Question #22890

Write an equation for the line in point/slope form and slope/intercept form that has the given condition.
6. Passes through (3,2) and is parallel to 2x-y=4
7. Passes through (-1,-1) and is perpendicular to y=5/2x+3

Expert's answer

The equation of the line is:

- in point/slope form y-y1=m(x-x1), where (x1,y1) - point on this line and m -

slope of this line;

- in slope/intercept form y=mx+b, where m - slope of this line and (0,b) -

coordinates of y-intercept

6. For this case: if two lines are parallel then their slopes are the same

=> m=2 and (x1,y1) = (3,2)

The equation for this line in point/slope form is y-2=2(x-3)

To get slope/intercept equation we have to transform the point/slope

equation to form y=mx+b:

y-2=2(x-3)

y=2x-6+2

y=2x-4

7. Similarly to previous case we have:

- in point/slope form: y+1=-2/5(x+1), 'cause for two perpendicular lines

m1=-1/m2, where m1 and m2 are the slopes of its lines;

- in slope/intercept form: y=-(2/5)x-2/5-1 y=-(2/5)x-7/5.

- in point/slope form y-y1=m(x-x1), where (x1,y1) - point on this line and m -

slope of this line;

- in slope/intercept form y=mx+b, where m - slope of this line and (0,b) -

coordinates of y-intercept

6. For this case: if two lines are parallel then their slopes are the same

=> m=2 and (x1,y1) = (3,2)

The equation for this line in point/slope form is y-2=2(x-3)

To get slope/intercept equation we have to transform the point/slope

equation to form y=mx+b:

y-2=2(x-3)

y=2x-6+2

y=2x-4

7. Similarly to previous case we have:

- in point/slope form: y+1=-2/5(x+1), 'cause for two perpendicular lines

m1=-1/m2, where m1 and m2 are the slopes of its lines;

- in slope/intercept form: y=-(2/5)x-2/5-1 y=-(2/5)x-7/5.

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