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Exercise 12.1 · Q3

Q.Find the equation of the bisector of the angle between the coordinate axes.

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✓ Free question

The xx-axis and yy-axis meet at the origin at 90∘90^\circ; their two angle bisectors are the lines making 45∘45^\circ with each axis.

Angle bisectors of two lines a1x+b1y+c1=0a_1x+b_1y+c_1=0 and a2x+b2y+c2=0a_2x+b_2y+c_2=0 are given by

a1x+b1y+c1a12+b12=±a2x+b2y+c2a22+b22\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}

  1. Write the two axes as lines. The xx-axis is y=0y=0 (i.e. 0⋅x+1⋅y+0=00\cdot x+1\cdot y+0=0); the yy-axis is x=0x=0 (i.e. 1⋅x+0⋅y+0=01\cdot x+0\cdot y+0=0).

  2. Apply the bisector formula.

0⋅x+1⋅y+002+12=±1⋅x+0⋅y+012+02\frac{0\cdot x+1\cdot y+0}{\sqrt{0^2+1^2}} = \pm\frac{1\cdot x+0\cdot y+0}{\sqrt{1^2+0^2}}

y1=±x1   ⟹   y=±x\frac{y}{1} = \pm\frac{x}{1} \ \implies\ y = \pm x

  1. State both bisectors.

y=xandy=−xy = x \qquad \text{and} \qquad y = -x

y=xy=x bisects the angle between the axes in the first/third quadrants (each 45∘45^\circ from both axes there); y=−xy=-x bisects the angle in the second/fourth quadrants.

  1. Self-check. Any point on y=xy=x, e.g. (1,1)(1,1), is equidistant from both axes: distance from xx-axis =∣1∣=1=|1|=1, distance from yy-axis =∣1∣=1=|1|=1 ✓. Similarly (1,−1)(1,-1) on y=−xy=-x: distance from xx-axis =1=1, from yy-axis =1=1 ✓. Also y=xy=x and y=−xy=-x are perpendicular to each other (slopes 11 and −1-1, product =−1=-1), as expected for bisectors of a right angle and its supplement.
✓Final answer

y=xy = x and y=−xy = -x — the two mutually perpendicular bisectors of the four right angles formed by the coordinate axes.

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