Line Equally Inclined with Two Lines

In this article, we will cover the concept of Line Equally Inclined with two lines. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of ten questions have been asked on JEE MAINS( 2013 to 2023) from this topic.

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This Story also Contains

1. What is Bisector?
2. Line Equally Inclined with two lines
3. Derivation of Line Equally Inclined with two lines
4. Solved Examples Based on Line Equally Inclined with two lines
5. Summary

What is Bisector?

The bisector is the locus of a point that moves in the plane of lines L₁ and L₂ such that lengths of perpendiculars drawn from it to the two given lines(L₁ and L₂) are equal.

Line Equally Inclined with two lines

A line equally inclined with two lines means the line which is the angle bisector of the angle made by two lines.

If the two lines with slope m1 and m2 are equally inclined to a line having slope m, then
(m1−m1+m1m)=−(m2−m1+m2m)

Derivation of Line Equally Inclined with two lines

Two lines with slopes m1 and m2 intersect at point A

As from the fig

∠PAQ=∠QAR=∠θtan⁡(∠PAQ)=m1−m1+m1 m=tan⁡θ and tan⁡(∠QAR)=m−m21+mm2=tan⁡θ

Hence,
(m1−m1+m1m)=−(m2−m1+m2m)

Recommended Video Based on Line Equally Inclined with Two Lines

Solved Examples Based on Line Equally Inclined with two lines

Example 1: Equations of line L1 is 3y−x+1=0 and Equation of line L2 is y−3x+1=0 then find the equation of Line which is equally inclined with both line and passing through the intersection point
Solution: If the two lines with slope m1 and m2 are equally inclined to a line having slope m, then
(m1−m1+m1m)=−(m2−m1+m2m)

Two lines with slopes m1 and m2 intersect at point A

As from the figure

∠PAQ=∠QAR=∠θtan⁡(∠PAQ)=m1−m1+m1 m=tan⁡θ and tan⁡(∠QAR)=m−m21+mm2=tan⁡θ

Hence,
(m1−m1+m1m)=−(m2−m1+m2m)3y−x+1=0⇒m1=13y−3x+1=0⇒m2=3yy−3x+1=0

If the two lines with slope m1 and m2 are equally inclined to a line having slope m, then (m1−m1+m1m)=−(m2−m1+m2m)
m=±1(13−m1+13m)=−(3−m1+3m)⇒m2=1m

For Intersection point
3y−x+1=0y−3x+1=0x=3−12y=32(1−3)

Equation of lines
y−32(1−3)=x−3−12L1:y=x+2−23y−32(1−3)=−1(x−3−12)L2:y=−x+1−3

Hence, the answer is y=x+2−23

Example 2: A line is passing through (1,1) and (2,1+3) and another line which is passing through (1,1) and (2,1+13) then find the equation of the line which is equally inclined with both lines.
Solution: If the two lines with slope m1 and m2 are equally inclined to a line having slope m, then
(m1−m1+m1m)=−(m2−m1+m2m)

Two lines with slopes m1 and m2 intersect at point A

As from the figure
∠PAQ=∠QAR=∠θtan⁡(∠PAQ)=m1−m1+m1 m=tan⁡θ
and
tan⁡(∠QAR)=m−m21+mm2=tan⁡θ

Hence,
(m1−m1+m1m)=−(m2−m1+m2m)

Slope of line L1=1+3−12−1=3=tan⁡60∘
Slope of line L2=1+13−12−1=13=tan⁡30∘
Slope of line which is equally inclined =tan⁡45∘=1 and another line which is also equally inclined,slope is =−1 Equations of lines
L1:y−1=x−1⇒y−x=0L2:y−1=−1(x−1)⇒y+x−2=0

Hence, the answer is y+x−2=0

Example 3: Two equal sides of an isosceles triangle have the equations 7x−y+3=0 and x+y=3 and its third side passes through the point ( −2,−1 ). Then the equation of the third side.
Solution: If the two lines with slope m1 and m2 are equally inclined to a line having slope m, then
Two lines with slopes m1 and m2 intersect at point A

Let Equation of AB is 7x−y+3=0 and equation of ACx+y=3
Slope of AB=7 and Slope of AC=−1
Let the equation of third side BC=y=mx+c
The angle between AB and BC= Angle between AC and BC
|7−m1+7m|=|−1−m1−m|−6m2−16m+6=03m2+8m−3=0m=−3 and m=1/3
equation of BC is
(y−(−1))=−3(x−(−2))⇒3x+y+7=0
or
(y−(−1))=13(x−(−2))⇒x−3y=1

Hence, the answer is x−3y=1

Example 4: If 5x2+λy2=20 represents a rectangular hyperbola, then λ equal
Solution: The general equation of the second degree represents a rectangular hyperbola if Δ≠0,h2>ab and coefficient of x2+ coefficient of y2=0.

The given equation represents a rectangular hyperbola if λ+5=0 i.e., λ=−5.
Hence, the answer is -5.

Example 5: A ray of light travelling along the line 2x−3y+5=0 after striking a plane mirror lying along the line x+y=2 gets reflected. Find the equation of the straight line containing the reflected ray

Solution: The point of intersection of the lines 2x−3y+5=0 and x+y=2 is (15,95).
(15,95) is the point of incidence.
The slope m of the normal to the mirror (i.e. normal to the line x+y=2 ) is 1
Now the incident ray and reflected ray both are equally inclined to the normal and are on opposite sides of it.
The slope of the incident ray
mI=23

Let the slope of the reflected ray be =m2
Then
m1−m1+m1 m=m−m21+m2m
23−11+23×1=1−m21+m2×1

∴m2=32,∴ the equation of the straight line containing the reflected ray is
y−95=32(x−15)
i.e. 3x−2y+3=0

Hence, the answer is 3x−2y+3=0

Summary

A line equally inclined with two given lines possesses a slope that is the negative reciprocal of the product of the slopes of those lines. This concept is fundamental in geometry and finds practical application in various fields, including engineering, architecture, and physics. Understanding the relationships between lines is crucial for problem-solving and analysis. Knowledge of the relation between lines helps us to solve complex problems.

This equation is added for testing

$\mathrm{D}(\mathrm{n})=\mathrm{n}!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\ldots+(-1)^{\mathrm{n}} \frac{1}{\mathrm{n}!}\right)$