the point - slope form of the equation of the line that passes through (-5, -1) and (10, -7) is (y +…

the point - slope form of the equation of the line that passes through (-5, -1) and (10, -7) is (y + 7=-\frac{2}{5}(x - 10)). what is the standard form of the equation for this line?\n2x - 5y=-15\n2x - 5y=-17\n2x + 5y=-15\n2x + 5y=-17

the point - slope form of the equation of the line that passes through (-5, -1) and (10, -7) is (y + 7=-\frac{2}{5}(x - 10)). what is the standard form of the equation for this line?\n2x - 5y=-15\n2x - 5y=-17\n2x + 5y=-15\n2x + 5y=-17

Answer

Explanation:

Step1: Expand the point - slope form

Start with $y + 7=-\frac{2}{5}(x - 10)$. Distribute $-\frac{2}{5}$ on the right - hand side: $y+7=-\frac{2}{5}x+4$.

Step2: Move all terms to one side

Add $\frac{2}{5}x$ to both sides and subtract 7 from both sides: $\frac{2}{5}x+y=4 - 7$. $\frac{2}{5}x+y=-3$.

Step3: Get rid of the fraction

Multiply every term in the equation by 5 to clear the fraction: $2x + 5y=-15$.

Answer:

C. $2x + 5y=-15$