Can I use the property that the orthocenter of $\triangle ABC$ is $H$ implies that $A$ is the orthocenter of $\triangle HBC$?

Absolutely! That is a brilliant observation. The orthocenter relationship is cyclic. If $H$ is the orthocenter of $\triangle ABC$, then $A$ is the orthocenter of $\triangle HBC$, $B$ is the orthocenter of $\triangle HAC$, and $C$ is the orthocenter of $\triangle HAB$. You could solve it that way, but it might involve more steps than the direct slope method we used here.

JEE Main 2019 (10 January)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Two vertices of a triangle are $(0, 2)$ and $(4, 3)$ . If its orthocentre is at the origin, then its third vertex lies in which quadrant ?

Q: Why do we use the orthocenter property $BH \perp AC$ instead of just finding the equation of the line $AB$?

Great question! While finding the equation of $AB$ is useful, it doesn't give us the coordinates of $C$ directly. The orthocenter is the 'meeting point' of all altitudes. By using the perpendicularity of $BH$ and $AC$, we create a system of two linear equations with two variables ($a$ and $b$), which is the most direct path to finding the unknown vertex.

Q: Why did we set the slope of $CH$ as $b/a$?

Because the orthocenter $H$ is at the origin $(0, 0)$ and the vertex $C$ is at $(a, b)$. The slope formula is $(y_2 - y_1) / (x_2 - x_1)$. So, $(b - 0) / (a - 0)$ simplifies perfectly to $b/a$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The orthocenter is the intersection of altitudes. If you know two vertices and the orthocenter, you are essentially looking for the point where the lines perpendicular to the sides meet. Think about the definition of an altitude: it is a line passing through a vertex and perpendicular to the opposite side.

Hint 2

Hint 2: Recall the condition for perpendicular lines in the Cartesian plane. If two lines with slopes

m_{1}

and

m_{2}

are perpendicular, their product must be

- 1

. You have two pairs of perpendicular lines here: the altitude from

C

A B

and the altitude from

B

A C

Hint 3

Hint 3: Start by defining the third vertex as

C (a, b)

. Express the slope of

A B

and the slope of the altitude

C H

(which passes through the origin). Use the condition

m_{C H} \cdot m_{A B} = - 1

to establish your first linear relationship between

a

and

b

Visualized Solution

Plotting the Given Points

Given vertices: $A (0, 2)$ and $B (4, 3)$
Orthocenter: $H (0, 0)$ (Origin)

The Unknown Vertex

Let the third vertex be $C (a, b)$
We need to find the coordinates of $C$ to determine its quadrant.

Orthocenter Property

The orthocenter $H$ is the intersection point of the triangle's altitudes.
Therefore, $C H ⊥ A B$ and $B H ⊥ A C$ .

Slope of Line $A B$

Slope formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Slope of $A B$ ( $m_{A B}$ ) $= \frac{3 - 2}{4 - 0} = \frac{1}{4}$

Slope of Altitude $C H$

Slope of $C H$ ( $m_{C H}$ ) $= \frac{b - 0}{a - 0} = \frac{b}{a}$

First Perpendicularity Condition

Since $C H ⊥ A B$ , $m_{C H} \cdot m_{A B} = - 1$
$\frac{b}{a} \cdot \frac{1}{4} = - 1$
$b = - 4 a$ --- (Equation 1)

Slope of Altitude $B H$

Slope of $B H$ ( $m_{B H}$ ) $= \frac{3 - 0}{4 - 0} = \frac{3}{4}$

Slope of Line $A C$

Slope of $A C$ ( $m_{A C}$ ) $= \frac{b - 2}{a - 0} = \frac{b - 2}{a}$

Second Perpendicularity Condition

Since $B H ⊥ A C$ , $m_{B H} \cdot m_{A C} = - 1$
$\frac{3}{4} \cdot \frac{b - 2}{a} = - 1$
$3 (b - 2) = - 4 a ⟹ 3 b - 6 = - 4 a$ --- (Equation 2)

Solving for Coordinate ' $a$ '

Substitute $b = - 4 a$ into Equation 2:
$3 (- 4 a) - 6 = - 4 a$
$- 12 a - 6 = - 4 a ⟹ - 8 a = 6$
$a = - \frac{3}{4}$

Solving for Coordinate ' $b$ '

Substitute $a = - \frac{3}{4}$ into Equation 1:
$b = - 4 (- \frac{3}{4}) = 3$
The third vertex is $C (- \frac{3}{4}, 3)$ .

Identifying the Quadrant

Vertex $C$ is $(- \frac{3}{4}, 3)$ .
Since $x < 0$ and $y > 0$ , vertex $C$ lies in the Second Quadrant.

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

We are given two vertices of a triangle,

A (0, 2)

and

B (4, 3)

, with the orthocenter

H

located at the origin

(0, 0)

. We seek the coordinates of the third vertex

C (a, b)

The orthocenter is the intersection of the triangle's altitudes. By definition, an altitude from a vertex is perpendicular to the opposite side. Therefore,

C H ⊥ A B

and

B H ⊥ A C

The Power of Perpendicularity

First, we calculate the slope of side

A B

using the formula

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

m_{A B} = \frac{3 - 2}{4 - 0} = \frac{1}{4}

The altitude

C H

passes through

C (a, b)

and

H (0, 0)

, so its slope is

m_{C H} = \frac{b}{a}

. Since

C H ⊥ A B

, the product of their slopes must be

- 1

\frac{b}{a} \cdot \frac{1}{4} = - 1 \Rightarrow b = - 4 a

The Second Bridge

Next, we examine the altitude

B H

, which connects

B (4, 3)

H (0, 0)

. Its slope is:

m_{B H} = \frac{3 - 0}{4 - 0} = \frac{3}{4}

This altitude is perpendicular to side

A C

. The slope of

A C

m_{A C} = \frac{b - 2}{a - 0} = \frac{b - 2}{a}

. Applying the perpendicularity condition

m_{B H} \cdot m_{A C} = - 1

\frac{3}{4} \cdot \frac{b - 2}{a} = - 1

This simplifies to the following linear relationship:

3 (b - 2) = - 4 a \Rightarrow 3 b - 6 = - 4 a

The Final Convergence

We now solve the system of two linear equations:

b = - 4 a

3 b - 6 = - 4 a

Substituting the first equation into the second yields:

3 (- 4 a) - 6 = - 4 a

- 12 a - 6 = - 4 a

- 8 a = 6 \Rightarrow a = - \frac{3}{4}

Using

b = - 4 a

, we find the value of

b

b = - 4 (- \frac{3}{4}) = 3

The third vertex

C

is located at

(- \frac{3}{4}, 3)

. Since the x-coordinate is negative and the y-coordinate is positive, the vertex

C

lies in the Second Quadrant.

Common Traps & FAQs

Why do we use the orthocenter property

B H ⊥ A C

instead of just finding the equation of the line

A B

Great question! While finding the equation of

A B

is useful, it doesn't give us the coordinates of

C

directly. The orthocenter is the 'meeting point' of all altitudes. By using the perpendicularity of

B H

and

A C

, we create a system of two linear equations with two variables (

a

and

b

), which is the most direct path to finding the unknown vertex.

Can I use the property that the orthocenter of

△ A B C

H

implies that

A

is the orthocenter of

△ H B C

Absolutely! That is a brilliant observation. The orthocenter relationship is cyclic. If

H

is the orthocenter of

△ A B C

, then

A

is the orthocenter of

△ H B C

B

is the orthocenter of

△ H A C

, and

C

is the orthocenter of

△ H A B

. You could solve it that way, but it might involve more steps than the direct slope method we used here.

Is it possible for the orthocenter to lie outside the triangle?

Yes, absolutely! In an obtuse-angled triangle, the orthocenter lies outside the triangle. In this specific problem, we don't need to worry about where it lies relative to the triangle's interior; we only need the algebraic definition of the orthocenter as the intersection of altitudes to find the coordinates.

Why did we set the slope of

C H

b / a

Because the orthocenter

H

is at the origin

(0, 0)

and the vertex

C

is at

(a, b)

. The slope formula is

(y_{2} - y_{1}) / (x_{2} - x_{1})

. So,

(b - 0) / (a - 0)

simplifies perfectly to

b / a

What if the slope of

A B

was zero or undefined?

If the slope were zero, the line would be horizontal, and the altitude would be a vertical line (undefined slope). If the slope were undefined (a vertical line), the altitude would be horizontal. Our slope product formula

m_{1} \cdot m_{2} = - 1

is a general tool, but always keep an eye on those special cases where lines are parallel to the axes!

Animated Solution for Mathematics - Straight Lines: Two vertices of a triangle are $(0, 2)$ and $(4, 3)$ . If its orthocentre is at the origin, then its third vertex lies in which quadrant ?

Select Answer:

Visualized Solution

Plotting the Given Points

The Unknown Vertex

Orthocenter Property

Slope of Line $A B$

Slope of Altitude $C H$

First Perpendicularity Condition

Slope of Altitude $B H$

Slope of Line $A C$

Second Perpendicularity Condition

Solving for Coordinate ' $a$ '

Solving for Coordinate ' $b$ '

Identifying the Quadrant

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

Analyzing the Setup

The Power of Perpendicularity

The Second Bridge

The Final Convergence

Similar Questions

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Select Answer:

Visualized Solution

Plotting the Given Points

The Unknown Vertex

Orthocenter Property

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

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

First Perpendicularity Condition

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

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

Second Perpendicularity Condition

Solving for Coordinate ' .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a'

Solving for Coordinate ' .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b'

Identifying the Quadrant

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

Analyzing the Setup

The Power of Perpendicularity

The Second Bridge

The Final Convergence

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