If quadrilateral JKLM had a translation where all the vertices were in the second quadrant, what would be the vertices?
2 answers:
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Answer:
See attachment below
Step-by-step explanation:
For the first option, we are given m = - 2 / 3, b = 3. This is likely in the point - slope form y = mx + b, where m = slope and b = y - intercept. Thus, the equation of the line in point - slope form given this information, should be y = - 2 / 3x + 3. It seems as if none of the following equations match this form, so let us interchange the equation a bit,
y = - 2 / 3x + 3,
y + 2 / 3x = 3,
2x + 3y = 9 - Option C
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Next we are given m = - 3 / 2, and that this line passes through the point ( 4, - 1 ). Substitute m as - 3 / 2, x as 4, y as - 1, knowing point ( 4, - 1 ), into the form y = mx + b - solving for b.
- 1 = ( - 3 / 2 )( 4 ) + b,
b = 5,
Thus the equation in point - slope form should be y = - 3 / 2x + 5. None of the choices match this, so let us again alter the equation,
y = - 3 / 2x + 5,
2y = - 3x + 10,
- 2y = 3x - 10 - Option A
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( 6, 3 ) and ( 3, 1 ) is given to lie on this line. The slope of the line should be as follows,
Slope = 3 - 1 / 6 - 3 = 2 / 3
Now let us determine the y - intercept ( b ) by substitute one of the points, say ( 6, 3 ). In this case x = 6, and y = 3,
3 = ( 2 / 3 )( 6 ) + b,
b = - 1,
y = 2 / 3x - 1 = Option D
Take a look at the attachment below for further help;
