Find the value of k such that the line (k - 2) x + (k + 3) y - 5 = 0 is :

(i) perpendicular to the line 2x - y + 7 = 0

(ii) parallel to it.

Open in App

Solution

(k - 2)x + (k + 3)y - 5 = 0 ....(1)

(k + 3)y = -(k - 2)x + 5

y = $open parentheses fraction numerator 2 minus k over denominator k plus 3 end fraction close parentheses x plus fraction numerator 5 over denominator k plus 3 end fraction$

Slope of this line = m₁ = $fraction numerator 2 minus k over denominator k plus 3 end fraction$

(i) 2x - y + 7 = 0

y = 2x + 7 = 0

Slope of this line = m₂ = 2

Line (1) is perpendicular to 2x - y + 7 = 0

$m subscript 1 cross times m subscript 2 equals negative 1 open parentheses fraction numerator 2 minus k over denominator k plus 3 end fraction close parentheses cross times 2 equals negative 1 4 minus 2 k equals negative k minus 3 k equals 7$

(ii) Line (1) is parallel to 2x - y + 7 = 0

$m subscript 1 equals m subscript 2 fraction numerator 2 minus k over denominator k plus 3 end fraction equals 2 k equals fraction numerator negative 4 over denominator 3 end fraction$

Suggest Corrections

Similar questions

k

2 x - k y - 5 = 0

2 x + 3 y - 6 = 0

k

k x + y - 3 = 0

2 x - 3 y + 4 = 0

2 x + 3 y + 4 + k (6 x - y + 12) = 0

7 x + 5 y - 4 = 0

k

2 x + k y - 9 = 0

2 x + 3 y - 1 = 0

(x + y + 1) + K (2 x - y - 1) = 0

2 x + 3 y - 8 = 0

Join BYJU'S Learning Program