to \: 2x - y + 7 = 0$

$\sf \: m {\tiny{1}}m {\tiny{2}} = - 1$

$➛ \sf \:( \frac{2 - k}{k + 3} )(2) = - 1$

$➛ \sf \: 4 - 2k = - k - 3 \\ ➛ \sf \: k = 7$

$\sf \: (ii)line(1)is \: paralel \: to \: 2x - y + 7 = 0$

$\sf \: therefore \: m{ \tiny{1}} = m {\tiny{2}}$

$➛ \sf \: \frac{2 - k}{k + 3} = 2 \\ ➛ \sf \: 2 - k = 2k + 6 \\ \sf➛3k = - 4 \\ \sf \red{➛k = \frac{ - 4}{3} }$