A triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^\circ$, and $ar(\Delta ABC) = 5\sqrt{5}$ sq. units, then the abscissa of the vertex $C$ is : Options: (A) $1 + \sqrt{5}$ (B) $1 + 2\sqrt{5}$ (C) $2\sqrt{5} - 1$ (D) $2 + \sqrt{5}$

### Analyzing the Setup We are given two points $A(1, 2)$ and $B(3, 1)$ in the Cartesian plane. We seek the third vertex $C$ of a right-angled triangle $ABC$ such that $\angle BAC = 90^{\circ}$ and $C$ lies in the first quadrant. First, we determine the slope of the line segment $AB$, denoted as $m_{AB}$: $$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 2}{3 - 1} = -\frac{1}{2}$$ Since $AC$ is perpendicular to $AB$, the slope of $AC$ ($m_{AC}$) must be the negative reciprocal of $m_{AB}$. Therefore: $$m_{AC} = 2$$ ### The Area Constraint The area of the right-angled triangle $ABC$ is given as $5\sqrt{5}$. The formula for the area is $\frac{1}{2} \times \text{base} \times \text{height}$, where the base is $AB$ and the height is $AC$. We calculate the length of $AB$ using the distance formula: $$AB = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{5}$$ Substituting the known values into the area equation: $$\frac{1}{2} \cdot \sqrt{5} \cdot AC = 5\sqrt{5}$$ By canceling $\sqrt{5}$ from both sides and multiplying by $2$, we find the length of $AC$: $$AC = 10$$ ### The Parametric Power We now have the starting point $A(1, 2)$, the distance $r = 10$, and the slope $m_{AC} = 2$. We use the parametric form of a line to find the coordinates of $C$. Given $\tan \theta = 2$, we construct a right triangle with an opposite side of $2$ and an adjacent side of $1$. The hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$. Thus, $\cos \theta = \frac{1}{\sqrt{5}}$. The abscissa $x_C$ is calculated as follows: $$x_C = x_A + r \cos \theta$$ $$x_C = 1 + 10 \cdot \left( \frac{1}{\sqrt{5}} \right)$$ Simplifying the expression, we obtain: $$x_C = 1 + \frac{10}{\sqrt{5}} = 1 + 2\sqrt{5}$$ The final abscissa of vertex $C$ is **$1 + 2\sqrt{5}$**.

JEE Main 2020 - 4 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: A triangle $A B C$ lying in the first quadrant has two vertices as $A (1, 2)$ and $B (3, 1)$ . If $∠ B A C = 9 0^{\circ}$ , and $a r (Δ A B C) = 55$ sq. units, then the abscissa of the vertex $C$ is :

Q: Why is the slope of $AC$ equal to $2$?

Since $AC \perp AB$, the product of their slopes must be $-1$. Given $m_{AB} = \frac{1 - 2}{3 - 1} = -\frac{1}{2}$, we have $m_{AC} \cdot (-\frac{1}{2}) = -1$, which gives $m_{AC} = 2$. It is a fundamental property of perpendicular lines in the Cartesian plane.

Q: Why can't we just use the distance formula for $C$?

You could, but it would involve two variables $(x_C, y_C)$ and two equations (slope and distance), leading to a system of equations. The parametric form is much more elegant and direct for finding coordinates at a specific distance along a line.

Select Answer:

Visualized Solution

Visualizing the Geometry

Given vertices: $A (1, 2)$ and $B (3, 1)$
Triangle $A B C$ lies in the first quadrant.

Right Angle Condition

Condition: $∠ B A C = 9 0^{\circ}$
This means $A C ⊥ A B$ .

Slope of $A B$

Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
We need to find the slope of line segment $A B$ .

Computing Slope of $A B$

$m_{A B} = \frac{1 - 2}{3 - 1}$
$m_{A B} = - \frac{1}{2}$

Perpendicularity Condition

Since $A C ⊥ A B$ , their slopes multiply to $- 1$ .
$m_{A C} \cdot m_{A B} = - 1$

Computing Slope of $A C$

$m_{A C} \cdot (- \frac{1}{2}) = - 1$
$m_{A C} = 2$

Distance Formula for $A B$

$A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
We need the base length to use the area formula.

Computing Length of $A B$

$A B = (3 - 1)^{2} + (1 - 2)^{2}$
$A B = 2^{2} + (- 1)^{2} = 5$

Area of Triangle $A B C$

Area of $Δ A B C = \frac{1}{2} \cdot A B \cdot A C$
Given Area = $55$ sq. units.

Computing Length of $A C$

$\frac{1}{2} \cdot 5 \cdot A C = 55$
$A C = 10$

Parametric Form of a Line

Coordinates of $C = (x_{A} + r cos θ, y_{A} + r sin θ)$
Here, distance $r = A C = 10$ .

Trigonometric Ratios

We know $tan θ = m_{A C} = 2$ .
From a right triangle with sides $2$ and $1$ , hypotenuse is $5$ .

Finding $cos θ$

$cos θ = \frac{1}{5}$
Positive value is taken because $C$ is in the first quadrant.

Calculating the Abscissa

Abscissa of $C$ ( $x_{C}$ ) = $x_{A} + r cos θ$
$x_{C} = 1 + 10 \cdot (\frac{1}{5})$

Final Simplification

$x_{C} = 1 + \frac{10}{5}$
$x_{C} = 1 + 25$

00:00 / 00:00

The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Setup

We are given two points

A (1, 2)

and

B (3, 1)

in the Cartesian plane. We seek the third vertex

C

of a right-angled triangle

A B C

such that

∠ B A C = 9 0^{\circ}

and

C

lies in the first quadrant.

First, we determine the slope of the line segment

A B

, denoted as

m_{A B}

m_{A B} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{1 - 2}{3 - 1} = - \frac{1}{2}

Since

A C

is perpendicular to

A B

, the slope of

A C

(

m_{A C}

) must be the negative reciprocal of

m_{A B}

. Therefore:

m_{A C} = 2

The Area Constraint

The area of the right-angled triangle

A B C

is given as

55

. The formula for the area is

\frac{1}{2} \times base \times height

, where the base is

A B

and the height is

A C

We calculate the length of

A B

using the distance formula:

A B = (3 - 1)^{2} + (1 - 2)^{2} = 2^{2} + (- 1)^{2} = 5

Substituting the known values into the area equation:

\frac{1}{2} \cdot 5 \cdot A C = 55

By canceling

5

from both sides and multiplying by

2

, we find the length of

A C

A C = 10

The Parametric Power

We now have the starting point

A (1, 2)

, the distance

r = 10

, and the slope

m_{A C} = 2

. We use the parametric form of a line to find the coordinates of

C

Given

tan θ = 2

, we construct a right triangle with an opposite side of

2

and an adjacent side of

1

. The hypotenuse is

2^{2} + 1^{2} = 5

. Thus,

cos θ = \frac{1}{5}

The abscissa

x_{C}

is calculated as follows:

x_{C} = x_{A} + r cos θ

x_{C} = 1 + 10 \cdot (\frac{1}{5})

Simplifying the expression, we obtain:

x_{C} = 1 + \frac{10}{5} = 1 + 25

The final abscissa of vertex

C

is $1 + 25$ .

Common Traps & FAQs

Why is the slope of

A C

equal to

2

Since

A C ⊥ A B

, the product of their slopes must be

- 1

. Given

m_{A B} = \frac{1 - 2}{3 - 1} = - \frac{1}{2}

, we have

m_{A C} \cdot (- \frac{1}{2}) = - 1

, which gives

m_{A C} = 2

. It is a fundamental property of perpendicular lines in the Cartesian plane.

Why can't we just use the distance formula for

C

You could, but it would involve two variables

(x_{C}, y_{C})

and two equations (slope and distance), leading to a system of equations. The parametric form is much more elegant and direct for finding coordinates at a specific distance along a line.

Why do we take the positive value for

cos θ

The problem states the triangle lies in the first quadrant. Since

A (1, 2)

and

C

must be in the first quadrant, and the slope is positive, the point

C

must move in the positive

x

direction from

A

to remain within the first quadrant.

Animated Solution for Mathematics - Straight Lines: A triangle $A B C$ lying in the first quadrant has two vertices as $A (1, 2)$ and $B (3, 1)$ . If $∠ B A C = 9 0^{\circ}$ , and $a r (Δ A B C) = 55$ sq. units, then the abscissa of the vertex $C$ is :

Select Answer:

Visualized Solution

Visualizing the Geometry

Right Angle Condition

Slope of $A B$

Computing Slope of $A B$

Perpendicularity Condition

Computing Slope of $A C$

Distance Formula for $A B$

Computing Length of $A B$

Area of Triangle $A B C$

Computing Length of $A C$

Parametric Form of a Line

Trigonometric Ratios

Finding $cos θ$

Calculating the Abscissa

Final Simplification

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Area Constraint

The Parametric Power

Similar Questions

Let $A B C$ be an isosceles triangle in which $A$ is at $(- 1, 0)$ , $∠ A = \frac{2 π}{3}$ , $A B = A C$ and $B$ is on the positive $x$ -axis. If $B C = 43$ and the line $B C$ intersects the line $y = x + 3$ at $(α, β)$ , then $\frac{β ^{4}}{α ^{2}}$ is :

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

Let the area of the triangle formed by a straight Line $L : x + b y + c = 0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line $L$ makes an angle of $4 5^{\circ}$ with the positive $x$ -axis, then the value of $b^{2} + c^{2}$ is:

The vertices of a triangle are $A (- 1, - 7), B (5, 1)$ and $C (1, 4)$ . The equation of the bisector of the angle $∠ A B C$ is .........

In a $△ A B C$ , suppose $y = x$ is the equation of the bisector of the angle $B$ and the equation of the side $A C$ is $2 x - y = 2$ . If $2 A B = B C$ and the point $A$ and $B$ are respectively $(4, 6)$ and $(α, β)$ , then $α + 2 β$ is equal to

Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2 x = 2$ such that $Δ A B C$ is an equilateral triangle. Then, the area of the $Δ A B C$ is

A ray of light coming from the point $(2, 23)$ is incident at an angle $3 0^{\circ}$ on the line $x = 1$ at the point $A$ . The ray gets reflected on the line $x = 1$ and meets $x$ -axis at the point $B$ . Then, the line $A B$ passes through the point:

Let a point A lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $A B C$ , where the points $B$ and $C$ lie on the lines $L_{1}$ and $L_{2}$ , respectively, is :

Select Answer:

Visualized Solution

Visualizing the Geometry

Right Angle Condition

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

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

Perpendicularity Condition

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

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Computing Length of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

Computing Length of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Parametric Form of a Line

Trigonometric Ratios

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθ

Calculating the Abscissa

Final Simplification

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Area Constraint

The Parametric Power

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. Then the equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. The equation of the bisector of the angle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vertices of a triangle are A(−1,−7),B(5,1) and C(1,4). The equation of the bisector of the angle ∠ABC is .........

Slope of $A B$

Computing Slope of $A B$

Computing Slope of $A C$

Distance Formula for $A B$

Computing Length of $A B$

Area of Triangle $A B C$

Computing Length of $A C$

Finding $cos θ$

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

The vertices of a triangle are $A (- 1, - 7), B (5, 1)$ and $C (1, 4)$ . The equation of the bisector of the angle $∠ A B C$ is .........