Question

The value of k such that (k - 2) x + (k+3) y-5=0 is parallel to the line 2x - y + 7 = 0
a) 4/3
b) -4/3
c) 1/3
d) -1/3​

Answer 1

EXPLANATION.

Value of k such that,

⇒ (k - 2)x + (k + 3)y - 5 = 0.

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

As we know that,

⇒ y = mx + c.

⇒ (k - 2)x + (k + 3)y - 5 = 0.

⇒ (k + 3)y = -(k - 2) + 5.

⇒ (k + 3)y = (-k + 2) + 5.

⇒ y = (2 - k)/(k + 3) + 5/(k + 3).

Now, the equation is in the form of y = mx + c.

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

As we know that,

Slope of the parallel line is = -a/b.

Slope = -(2)/-1 = 2.

⇒ M₁ = M₂.

⇒ (2 - k)/(k + 3) = 2.

⇒ 2 - k = 2(k + 3).

⇒ 2 - k = 2k + 6.

⇒ 2 - 6 = 2k + k.

⇒ -4 = 3k.

⇒ k = -4/3.

Option [B] is correct answer.

                                                                                                                 

MORE INFORMATION.

(1) = Equation of straight lines through (x₁, y₁) and making an angle α with y = mx + c.

(y - y₁) = m ± tanα/1 ± m tanα (x - x₁).

(2) = Lengths of perpendicular.

From (x₁, y₁) to the straight line ax + by + c = 0 then,

P = | ax₁ + by₁ + c |/√a² + b².

(3) = Distance between two parallel lines.

ax + by + c₁ = 0 & ax + by + c₂ = 0 then,

d = | c₁ - c₂ |/a² + b².

Answer 2

Answer:

Given :-

such that (k - 2) x + (k+3) y-5=0 is parallel to the line 2x - y + 7 = 0

To Find :-

k

Solution :-

$\sf \: (k - 2)x + (k + 3)y - 5 = 0.$

$\sf \:  (k + 3)y = -(k - 2) + 5.$

$\sf \: (k + 3)y = (-k + 2) + 5.$

$\sf \: y = \dfrac{(2 - k)}{(k + 3)} + \dfrac{5}{k + 3}$

Now,

Slope = -(2)/-1

= -2/-1

= -2

M₁ = M₂.

$\sf \: \dfrac{2 - k}{k + 3} = 2$

By Cross Multiplication

$\sf \: 2(k + 3) = 2 - k$

$\sf \: (2 \times k) + (2 \times 3) = 2 - k$

$\sf \: 2k + 6 = 2 - k$

$\sf \: 2k + k \: = 2 - 6$

$\sf \: 3k = 4$

$\sf \: k \: = \dfrac{3}{4}$