JEE Advanced 1985

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Animated Solution for Mathematics - Straight Lines: 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 $A$ .

Q: Why is the slope of $AP$ the same as the slope of the diagonal?

Because $A$ is a vertex and $P$ is the intersection of the diagonals. By definition, the segment $AP$ is a part of the diagonal line. Therefore, the slope of the segment $AP$ must be identical to the slope of the entire diagonal line passing through $P$. You are simply using a segment of the line to define the line itself.

Visualized Solution

Visualizing the Rhombus Geometry

Diagonals of rhombus $A B C D$ intersect at $P (1, 2)$ .
Vertex $A$ lies on the y-axis, so $A = (0, a)$ .

Slopes of the Rhombus Sides

Sides are parallel to $y = x + 2$ and $y = 7 x + 3$ .
Slope of first side: $m_{1} = 1$ .
Slope of second side: $m_{2} = 7$ .

The Angle Bisector Property

Key Property: In a rhombus, the diagonals bisect the interior angles.
Therefore, the diagonals are parallel to the angle bisectors of the sides.

Angle Bisector Formula

Let $M$ be the slope of a diagonal.
Angle between diagonal and side 1 = Angle between diagonal and side 2.
$\frac{M - m _{1}}{1 + M m _{1}} = \frac{M - m _{2}}{1 + M m _{2}}$
Substituting $m_{1} = 1, m_{2} = 7$ :
$\frac{M - 1}{1 + M} = \frac{M - 7}{1 + 7 M}$

Solving the Modulus (Positive Case)

Case 1: $\frac{M - 1}{1 + M} = \frac{M - 7}{1 + 7 M}$
Cross-multiplying: $(M - 1) (1 + 7 M) = (M - 7) (1 + M)$
$7 M^{2} - 6 M - 1 = M^{2} - 6 M - 7$
$6 M^{2} = - 6 ⟹ M^{2} = - 1$
No real solution for $M$ .

Solving the Modulus (Negative Case)

Case 2: $\frac{M - 1}{1 + M} = - (\frac{M - 7}{1 + 7 M})$
Cross-multiplying: $(M - 1) (1 + 7 M) = - (M - 7) (1 + M)$
$7 M^{2} - 6 M - 1 = - (M^{2} - 6 M - 7)$
$7 M^{2} - 6 M - 1 = - M^{2} + 6 M + 7$

Finding Diagonal Slopes

Bringing all terms to one side: $8 M^{2} - 12 M - 8 = 0$
Dividing by 4: $2 M^{2} - 3 M - 2 = 0$
Factorizing: $(2 M + 1) (M - 2) = 0$
Possible slopes: $M = 2$ or $M = - \frac{1}{2}$

Slope of Diagonal $A P$

We know vertex $A$ is $(0, a)$ and center $P$ is $(1, 2)$ .
The line segment $A P$ lies on one of the diagonals.
Slope of $A P = \frac{2 - a}{1 - 0} = 2 - a$

Equating Slopes (Case 1)

Equating the slope of $A P$ to the first possible diagonal slope:
$2 - a = 2$
$a = 0$
First possible coordinate for $A$ is $(0, 0)$ .

Equating Slopes (Case 2)

Equating the slope of $A P$ to the second possible diagonal slope:
$2 - a = - \frac{1}{2}$
$a = 2 + \frac{1}{2} = \frac{5}{2}$
Second possible coordinate for $A$ is $(0, \frac{5}{2})$ .

Final Coordinates of Vertex $A$

The possible coordinates for vertex $A$ are $(0, 0)$ or $(0, \frac{5}{2})$ .
Key Takeaway: The diagonals of a rhombus are the angle bisectors of its sides.

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The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

We are given a rhombus

A B C D

where two sides are parallel to the lines

y = x + 2

and

y = 7 x + 3

. The diagonals of this rhombus intersect at the point

P (1, 2)

Vertex

A

lies on the y-axis, meaning its coordinates can be expressed as

A (0, a)

. Our objective is to determine the possible values of

a

From the given lines, we identify the slopes of the sides as

m_{1} = 1

and

m_{2} = 7

. These slopes define the orientation of the rhombus.

The Master Equation

In any rhombus, the diagonals act as the angle bisectors of the interior angles. Let

M

represent the slope of a diagonal.

Using the angle bisector formula, the angle between the diagonal and the sides must be equal, leading to:

\frac{M - m _{1}}{1 + M m _{1}} = \frac{M - m _{2}}{1 + M m _{2}}

Substituting our known slopes

m_{1} = 1

and

m_{2} = 7

, we obtain:

\frac{M - 1}{1 + M} = \frac{M - 7}{1 + 7 M}

Solving for Diagonal Slopes

We must consider two cases based on the modulus.

Case 1: The positive sign Equating the expressions directly:

(M - 1) (1 + 7 M) = (M - 7) (1 + M)

7 M^{2} - 6 M - 1 = M^{2} - 6 M - 7

6 M^{2} = - 6 \Rightarrow M^{2} = - 1

This yields no real solution, indicating this specific bisector does not exist in the real plane.

Case 2: The negative sign Equating with a negative sign:

\frac{M - 1}{1 + M} = - (\frac{M - 7}{1 + 7 M})

(M - 1) (1 + 7 M) = - (M - 7) (1 + M)

7 M^{2} - 6 M - 1 = - M^{2} + 6 M + 7

Rearranging the terms results in the quadratic equation:

8 M^{2} - 12 M - 8 = 0

Dividing by 4, we get:

2 M^{2} - 3 M - 2 = 0

Factoring the quadratic:

(2 M + 1) (M - 2) = 0

The possible slopes for the diagonals are

M = 2

and

M = - \frac{1}{2}

. Note that their product is

- 1

, confirming the diagonals are perpendicular.

Final Calculation

We know vertex

A

(0, a)

and the intersection point

P

(1, 2)

. The slope of the diagonal passing through

A

is given by:

Slope = \frac{2 - a}{1 - 0} = 2 - a

Equating this to our calculated slopes:

1. For

M = 2

2 - a = 2 \Rightarrow a = 0

2. For

M = - \frac{1}{2}

2 - a = - \frac{1}{2} \Rightarrow a = 2 + \frac{1}{2} = \frac{5}{2}

The possible coordinates for vertex

A

are $(0, 0)$ and $(0, \frac{5}{2})$ .

Common Traps & FAQs

Wait, why are the diagonals angle bisectors? Can't I just use the midpoint formula?

The midpoint formula tells you where they meet, but it doesn't tell you their direction. In a rhombus, all four sides are equal in length. This geometric constraint forces the triangles formed by the diagonals to be congruent, which mathematically necessitates that the diagonals bisect the interior angles. It is a fundamental property of the rhombus, not just a coincidence!

Why did the first case of the modulus equation give

M^{2} = - 1

? Did I make a mistake?

You didn't make a mistake! That is the beauty of the algebra. The equation

\frac{M - 1}{1 + M} = \frac{M - 7}{1 + 7 M}

represents two lines: the internal angle bisector and the external angle bisector. The

M^{2} = - 1

result simply means that for this specific configuration, one of those bisectors is vertical or undefined in a way that doesn't yield a real slope in the standard Cartesian plane. Trust the algebra; it is guiding you to the valid solutions.

Why is the slope of

A P

the same as the slope of the diagonal?

Because

A

is a vertex and

P

is the intersection of the diagonals. By definition, the segment

A P

is a part of the diagonal line. Therefore, the slope of the segment

A P

must be identical to the slope of the entire diagonal line passing through

P

. You are simply using a segment of the line to define the line itself.

Animated Solution for Mathematics - Straight Lines: 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 Rhombus Geometry

Slopes of the Rhombus Sides

The Angle Bisector Property

Angle Bisector Formula

Solving the Modulus (Positive Case)

Solving the Modulus (Negative Case)

Finding Diagonal Slopes

Slope of Diagonal $A P$

Equating Slopes (Case 1)

Equating Slopes (Case 2)

Final Coordinates of Vertex $A$

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Master Equation

Solving for Diagonal Slopes

Final Calculation

Similar Questions

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Visualized Solution

Visualizing the Rhombus Geometry

Slopes of the Rhombus Sides

The Angle Bisector Property

Angle Bisector Formula

Solving the Modulus (Positive Case)

Solving the Modulus (Negative Case)

Finding Diagonal Slopes

Slope of Diagonal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AP

Equating Slopes (Case 1)

Equating Slopes (Case 2)

Final Coordinates of Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Master Equation

Solving for Diagonal Slopes

Final Calculation

Similar Questions

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