3. The of each of the points lines on the parallel to y axis is equal.
Answers
Step-by-step explanation:
Generally the line equation is y=mx+c so equation of y-axis is x=0.
Answer:
Step-by-step explanation:
Equation of line in two point form is given by
Equation of line in two point form is given by x−x
Equation of line in two point form is given by x−x 1
Equation of line in two point form is given by x−x 1
Equation of line in two point form is given by x−x 1
Equation of line in two point form is given by x−x 1 y−y
Equation of line in two point form is given by x−x 1 y−y 1
Equation of line in two point form is given by x−x 1 y−y 1
Equation of line in two point form is given by x−x 1 y−y 1
Equation of line in two point form is given by x−x 1 y−y 1 =
Equation of line in two point form is given by x−x 1 y−y 1 = x
Equation of line in two point form is given by x−x 1 y−y 1 = x 2
Equation of line in two point form is given by x−x 1 y−y 1 = x 2
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3 =
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3 = 2−2
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3 = 2−23−1
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3 = 2−23−1
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3 = 2−23−1 ⟹x−2=0 i.e. x=2
Equation of line in two point form is given by x−x 1 y−y 1 = x 2 −x 1 y 2 −y 1 ∴ equation of line for (2,3) and (2,1) is x−2y−3 = 2−23−1 ⟹x−2=0 i.e. x=2Here the abscissa for P as well as Q is 2 i.e same. So both the points lie on the line x=y