JEE Main 2019 (9 January)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: Let the equations of two sides of a triangle be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ . If the orthocentre of this triangle is at (1,1), then the equation of its third side is :

Q: Wait, why do we need to find the vertices $B$ and $C$ individually? Can't we just use the orthocenter property directly?

That is a great question! While the orthocenter $H$ is a powerful tool, it defines the altitudes, not the side $BC$ directly. To find the equation of the third side $BC$, we need two points on that line. Since we know $B$ lies on $AB$ and $C$ lies on $AC$, and both are connected to $H$ via altitudes, finding their coordinates is the most reliable and systematic way to define the line $BC$.

Q: I got a different slope for the third side. Did I make a mistake with the fractions?

Fractions can be tricky, especially with coordinates like $(\frac{35}{2}, -10)$ and $(-13, -\frac{33}{2})$. Double-check your subtraction in the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$. It is very easy to lose a negative sign when dealing with these values. Take a deep breath, re-calculate the slope, and ensure your final equation is in the form $Ax + By + C = 0$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the triangle

A B C

. You are given two sides,

A B

and

A C

, and the orthocenter

H (1, 1)

. Remember that the orthocenter is the intersection point of all altitudes. This means the altitude from vertex

C

to side

A B

must pass through

H

, and the altitude from vertex

B

to side

A C

must also pass through

H

Hint 2

Hint 2: The Arsenal. Use the property of perpendicular lines. If a line has a slope

m

, any line perpendicular to it has a slope of

- \frac{1}{m}

. Since the altitude from

C

is perpendicular to

A B

, find the slope of

A B

first, then determine the slope of the altitude, and use the point-slope form with

H (1, 1)

to find the equation of the altitude.

Hint 3

Hint 3: The Execution. Start by finding the slope of

A B

from

3 x - 2 y + 6 = 0

. The slope is

\frac{3}{2}

. The altitude from

C

A B

has a slope of

- \frac{2}{3}

and passes through

(1, 1)

. Its equation is

y - 1 = - \frac{2}{3} (x - 1)

, which simplifies to

2 x + 3 y - 5 = 0

. Now, find vertex

C

by intersecting this altitude with the given side

A C

(

4 x + 5 y - 20 = 0

Visualized Solution

Visualizing the Setup

Side $A B$ : $3 x - 2 y + 6 = 0$
Side $A C$ : $4 x + 5 y - 20 = 0$
Orthocenter $H$ : $(1, 1)$

The Orthocenter Strategy

Altitude from a vertex to the opposite side passes through the orthocenter $H (1, 1)$ .

Slope of Side $A B$

Equation of $A B$ : $3 x - 2 y + 6 = 0$
$2 y = 3 x + 6 ⟹ y = \frac{3}{2} x + 3$
Slope of $A B$ ( $m_{1}$ ) = $\frac{3}{2}$

Altitude from $C$ to $A B$

Altitude $h_{1} ⊥ A B$ , passes through $H (1, 1)$ .
Slope $m_{h 1} = - \frac{1}{m _{1}} = - \frac{2}{3}$
Equation: $y - 1 = - \frac{2}{3} (x - 1) ⟹ 2 x + 3 y - 5 = 0$

Finding Vertex $C$

Vertex $C$ is the intersection of $A C$ and $h_{1}$ .
$A C : 4 x + 5 y = 20$
$h_{1} : 2 x + 3 y = 5$

Coordinates of Vertex $C$

Multiply $h_{1}$ by 2: $4 x + 6 y = 10$
Subtract from $A C$ : $(4 x + 5 y) - (4 x + 6 y) = 20 - 10$
$- y = 10 ⟹ y = - 10$
Substitute $y = - 10$ : $2 x - 30 = 5 ⟹ x = \frac{35}{2}$
Vertex $C = (\frac{35}{2}, - 10)$

Slope of Side $A C$

Equation of $A C$ : $4 x + 5 y - 20 = 0$
$5 y = - 4 x + 20 ⟹ y = - \frac{4}{5} x + 4$
Slope of $A C$ ( $m_{2}$ ) = $- \frac{4}{5}$

Altitude from $B$ to $A C$

Altitude $h_{2} ⊥ A C$ , passes through $H (1, 1)$ .
Slope $m_{h 2} = - \frac{1}{m _{2}} = \frac{5}{4}$
Equation: $y - 1 = \frac{5}{4} (x - 1) ⟹ 5 x - 4 y - 1 = 0$

Finding Vertex $B$

Vertex $B$ is the intersection of $A B$ and $h_{2}$ .
$A B : 3 x - 2 y = - 6$
$h_{2} : 5 x - 4 y = 1$

Coordinates of Vertex $B$

Multiply $A B$ by 2: $6 x - 4 y = - 12$
Subtract $h_{2}$ : $(6 x - 4 y) - (5 x - 4 y) = - 12 - 1$
$x = - 13$
Substitute $x = - 13$ : $5 (- 13) - 4 y = 1 ⟹ y = - \frac{33}{2}$
Vertex $B = (- 13, - \frac{33}{2})$

Slope of the Third Side $B C$

Points $B (- 13, - \frac{33}{2})$ and $C (\frac{35}{2}, - 10)$ .
Slope $m_{B C} = \frac{- 10 - ( - \frac{33}{2} )}{\frac{35}{2} - ( - 13 )}$
$m_{B C} = \frac{13/2}{61/2} = \frac{13}{61}$

Final Equation of $B C$

Equation of $B C$ : $y - (- 10) = \frac{13}{61} (x - \frac{35}{2})$
$122 (y + 10) = 13 (2 x - 35)$
$122 y + 1220 = 26 x - 455$
$26 x - 122 y - 1675 = 0$

00:00 / 00:00

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Solution Diagram

Analyzing the Setup

We are given two sides of a triangle:

L_{1} : 3 x - 2 y + 6 = 0

and

L_{2} : 4 x + 5 y - 20 = 0

. We are also given the orthocenter

H (1, 1)

Our objective is to determine the equation of the third side,

B C

. The orthocenter is defined as the intersection point of the altitudes of a triangle.

The Altitude Hunt

Consider side

L_{1}

(

3 x - 2 y + 6 = 0

). Its slope is

m_{1} = \frac{3}{2}

The altitude from the opposite vertex to

L_{1}

must be perpendicular to

L_{1}

. Therefore, the slope of this altitude is

m_{h 1} = - \frac{2}{3}

Since this altitude passes through

H (1, 1)

, its equation is:

y - 1 = - \frac{2}{3} (x - 1)

2 x + 3 y - 5 = 0

Locating Vertex C

Vertex

C

is the intersection of side

L_{2}

(

4 x + 5 y - 20 = 0

) and the altitude

2 x + 3 y - 5 = 0

. Solving this system:

4 x + 5 y = 20

4 x + 6 y = 10

Subtracting the equations yields

y = - 10

. Substituting back, we find

x = \frac{35}{2}

. Thus,

C = (\frac{35}{2}, - 10)

Locating Vertex B

Now, consider side

L_{2}

(

4 x + 5 y - 20 = 0

). Its slope is

m_{2} = - \frac{4}{5}

The altitude from the opposite vertex to

L_{2}

must have a slope of

m_{h 2} = \frac{5}{4}

. Using

H (1, 1)

, the equation is:

y - 1 = \frac{5}{4} (x - 1)

5 x - 4 y - 1 = 0

Vertex

B

is the intersection of

L_{1}

(

3 x - 2 y + 6 = 0

) and

5 x - 4 y - 1 = 0

. Solving this system:

6 x - 4 y = - 12

5 x - 4 y = 1

Subtracting these gives

x = - 13

. Substituting back, we find

y = - \frac{33}{2}

. Thus,

B = (- 13, - \frac{33}{2})

Final Calculation

We now have the coordinates of

B (- 13, - \frac{33}{2})

and

C (\frac{35}{2}, - 10)

. The slope of line

B C

is:

m_{B C} = \frac{- 10 - ( - \frac{33}{2} )}{\frac{35}{2} - ( - 13 )} = \frac{\frac{13}{2}}{\frac{61}{2}} = \frac{13}{61}

Using the point-slope form with point

C

y + 10 = \frac{13}{61} (x - \frac{35}{2})

Multiplying by

122

to clear denominators:

122 y + 1220 = 26 x - 455

The final equation of the third side is $26 x - 122 y - 1675 = 0$ .

Common Traps & FAQs

Wait, why do we need to find the vertices

B

and

C

individually? Can't we just use the orthocenter property directly?

That is a great question! While the orthocenter

H

is a powerful tool, it defines the altitudes, not the side

B C

directly. To find the equation of the third side

B C

, we need two points on that line. Since we know

B

lies on

A B

and

C

lies on

A C

, and both are connected to

H

via altitudes, finding their coordinates is the most reliable and systematic way to define the line

B C

Why do we use the negative reciprocal for the slope of the altitude?

Remember that an altitude is, by definition, perpendicular to the opposite side. In coordinate geometry, if two lines are perpendicular, the product of their slopes is

- 1

. Thus, if the side has a slope

m

, the altitude must have a slope of

- \frac{1}{m}

. It is a fundamental geometric relationship that unlocks the entire problem!

I got a different slope for the third side. Did I make a mistake with the fractions?

Fractions can be tricky, especially with coordinates like

(\frac{35}{2}, - 10)

and

(- 13, - \frac{33}{2})

. Double-check your subtraction in the slope formula

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

. It is very easy to lose a negative sign when dealing with these values. Take a deep breath, re-calculate the slope, and ensure your final equation is in the form

A x + B y + C = 0

Animated Solution for Mathematics - Straight Lines: Let the equations of two sides of a triangle be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ . If the orthocentre of this triangle is at (1,1), then the equation of its third side is :

Select Answer:

Visualized Solution

Visualizing the Setup

The Orthocenter Strategy

Slope of Side $A B$

Altitude from $C$ to $A B$

Finding Vertex $C$

Coordinates of Vertex $C$

Slope of Side $A C$

Altitude from $B$ to $A C$

Finding Vertex $B$

Coordinates of Vertex $B$

Slope of the Third Side $B C$

Final Equation of $B C$

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Altitude Hunt

Locating Vertex C

Locating Vertex B

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Setup

The Orthocenter Strategy

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

Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

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

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

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

Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

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

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

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

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

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Altitude Hunt

Locating Vertex C

Locating Vertex B

Final Calculation

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Coordinates of Vertex $B$

Slope of the Third Side $B C$

Final Equation of $B C$

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The equations of the sides $A B, B C$ and $C A$ of a triangle $A B C$ are $2 x + y = 0$ , $x + p y = 15 a$ and $x - y = 3$ respectively. If its orthocentre is $(2, a)$ , $- \frac{1}{2} < a < 2$ , then $p$ is equal to

If the orthocentre of the triangle formed by the lines $y = x + 1, y = 4 x - 8$ and $y = m x + c$ is at $(3, - 1)$ , then $m - c$ is :

If $P (6, 1)$ be the orthocentre of the triangle whose vertices are $A (5, - 2)$ , $B (8, 3)$ and $C (h, k)$ , then the point $C$ lies on the circle:

Let k be an integer such that the triangle with vertices $(k, - 3 k)$ , $(5, k)$ and $(- k, 2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point:

The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2 y + 1 = 0$ and $2 x - y - 1 = 0$ and whose orthocenter is $(\frac{7}{3}, \frac{7}{3})$ is:

Orthocentre of triangle with vertices $(0, 0), (3, 4)$ and $(4, 0)$ is