  Question

# Find the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3).

Solution

## The given piont are A (1, 0) and B (2, 3)Let M be the midpoint of AB.Coordinates of M=(1+22,0+32)=(32,32) And, slope of AB=3−02−1=3Let m be the slope of the perpendicular bisector of the line joining the points A (1, 0) and B (2, 3)∴ m× Slope of AB=−1⇒ m×3=−1⇒ m=−13So, the equation of the line that passesthrough M(32,32) and has slope−13 isy−32=−13(x−32)⇒ x+3y.−6=0Hence, the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3) is x + 3y - 6 = 0P.Q. Find the equation of the line passing through (1, 2) and making angle of 30∘ with y=axis..Equation of the line passing through (x1, y1) and making angle θ with the x-axis is,(y−y1)=tanθ (x−x1)Here, (x1,y1)=(1, 2), angle with y-axis is 30∘∴ angle with x-axis isθ=90∘−30∘=60∘(y−y1)=tanθ (x−x1)(y−2)=(tan 60∘)(x−1)y−2=√3x−√3√3x−y+2−√3=0  Suggest corrections  Similar questions
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