Answered

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Find the value of k such that the line through (k+3,7) and (-5,3 – k) has a
slope of 5/3.

Sagot :

Leora03

Answer:

k= -14

Step-by-step explanation:

[tex]\boxed{slope = \frac{y1 - y2}{x1 - x2} }[/tex]

[tex] \frac{5}{3} = \frac{7 - (3 - k)}{k + 3 - ( - 5)} [/tex]

[tex] \frac{5}{3} = \frac{7 - 3 + k}{k + 3 + 5} [/tex]

[tex] \frac{5}{3} = \frac{4 + k}{k + 8} [/tex]

Cross multiply:

3(4 +k)= 5(k +8)

Expand:

12 +3k= 5k +40

Bring k terms to one side, constant to the other:

5k -3k= 12 -40

Simplify:

2k= -28

Divide both sides by 2:

k= -14

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