Question

Write the standard form of the line that passes through (-1,-3) and (2,1).

Answer 1

To find the standard form of the line passing through (-1,-3) and (2,1), you calculate the slope, then use the point-slope form with one of the points to plot the equation before converting to standard form. The equation of the line in standard form is -4x + 3y = -5.

Step-by-step explanation:

To write the standard form of a line that passes through two points, we first need to calculate the slope of the line using the coordinates of the points (-1,-3) and (2,1). The slope formula (m) is given by:

m = (y2 - y1) / (x2 - x1)

For our points, the slope will be:

m = (1 - (-3)) / (2 - (-1)) = 4 / 3

Now that we have the slope, we can use the point-slope form to find the equation of the line. The point-slope form is:

y - y1 = m (x - x1)

Using point (-1, -3) and slope 4/3:

y - (-3) = (4/3)(x - (-1))

Simplify:

y + 3 = (4/3)x + (4/3)

To convert to standard form, which is Ax + By = C, we need to clear the fractions and get x and y terms on one side of the equation:

3y + 9 = 4x + 4

Subtract 4x from both sides:

-4x + 3y = -5

The standard form of the line is then:

-4x + 3y = -5

Answer 2

Step-by-step explanation:

`\mathrm{Line\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Find\:the\:line\:}\\\mathbf{y=mx+b}\mathrm{\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Compute\:the\:slope\:}\left(-1,\:-3\right),\:\left(2,\:1\right):\quad \\m=(4)/(3)\\\mathrm{Slope}=(y_2-y_1)/(x_2-x_1)\\\\\left(x_1,\:y_1\right)=\left(-1,\:-3\right),\:\left(x_2,\:y_2\right)\\\left\\`

`\\m=(1-\left(-3\right))/(2-\left(-1\right))\\\\Refine\\\\m=(4)/(3)\\\\\mathrm{Compute\:the\:}y\mathrm{\:intercept}:\quad b=-(5)/(3)\\\\y=(4)/(3)x-(5)/(3)\\`

Convert equation to standard form

`y - (4)/(3) x + (5)/(3) = 0\\\\Multiply \:through\: by\: 3;\\\\3* (y - (4)/(3) x + (5)/(3) )=0\\\\3y -4x+5 = 0\\Arrange\:n\:form\:of \:;\\Ax +By = C\\\\-4x +3y =-5`