Question

find the value of k so that the line passing through each pair of points has the given slope please help and please try

Answer 1

• a) Given the points:

(x1, y1) ==> (k, 2)

(x2, y2) ==> (11, 14)

Slope, m = 2

To find the missing coordinate, k, use the slope formula below:

$m=\frac{y2-y1}{x2-x1}$

Input values into the formula to find k:

$2=\frac{14-2}{11-k}$

Let's solve for k.

Cross multiply:

$\begin{gathered} 2(11-k)=14-2 \\ \\ 22-2k=14-2 \\ \\ -2k=14-2-22 \\ \\ -2k=-10 \\ \\ \frac{-2k}{-2}=\frac{-10}{-2} \\ \\ k=5 \end{gathered}$

b) Given the points:

(x1, y1) ==> (1, k)

(x2, y2) ==> (4, 1)

slope = -2

Let's use the method in question (a) to find k:

$\begin{gathered} -2=\frac{1-k}{4-1} \\ \\ -2(4-1)=1-k \\ \\ -8+2=1-k \\ \\ -8+2-1=-k \\ \\ -7=-k \\ \\ k=7 \end{gathered}$

c) Given the points:

(x1, y1) ==> (3, 5)

(x2, y2) ==> (k, 9)

slope = 1/2

Let's use the method above to solve for k:

$\begin{gathered} \frac{1}{2}=\frac{9-5}{k-3} \\ \\ 1(k-3)=2(9-5) \\ \\ k-3=18-10 \\ \\ k=18-10+3 \\ \\ k=11 \end{gathered}$

d) Given the points:

(x1, y1) ==> (-1, 4)

(x2, y2) ==> (-3, k)

slope = -1/2

Solve for k:

$\begin{gathered} -\frac{1}{2}=\frac{k-4}{-3--1} \\ \\ -\frac{1}{2}=\frac{k-4}{-3+1} \\ \\ 2(k-4)=-1(-3+1) \\ \\ 2k-8=3-1 \\ \\ 2k-8=2 \\ \\ 2k-8+8=2+8 \\ \\ 2k=10 \\ \\ \frac{2k}{2}=\frac{10}{2} \\ \\ k=5 \end{gathered}$

ANSWER:

a) k = 5

b) k = 7

c) k = 11

d) k = 5