JEE Advanced 1990

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Straight lines $3 x + 4 y = 5$ and $4 x - 3 y = 15$ intersect at the point $A$ . Points $B$ and $C$ are chosen on these two lines such that $A B = A C$ . Determine the possible equations of the line $B C$ passing through the point $(1, 2)$ .

Q: Is there a way to solve this without using the angle bisector property? It feels like a shortcut.

You could technically use the distance formula $AB = AC$ by setting coordinates for $B$ and $C$ as variables, but that leads to a massive algebraic nightmare. The beauty of JEE problems lies in spotting the geometric symmetry—the angle bisector property is the elegant path, not just a shortcut.

Q: Wait, the original lines $3x + 4y = 5$ and $4x - 3y = 15$ are perpendicular. Does that make the problem easier?

Absolutely! Since the product of their slopes is $(-\frac{3}{4}) \times (\frac{4}{3}) = -1$, the lines are perpendicular. This means the angle bisectors are at $45^\circ$ to the lines. It simplifies the math significantly, confirming that our bisector slopes will also be perpendicular to each other.

Progressive Hints

Hint 1

Visualize the geometry. Since

△ A B C

is isosceles with

A B = A C

, the line

B C

acts as the base. Recall a fundamental property of isosceles triangles: the altitude from the vertex

A

to the base

B C

is also the angle bisector of the angle formed by the two lines. How does this constrain the slope of

B C

Hint 2

You need the slopes of the angle bisectors of the lines

3 x + 4 y - 5 = 0

and

4 x - 3 y - 15 = 0

. Use the standard formula:

\frac{a _{1} x + b _{1} y + c _{1}}{a _{1}^{2} + b _{1}^{2}} = \pm \frac{a _{2} x + b _{2} y + c _{2}}{a _{2}^{2} + b _{2}^{2}}

. This will give you two distinct lines.

Hint 3

Once you find the slopes of the two angle bisectors, say

m_{1}

and

m_{2}

, remember that the line

B C

must be perpendicular to these bisectors. Therefore, the slopes of your required lines will be

- 1/ m_{1}

and

- 1/ m_{2}

. Finally, apply the point-slope form

y - 2 = m (x - 1)

for both cases.

Visualized Solution

Visualizing the Geometry

Given lines: $L_{1} : 3 x + 4 y - 5 = 0$ and $L_{2} : 4 x - 3 y - 15 = 0$
Intersection point is $A$
Condition: $A B = A C$ , making $△ A B C$ an isosceles triangle

The Isosceles Triangle Property

In $△ A B C$ , since $A B = A C$ , the line $B C$ must be perpendicular to the angle bisectors of the lines $L_{1}$ and $L_{2}$
There are two angle bisectors for any two intersecting lines

Equation of Angle Bisectors

Equation of angle bisectors: $\frac{3 x + 4 y - 5}{3 ^{2} + 4 ^{2}} = \pm \frac{4 x - 3 y - 15}{4 ^{2} + ( - 3 ) ^{2}}$

Simplifying the Bisector Equation

Simplifying the denominators: $3^{2} + 4^{2} = 5$ and $4^{2} + (- 3)^{2} = 5$
The equation simplifies to: $\frac{3 x + 4 y - 5}{5} = \pm \frac{4 x - 3 y - 15}{5}$

First Angle Bisector ( $B_{1}$ )

Taking the positive sign: $3 x + 4 y - 5 = 4 x - 3 y - 15$
Rearranging terms: $x - 7 y - 10 = 0$
Slope of first bisector ( $m_{1}$ ): $\frac{1}{7}$

Second Angle Bisector ( $B_{2}$ )

Taking the negative sign: $3 x + 4 y - 5 = - (4 x - 3 y - 15)$
Simplifying: $3 x + 4 y - 5 = - 4 x + 3 y + 15$
Rearranging terms: $7 x + y - 20 = 0$
Slope of second bisector ( $m_{2}$ ): $- 7$

Determining the Slopes of $B C$

Since $B C ⊥$ bisectors, the possible slopes ( $m_{B C}$ ) are:
Case 1: $m = - \frac{1}{1/7} = - 7$
Case 2: $m = - \frac{1}{- 7} = \frac{1}{7}$

First Possible Equation of $B C$

Using point-slope form: $y - y_{1} = m (x - x_{1})$ with $P (1, 2)$ and $m = - 7$
Equation: $y - 2 = - 7 (x - 1)$
Simplifying: $y - 2 = - 7 x + 7 ⟹ 7 x + y - 9 = 0$

Second Possible Equation of $B C$

Using point-slope form with $P (1, 2)$ and $m = \frac{1}{7}$
Equation: $y - 2 = \frac{1}{7} (x - 1)$
Simplifying: $7 y - 14 = x - 1 ⟹ x - 7 y + 13 = 0$

Final Conclusion

Final Equations of line $B C$ :
1. $7 x + y - 9 = 0$
2. $x - 7 y + 13 = 0$
Key Takeaway: The base of an isosceles triangle is always perpendicular to the angle bisector of the vertex angle.

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The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

Imagine you are standing at the intersection point

A

of two roads, represented by the lines

3 x + 4 y - 5 = 0

and

4 x - 3 y - 15 = 0

. You are tasked with drawing a third road,

B C

, such that the triangle formed,

△ A B C

, is perfectly isosceles with

A B = A C

In any isosceles triangle, the altitude from the vertex to the base is the axis of symmetry. This altitude is none other than the angle bisector of the vertex angle. Therefore, our line

B C

must be perpendicular to the angle bisectors of the given lines.

The Algebraic Engine

To find these bisectors, we use the standard locus definition: the set of points equidistant from two lines. We set up the equation:

\frac{3 x + 4 y - 5}{3 ^{2} + 4 ^{2}} = \pm \frac{4 x - 3 y - 15}{4 ^{2} + ( - 3 ) ^{2}}

Notice the beauty here? The denominators are both

25

, which is

5

. They cancel out instantly, leaving us with:

3 x + 4 y - 5 = \pm (4 x - 3 y - 15)

This is the moment where the complexity collapses into simplicity.

The Two Paths

Taking the positive sign, we get

3 x + 4 y - 5 = 4 x - 3 y - 15

, which simplifies to

x - 7 y - 10 = 0

. The slope of this bisector is

m_{1} = \frac{1}{7}

Taking the negative sign, we get

3 x + 4 y - 5 = - 4 x + 3 y + 15

, which simplifies to

7 x + y - 20 = 0

. The slope of this second bisector is

m_{2} = - 7

The Final Leap

We know that

B C

must be perpendicular to these bisectors. If

B C ⊥

bisector 1, its slope is the negative reciprocal of

\frac{1}{7}

, which is

- 7

B C ⊥

bisector 2, its slope is the negative reciprocal of

- 7

, which is

\frac{1}{7}

Now, we use the point-slope form

y - y_{1} = m (x - x_{1})

with the intersection point

A (1, 2)

For

m = - 7

, we get

y - 2 = - 7 (x - 1)

, leading to $7 x + y - 9 = 0$ .

For

m = \frac{1}{7}

, we get

y - 2 = \frac{1}{7} (x - 1)

, leading to $x - 7 y + 13 = 0$ .

You have just navigated the geometry of symmetry to find both possible lines. Remember, in JEE Advanced, always look for the geometric property before diving into the algebra.

Common Traps & FAQs

Why do we have two possible equations? Did I miss a condition?

You haven't missed anything! In coordinate geometry, when we talk about the 'angle bisector' of two intersecting lines, we are referring to both the internal and external bisectors. Since the problem doesn't specify which one, both are valid geometric solutions.

Is there a way to solve this without using the angle bisector property? It feels like a shortcut.

You could technically use the distance formula

A B = A C

by setting coordinates for

B

and

C

as variables, but that leads to a massive algebraic nightmare. The beauty of JEE problems lies in spotting the geometric symmetry—the angle bisector property is the elegant path, not just a shortcut.

Wait, the original lines

3 x + 4 y = 5

and

4 x - 3 y = 15

are perpendicular. Does that make the problem easier?

Absolutely! Since the product of their slopes is

(- \frac{3}{4}) \times (\frac{4}{3}) = - 1

, the lines are perpendicular. This means the angle bisectors are at

4 5^{\circ}

to the lines. It simplifies the math significantly, confirming that our bisector slopes will also be perpendicular to each other.

Animated Solution for Mathematics - Straight Lines: Straight lines $3 x + 4 y = 5$ and $4 x - 3 y = 15$ intersect at the point $A$ . Points $B$ and $C$ are chosen on these two lines such that $A B = A C$ . Determine the possible equations of the line $B C$ passing through the point $(1, 2)$ .

Visualized Solution

Visualizing the Geometry

The Isosceles Triangle Property

Equation of Angle Bisectors

Simplifying the Bisector Equation

First Angle Bisector ( $B_{1}$ )

Second Angle Bisector ( $B_{2}$ )

Determining the Slopes of $B C$

First Possible Equation of $B C$

Second Possible Equation of $B C$

Final Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Algebraic Engine

The Two Paths

The Final Leap

Similar Questions

Find the equation of the line passing through the point $(2, 3)$ and making intercept of length 2 units between the lines $y + 2 x = 3$ and $y + 2 x = 5$ .

The equation of one of the straight lines which passes through the point (1,3) and makes an angles $tan^{- 1} (2)$ with the straight line, $y + 1 = 32 x$ is

The combined equation of the two lines $a x + b y + c = 0$ and $a^{'} x + b^{'} y + c^{'} = 0$ can be written as $(a x + b y + c) (a^{'} x + b^{'} y + c^{'}) = 0$ . The equation of the angle bisectors of the lines represented by the equation $2 x^{2} + x y - 3 y^{2} = 0$ is

The two lines $x = a y + b, z = cy + d$ ; and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

Let $P QR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

Two equal sides of an isosceles triangle are given by the equations $7 x - y + 3 = 0$ and $x + y - 3 = 0$ and its third side passes through the point $(1, - 10)$ . Determine the equation of the third side.

The two lines $x = a y + b, z = cy + d$ and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ will be perpendicular, if and only if

Two sides of a rhombus $A B C D$ are parallel to the lines $y = x + 2$ and $y = 7 x + 3$ . If the diagonals of the rhombus intersect at the point $(1, 2)$ and the vertex $A$ is on the y-axis, find possible co-ordinates of $A$ .

Visualized Solution

Visualizing the Geometry

The Isosceles Triangle Property

Equation of Angle Bisectors

Simplifying the Bisector Equation

First Angle Bisector ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B1​)

Second Angle Bisector ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2​)

Determining the Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

First Possible Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Second Possible Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Final Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Algebraic Engine

The Two Paths

The Final Leap

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equation of the line passing through the point (2,3) and making intercept of length 2 units between the lines y+2x=3 and y+2x=5.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of one of the straight lines which passes through the point (1,3) and makes an angles tan−1(2​) with the straight line, y+1=32​x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The two lines x=ay+b,z=cy+d; and x=a′y+b′,z=c′y+d′ are perpendicular to each other if

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let PQR be a right angled isosceles triangle, right angled at P(2,1). If the equation of the line QR is 2x+y=3, then the equation representing the pair of lines PQ and PR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A(1,0),B(2,−1) and C(37​,34​) be three points. If the equation of the bisector of the angle ABC is αx+βy=5, then the value of α2+β2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The two lines x=ay+b,z=cy+d and x=a′y+b′,z=c′y+d′ will be perpendicular, if and only if

First Angle Bisector ( $B_{1}$ )

Second Angle Bisector ( $B_{2}$ )

Determining the Slopes of $B C$

First Possible Equation of $B C$

Second Possible Equation of $B C$

Find the equation of the line passing through the point $(2, 3)$ and making intercept of length 2 units between the lines $y + 2 x = 3$ and $y + 2 x = 5$ .

The equation of one of the straight lines which passes through the point (1,3) and makes an angles $tan^{- 1} (2)$ with the straight line, $y + 1 = 32 x$ is

The two lines $x = a y + b, z = cy + d$ ; and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

Let $P QR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

The two lines $x = a y + b, z = cy + d$ and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ will be perpendicular, if and only if