Question

find the equation of the line which is parallel to the line 3x-4y+10=0 and passing through the point(6,-3)​

Answer 1

Answer:

3x+4y−11=0.

Step-by-step explanation:

Two parallel lines have the same slope, put 3x+4y=−7 in slope-intercept form:

3x+4y=−7

4y=−3x−7

y=−

4

3

x−

4

7

The slope is −

4

3

, so a line parallel to this has a slope of −

4

3

also.

Now since the equation of a line is y−y

1

=m(x−x

1

), where m is the slope of the line, therefore with point (1,2)

We have, (y−2)=−

4

3

(x−1)

⇒4(y−2)=−3(x−1)

⇒4y−8=−3x+3

⇒3x+4y−8−3=0

⇒3x+4y−11=0

Hence, resulting linear equation is 3x+4y−11=0.

Answer 2

m=y-c/x this formula wpply it