Answer: [tex]\bold{k=\dfrac{48}{11}}[/tex]
Step-by-step explanation:
Use the slope formula: [tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Let (x₁, y₁) = (k + 3, 12) . and (x₂, y₂) = (8, 14 + k) then
[tex]m=\dfrac{14 + k - (12)}{8 - (k - 3)}= \dfrac{2+k}{5 - k}[/tex]
Since, slope (m) is 10, set the slope (above) equal to 10 and solve for k:
[tex]\dfrac{2+k}{5-k}=10\\\\2+k=10(5-k)\qquad \rightarrow \ \text{cross multiplied}\\\\2+k=50-10k\qquad \rightarrow \ \text{distributed 10 into (5-k)}\\\\2+11k=50\qquad \rightarrow \ \text{added 10k to both sides}\\\\11k=48\qquad \rightarrow \ \text{subtracted 2 from both sides}\\\\k=\dfrac{48}{11}\qquad \rightarrow \ \text{divided 11 from both sides}[/tex]
