If the straight lines 2x + 3y - 3 = 0 and x + ky + 7 = 0 are perpendic

1.	If the straight lines 2x + 3y - 3 = 0 and x + ky + 7 = 0 are perpendicular, then the value of k is1. -3/22. -2/33. 3/24. 2/3
Answer» Correct Answer - Option 2 : -2/3 Concept: Let the one line has slope m₁ and the second line has slope m₂. If two straight lines are perpendicular then the multiplication of their slopes will be -1, that is "m₁m₂= -1". Calculation: Compare both the given equation with the standard form, y = mx + c. The slope, m₁ for the line, 2x + 3y - 3 = 0 is, \(- \frac{2}{3}\). The slope, m2 for the line, x + ky + 7 = 0 is, \( \rm- \frac{1}{k}\). As both the equations are perpendicular, so m1m2 = -1 \(\rm\left( { - \frac{2}{3}} \right)\left( { - \frac{1}{k}} \right) = - 1\) ⇒ \(\rm \frac{2}{{3k}} = - 1\) ⇒ \(\rm k = - \frac{2}{3}\)

If the straight lines 2x + 3y - 3 = 0 and x + ky + 7 = 0 are perpendicular, then the value of k is1. -3/22. -2/33. 3/24. 2/3

Answer» Correct Answer - Option 2 : -2/3

Concept:

Let the one line has slope m₁ and the second line has slope m₂.

If two straight lines are perpendicular then the multiplication of their slopes will be -1, that is "m₁m₂= -1".

Calculation:

Compare both the given equation with the standard form, y = mx + c.

The slope, m₁ for the line, 2x + 3y - 3 = 0 is, \(- \frac{2}{3}\).

The slope, m2 for the line, x + ky + 7 = 0 is, \( \rm- \frac{1}{k}\).

As both the equations are perpendicular, so m1m2 = -1

\(\rm\left( { - \frac{2}{3}} \right)\left( { - \frac{1}{k}} \right) = - 1\)

⇒ \(\rm \frac{2}{{3k}} = - 1\)

⇒ \(\rm k = - \frac{2}{3}\)