(i) Lines 2x - by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
$[1 M a r k]$

(ii) Lines mx + 3y + 7 = 0 and 5x - ny - 3 = 0 are perpendicular to each other. Find the relation connecting m and n.

$[1 M a r k]$

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Solution

(i) 2x - by + 3 = 0

by = 2x + 3

y =

Slope of this line =

ax + 3y = 2

3y = -ax + 2

y = $fraction numerator negative a over denominator 3 end fraction x plus 2 over 3$

Slope of this line = $fraction numerator negative a over denominator 3 end fraction$

Since, the lines are parallel, so the slopes of the two lines are equal.

$2 over b equals fraction numerator negative a over denominator 3 end fraction a b equals negative 6$

$[1 M a r k]$

(ii) mx + 3y + 7 = 0

3y = -mx - 7

y = $fraction numerator negative m over denominator 3 end fraction x minus 7 over 3$

Slope of this line = $fraction numerator negative m over denominator 3 end fraction$

5x - ny - 3 = 0

ny = 5x - 3

y =

Slope of this line =

Since, the lines are perpendicular; the product of their slopes is -1.

$open parentheses fraction numerator negative m over denominator 3 end fraction close parentheses open parentheses 5 over n close parentheses equals negative 1 5 m equals 3 n$
$[1 M a r k]$

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(i) Lines 2x - by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.

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