The vertices of a triangle are (2,1), (5,2), and (4,4). Find the lengths of the perpendiculars from these vertices to their opposite sides. [IIT 1962]

Triangle Perpendiculars Problem Solution

Problem Statement

The vertices of a triangle are $(2,1)$, $(5,2)$, and $(4,4)$. Find the lengths of the perpendiculars from these vertices to their opposite sides. [IIT 1962]

Options:

a) $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{6}}$ b) $\frac{7}{\sqrt{6}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{10}}$ c) $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{15}}$ d) $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}}$

Detailed Solution

We need to find the lengths of the perpendiculars (altitudes) from each vertex to the line containing the opposite side in a triangle with vertices $A(2,1)$, $B(5,2)$, and $C(4,4)$. This is a coordinate geometry problem, and we’ll use the perpendicular distance formula. Let’s dive in!

Step 1: Understand the Problem

For a triangle $ABC$: - Perpendicular from $A(2,1)$ to side $BC$. - Perpendicular from $B(5,2)$ to side $AC$. - Perpendicular from $C(4,4)$ to side $AB$.

We’ll find the equations of lines $BC$, $AC$, and $AB$, then calculate the perpendicular distances from the opposite vertices.

Step 2: Formula for Perpendicular Distance

The distance from a point $(x_0,y_0)$ to a line $ax+by+c=0$ is:

$$\text{Distance}=\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$$

We’ll use this for each calculation.

Step 3: Perpendicular from $A(2,1)$ to $BC$

Find the equation of line $BC$:

Points: $B(5,2)$, $C(4,4)$.

Slope of $BC$:

$$m_{BC}=\frac{4-2}{4-5}=\frac{2}{-1}=-2$$

Using point-slope form with $B(5,2)$:

$$y-2=-2(x-5)$$ $$y-2=-2x+10$$ $$y=-2x+12$$

Convert to standard form:

$$2x+y-12=0$$

Verify: For $C(4,4)$: $2(4)+4-12=8+4-12=0$. It holds.

Distance from $A(2,1)$ to $2x+y-12=0$:

Here, $a=2$, $b=1$, $c=-12$, $(x_0,y_0)=(2,1)$.

$$\text{Distance}=\frac{|2\cdot 2+1\cdot 1-12|}{\sqrt{2^2+1^2}}=\frac{|4+1-12|}{\sqrt{4+1}}=\frac{|-7|}{\sqrt{5}}=\frac{7}{\sqrt{5}}$$

So, the perpendicular from $A$ to $BC$ is $\frac{7}{\sqrt{5}}$.

Step 4: Perpendicular from $B(5,2)$ to $AC$

Find the equation of line $AC$:

Points: $A(2,1)$, $C(4,4)$.

Slope of $AC$:

$$m_{AC}=\frac{4-1}{4-2}=\frac{3}{2}$$

Using point-slope form with $A(2,1)$:

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

Standard form (multiply by 2 to clear fraction):

$$3x-2y-4=0$$

Verify: For $C(4,4)$: $3(4)-2(4)-4=12-8-4=0$. Correct.

Distance from $B(5,2)$ to $3x-2y-4=0$:

Here, $a=3$, $b=-2$, $c=-4$, $(x_0,y_0)=(5,2)$.

$$\text{Distance}=\frac{|3\cdot 5+(-2)\cdot 2-4|}{\sqrt{3^2+(-2{)}^2}}=\frac{|15-4-4|}{\sqrt{9+4}}=\frac{|7|}{\sqrt{13}}=\frac{7}{\sqrt{13}}$$

So, the perpendicular from $B$ to $AC$ is $\frac{7}{\sqrt{13}}$.

Step 5: Perpendicular from $C(4,4)$ to $AB$

Find the equation of line $AB$:

Points: $A(2,1)$, $B(5,2)$.

Slope of $AB$:

$$m_{AB}=\frac{2-1}{5-2}=\frac{1}{3}$$

Using point-slope form with $A(2,1)$:

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

Standard form (multiply by 3):

$$x-3y+1=0$$

Verify: For $B(5,2)$: $5-3(2)+1=5-6+1=0$. Correct.

Distance from $C(4,4)$ to $x-3y+1=0$:

Here, $a=1$, $b=-3$, $c=1$, $(x_0,y_0)=(4,4)$.

$$\text{Distance}=\frac{|1\cdot 4+(-3)\cdot 4+1|}{\sqrt{1^2+(-3{)}^2}}=\frac{|4-12+1|}{\sqrt{1+9}}=\frac{|-7|}{\sqrt{10}}=\frac{7}{\sqrt{10}}$$

So, the perpendicular from $C$ to $AB$ is $\frac{7}{\sqrt{10}}$.

Step 6: Verification with Area (Optional Check)

Let’s confirm using the triangle’s area. Area formula with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$:

$$\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$$

For $A(2,1)$, $B(5,2)$, $C(4,4)$:

$$\text{Area}=\frac{1}{2}|2(2-4)+5(4-1)+4(1-2)|=\frac{1}{2}|2(-2)+5(3)+4(-1)|=\frac{1}{2}|-4+15-4|=\frac{1}{2}\cdot 7=\frac{7}{2}$$

Area = $\frac{7}{2}$. Now, $\text{Area}=\frac{1}{2}\cdot \text{base}\cdot \text{height}$:

• Base $BC=\sqrt{(5-4{)}^2+(2-4{)}^2}=\sqrt{5}$, height = $\frac{7}{\sqrt{5}}$:
$$\frac{1}{2}\cdot \sqrt{5}\cdot \frac{7}{\sqrt{5}}=\frac{7}{2}$$
• Base $AC=\sqrt{(2-4{)}^2+(1-4{)}^2}=\sqrt{13}$, height = $\frac{7}{\sqrt{13}}$:
$$\frac{1}{2}\cdot \sqrt{13}\cdot \frac{7}{\sqrt{13}}=\frac{7}{2}$$
• Base $AB=\sqrt{(2-5{)}^2+(1-2{)}^2}=\sqrt{10}$, height = $\frac{7}{\sqrt{10}}$:
$$\frac{1}{2}\cdot \sqrt{10}\cdot \frac{7}{\sqrt{10}}=\frac{7}{2}$$

All match the area, confirming our distances!

Step 7: Match with Options

Our perpendiculars are:

• From $A$ to $BC$: $\frac{7}{\sqrt{5}}$
• From $B$ to $AC$: $\frac{7}{\sqrt{13}}$
• From $C$ to $AB$: $\frac{7}{\sqrt{10}}$

Comparing with options:

• a) $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{6}}$ – Third term incorrect.
• b) $\frac{7}{\sqrt{6}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{10}}$ – First two incorrect.
• c) $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{15}}$ – Last two incorrect.
• d) $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}}$ – Matches exactly!

Correct Answer: Option d