Question

The point-slope form of the equation of the line that passes through (–5, –1) and (10, –7) is y plus 7 equals negative StartFraction 2 over 5 EndFraction left-parenthesis x minus 10 right-parenthesis.. What is the standard form of the equation for this line?

Answer 1

$\huge \underline \green{\sf{ Answer :- }}$

$\small \underline {\sf{ (-5, -1), (10, -7).}}$

$\small \underline {\sf{Y = mx + b. }}$

$\small \underline {\sf{ m = (-7+1)/(10+5) = -6/15 = -2/5.}}$

$\small \underline {\sf{ b = y-mx = -1-(-2/5(-5) = -1 - 2 = -3.}}$

$\small \underline {\sf{ Y = -2x/5 - 3.}}$

$\small \underline {\sf{2x/5 + y = -3, }}$

$\small \underline {\sf{2x + 5y = -15 }}$

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Answer 2

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$\small \underline {\sf{ (-5, -1), (10, -7).}}$

$\small \underline {\sf{Y = mx + b. }}$

$\small \underline {\sf{ m = (-7+1)/(10+5) = -6/15 = -2/5.}}$

$\small \underline {\sf{ b = y-mx = -1-(-2/5(-5) = -1 - 2 = -3.}}$

$\small \underline {\sf{ Y = -2x/5 - 3.}}$

$\small \underline {\sf{2x/5 + y = -3, }}$

$\small \underline {\sf{2x + 5y = -15 }}$