The point-slope form of the equation of the line that passes through (–5, –1) and (10, –7) is y + 7 = (x – 10). what is the standard form of the equation for this line?

2x – 5y = –15
2x – 5y = –17
2x + 5y = –15
2x + 5y = –17

Option C. 2x + 5y = -15

Step-by-step explanation:

A line which passes through two points (-5, -1) and (10, -7) will have the slope = (y - y')/(x - x') = (-1 + 7)/(-10 - 5) = 6/(-15) = (-2/5)

Now the equation will be y = (-2/5)x + c

Since this line passes through (10, -7) therefore the equation of the line will be

-7 = (-2/5)(10) + c

-7 = -4 + c

c = -7 + 4 = -3

Then the final equation of the line will be

y = (-2/5)x - 3

Multiplying with 5 on both the sides of the equation

5y = (-2/5)×5 - 3×5

5y = -2x - 15

2x + 5y = -15

Therefore option C is the answer.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116