In what ratio is the segment joining points (2, 3) and (-2, 1) divided by the Y-axis?
A. 1 : 2
B. 1 : 1
C. 3 : 1
D. 2 : 3
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & \frac{{k:1}}{{{\text{A}}\left( { - 2,\,1} \right)\,\,\,\,\,\,\,\,\,\,{\text{C}}\left( {x,\,y} \right)\,\,\,\,\,\,\,\,\,\,{\text{B}}\left( {2,\,3} \right)}} \cr & \Rightarrow x = \frac{{ - 2k + 2}}{{k + 1}} \cr & {\text{At }}y{\text{ - axis}}\,\boxed{x = 0} \cr & \therefore 0 = \frac{{ - 2k + 2}}{{k + 1}} \cr & \Rightarrow \boxed{k = 1} \cr & {\text{Ratio}} = 1:1 \cr} $$Related Questions on Coordinate Geometry
In what ratio does the point T(x, 0) divide the segment joining the points S(-4, -1) and U(1, 4)?
A. 1 : 4
B. 4 : 1
C. 1 : 2
D. 2 : 1
A. 2x - y = 1
B. 3x + 2y = 3
C. 2x + y = 2
D. 3x + 5y = 1
If a linear equation is of the form x = k where k is a constant, then graph of the equation will be
A. a line parallel to x-axis
B. a line cutting both the axes
C. a line making positive acute angle with x-axis
D. a line parallel to y-axis

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