Aline passes through (3, –2) and (6, 2). a. write an equation for the line in point-slope form. b. rewrite the equation in standard form using integers. a. y+2=4/3(x-3); -4x-3y=-18 b. y-2=4/3(x-3); -4x-3y=18 c. y+2=4/3(x+3); -4x+3y=-18 d. y-3=4/3(x+2); -4x+3y=17 i need !

First thing you have to do if you're going to write this line's equation in point-slope form is to find the slope. Use the points in the slope formula as follows: . So the slope is 4/3. The point-slope form of a line is where x1 is the x-coordinate from either point, and y1 is the y coordinate from the SAME point as the x. I'll use (3, -2) since that's the point they used in the solutions above. x is 3 and y is -2. Filling in accordingly, we have . That is point-slope form. Now they want us to rewrite in standard form which is Ax + By = C, no fractions allowed and the x and the y have to be on the same side of the equals sign. However, before we can do that successfully, we need to rewrite the point-slope into slope-intercept. THEN we can do standard. In slope intercept, we will set the equation equal to y. . Simplifying a bit we get or . Now in order to get rid of the fraction we will multiply everything, every term, by 3. Doing that gives us . Getting x and y on the same side is . I am not seeing this exact combination of point-slope and standard within the same answer. I checked my work several times, perhaps you can check how you posted the possible solutions?

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