Question

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

Answer 1

Answer:

2x + 5y = –15.

Step-by-step explanation:

We are given coordinates of the line passed through (–5, –1) and (10, –7) .

Applying slope formula,

$Slope=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-5,\:-1\right),\:\left(x_2,\:y_2\right)=\left(10,\:-7\right)$

Therefore,

$m=\frac{-7-\left(-1\right)}{10-\left(-5\right)}$

$m=-\frac{2}{5}$

Therefore, slope is $m=-\frac{2}{5}$.

Applying point-slope form $y-y_1=m(x-x_1),$ we get

$y+7 = -\frac{2}{5}(x-10)$

$y+7=-\frac{2}{5}(x-10)$

On multiplying both sides by 5, we get

$5(y+7)=5\times-\frac{2}{5}(x+1)$

5y+35=-2(x-10)

5y+35=-2x+20

Adding 2x on both sides, we get

5y+25+2x=-2x+20+2x

2x+5y+35=20

Subtracting 35 from both sides, we get

2x+5y+35-35=20-35

2x+5y=-15.

Therefore, required equation is :

2x + 5y = –15.