**Combined Equations of Angle Bisectors of Lines**

**Combined equations of the angle bisectors of the lines represented by ax ^{2} + 2hxy +by^{2} = 0**

_{1}x = 0 and y – m

_{2}x = 0

_{1}α)/√1+m

_{1}

^{2}= + ß–m2α/√1+m

_{2}

^{2}

_{2}

^{2}) (ß – m

_{1}α)

^{2}– (1 + m

_{1}

^{2}) (ß – m

_{2}α)

^{2}= 0

^{2}– y

^{2}= 2hxy 1–m

_{1}m

_{2}/m

_{1}+m

_{1}

_{2}–y

_{2}/a–b = xy/h; is required equation of angle bisectors …… (1)

**Note:**

^{2}– y

^{2}= 0 i.e. x – y = 0, x + y = 0

t of x

^{2}+ coefficient of 2 = 0, then the two bisectors are always perpendicular to each other

**Illustration:**

ax

^{2}+ 2hxy + by

^{2}= 0 and one of the lines given by ax

^{2}+ 2hxy+ by

^{2}+ K(x

^{2}+ y

^{2}) = 0 is equal to the angle between the other two lines of the system.

**Solution:**

_{1}and L

_{2}be one pair and L

_{3}and L

_{4}be the other pair of lines.

_{1}and L

_{3}is equal to the angle between L

_{2}and L

_{4}then pair of bisectors of L

_{1}and L

_{2}would be same as that of L

_{3}and L

_{4}. Pair of bisectors of L

_{3}and L

_{4}is

_{2}–y

_{2}/(a+k)–(b+k) = xy/h

_{2}–y

_{2}/a–b = xy/h

_{1}and L

_{2}.

^{2}+ 2hxy + by

^{2}+ 2gx + 2fy + c = 0 (i) are given by (x–x

_{0})

^{2}–(y–y

_{0})

^{2}/a–b = (x–x

_{0})(y–y

_{0})/h where (x

_{0}, y

_{0}) is the point of intersection of (i)

**Enquiry: If we shift the origin of the coordinate system how can the coordinates of a point be known in the new system? What will happen if we rotate the axis?**

P(x, y) in the new system are

**Illustration:**

**Solution:**

x = 3 + X and y = 1 + Y

**Note:**

^{2}+ 2hxy + by

^{2}+ 2gx + 2fy + c=0 represents a pair of straight lines if = 0.

^{2}– bg

^{2}– ch

^{2}= 0 and h

^{2}> ab.

^{2}+ 2hxy + by

^{2}= 0 represents a pair of straight lines through the origin.

^{2}+ 2hxy + by

^{2}= 0, are y = m

_{1}x and y = m

_{2}x, then y

^{2}– (m

_{1}+ m

_{2})xy + m

_{1}m

_{2}x

^{2}= 0 and y

^{2}+ 2h/b xy + a/b x

^{2}= 0 are identical. If θ is the angle between the two lines, then tanθ = + √(m

_{1}+m

_{2})

^{2}–4m

_{1}m

_{2}/1+m

_{1}m

_{2}= + 2√h

^{2}–ab/a+b.

The lines are perpendicular if a + b = 0 and coincident if h^{2} = ab.

**To read more, Buy study materials of ****Straight Lines**** ****comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics ****here**.