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Write the equation of the line described in point-slope form. Then, change to Standard Form. Contains points (3/2,-1/2) and (-1/2,5/2). ?

Point slope form is y−y1 =m(x−x1) and standard form is Ax + By = C. Please explain.

User Oranutachi
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Slope point form :
To put in slope point form, label any of the points as either X1,y1 and X and y, then plug in those values into the following equation form.
Y - y1 = m(X-X1)

But before, we must solve for the m value or slope.
M = y2-y1/x2-X1
M = 5/2 - -1/2 / -1/2 - 3/2.
M = 5/2 + 1/2 / -(1/2+3/2)
M = 6/2 / -(4/2)
M = 3/-2

Now we can place it in slope point and also in standard form of a line.

Y-y1 = m(X -X1)

Y - -1/2 = -3/2(X - 3/2)

Y + 1/2 = -3/2(X - 3/2)

This is in slope point form.

Y + 1/2 = -3/2x + 9/4
Y + 1/2 - 1/2 = -3/2x + 9/4 - 1/2
1/2 = 2/4
Y = -3/2x + 7/4
-3/2x = -6/4x
Y = -6/4x + 7/4
Y • 4 = 4( -6/4 X + 7/4)
4y = -6x + 7
4y + 6x = -6x + 6x +7

6x + 4y = 7

This is in standard form. If you have any questions of the steps just ask.
User Soojin
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