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.

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Answer 1

Answer:

i am fool for u

Explanation:

not for others....lol

Answer 2

Answer:

$Given \: line \: equation \\ </p><p></p><p>(k−2)x+(k+3)y−5=0 (1) \\ </p><p></p><p>(k+3)y=−(k−2)x+5</p><p></p><p>$

$y = ( \frac{2 - k}{k + 3} )x + \frac{5}{k + 3}$

$Slope of this line =m1=( \frac{(2 - k)}{(k + 2)}$

$Given, 2x−y+7=0 \\ </p><p>y=2x+7=0 \\ </p><p></p><p>Slope of this line =m2=2 \\ </p><p></p><p></p><p>Given that, line (1) is \\ perpendicular to \\ \: 2x−y+7=0 \\ </p><p></p><p>m1×m2=−1 \\ </p><p></p><p>$

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

$4−2k=−k−3 \\ </p><p></p><p>k=7</p><p></p><p>$

$hope \: it \: helps \: you$

$a \: request \: = \: could \: you \: \\ please \: thank \: my \: answers \: plz \:$