Question

iii) If slope of the line joining points P(k, O) and Q (-3, -2) is 2/7 then find k​

Answer 1

The value of k is 4 if the slope of the line joining the points P(k, 0) and Q( - 3, - 2) is 2/7

Slope of a line joining (x₁, y₁), (x₂, y₂) is given by,

$m = \dfrac{y _2 -y _1}{x_2 -x _1}$

Given,

The Slope of the line joining the points P(k, 0) and Q( - 3, - 2) is 2/7

Slope of the line PQ is,

$m = \frac{ - 2 - 0}{ - 3 - k} \\ \\ m = \frac{2}{3 + k}$

Given Slope = 2/7

$\frac{2}{7} = \frac{2}{3 + k} \\ \\ 7 = 3 + k \\ \\ k = 7 - 3 \\ \\ k = 4$

Therefore, The value of k is 4.

Another method to solve :

The equation of line passing through (h, k) and slope m is,

⇒y - k = m ( h - x)

Given, Slope of line PQ is 2/7

⇒ m = 2/7

Also, points P(k, 0) and Q( - 3, - 2)

By Point - Slope form,

⇒ Equation of line PQ

⇒ y + 2 = 2/7 ( x + 3)

⇒ 7 ( y + 2) = 2 ( x + 3)

⇒ 7y + 14 = 2x + 6

⇒ 2x - 7y + 6 - 14 = 0

⇒ 2x - 7y - 8 = 0

Since, P lies on the line, it should satisfy the above equation.

P(k, 0) lies on 2x - 7y - 8 = 0

⇒ 2(k) - 7(0) - 8 = 0

⇒ 2k - 8 = 0

⇒ 2k = 8

⇒ k = 4

Therefore, The value of k is 4

Answer 2

Answer:

k = 4.

Step-by-step explanation:

Apply the formula of slope:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Let,

$P(k,0) = (x_{1},y_{1})$

and,

$Q(-3,-2) = (x_{2},y_{2})$

Now, check out the attachment for the solution.

Thanks!