Question Statement

Prove that (a) altitudes, (b) right bisectors, and (c) medians of the following triangles are concurrent:

(i) $A (4, 6)$ , $B (7, 2)$ , $C (2, 3)$

(ii) $P (- 4, 0)$ , $Q (2, 0)$ , $R (0, 3)$

Background and Explanation

Three lines are concurrent if they all intersect at a single common point. For three lines given by $a_{1} x + b_{1} y + c_{1} = 0$ , $a_{2} x + b_{2} y + c_{2} = 0$ , and $a_{3} x + b_{3} y + c_{3} = 0$ , concurrency can be verified by showing that the determinant of their coefficients equals zero. This problem demonstrates the classic geometry theorems regarding the orthocenter (altitudes), circumcenter (right bisectors), and centroid (medians).

Solution

Part (i) — Triangle $A B C$ with $A (4, 6)$ , $B (7, 2)$ , $C (2, 3)$

(a) Altitudes

First, we calculate the slopes of the sides of $△ A B C$ :

$Slope of \overline{A B} = \frac{6 - 2}{4 - 7} = \frac{- 4}{3}$

$Slope of \overline{B C} = \frac{2 - 3}{7 - 2} = \frac{- 1}{5}$

$Slope of \overline{C A} = \frac{6 - 3}{4 - 2} = \frac{3}{2}$

Altitude $\overline{A E}$ from $A$ to $\overline{B C}$ :

Since $\overline{A E} ⊥ \overline{B C}$ :

$Slope of \overline{A E} = \frac{- 1}{Slope of B C} = \frac{- 1}{\frac{- 1}{5}} = 5$

Using point-slope form through $A (4, 6)$ :

$y - 6 = 5 (x - 4)$ $y - 6 = 5 x - 20$ $\Rightarrow 5 x - y - 14 = 0$

Altitude $\overline{B F}$ from $B$ to $\overline{C A}$ :

Since $\overline{B F} ⊥ \overline{C A}$ :

$Slope of \overline{B F} = \frac{- 1}{\frac{3}{2}} = \frac{- 2}{3}$

Equation through $B (7, 2)$ :

$y - 2 = - \frac{2}{3} (x - 7)$ $3 y - 6 = - 2 x + 14$ $\Rightarrow 2 x + 3 y - 20 = 0$

Altitude $\overline{C D}$ from $C$ to $\overline{A B}$ :

Since $\overline{C D} ⊥ \overline{A B}$ :

$Slope of \overline{C D} = \frac{- 1}{\frac{- 4}{3}} = \frac{3}{4}$

Equation through $C (2, 3)$ :

$y - 3 = \frac{3}{4} (x - 2)$ $4 y - 12 = 3 x - 6$ $\Rightarrow 3 x - 4 y + 6 = 0$

Proving Concurrency:

To verify that the three altitudes meet at one point, we evaluate the determinant of their coefficients:

$523 - 1 3 - 4 - 14 - 20 6$

Expanding along the first row:

$= 5 3 - 4 - 20 6 + 1 23 - 20 6 - 14 23 3 - 4$

$= 5 (18 - 80) + 1 (12 + 60) - 14 (- 8 - 9)$ $= 5 (- 62) + 72 - 14 (- 17)$ $= - 310 + 72 + 238 = 0$

Since the determinant equals zero, the altitudes are concurrent.

(b) Right Bisectors

Let $M_{1}$ , $M_{2}$ , $M_{3}$ be the midpoints of sides $\overline{A B}$ , $\overline{B C}$ , and $\overline{C A}$ respectively:

$M_{1} = (\frac{4 + 7}{2}, \frac{6 + 2}{2}) = (\frac{11}{2}, 4)$

$M_{2} = (\frac{2 + 7}{2}, \frac{3 + 2}{2}) = (\frac{9}{2}, \frac{5}{2})$

$M_{3} = (\frac{2 + 4}{2}, \frac{3 + 6}{2}) = (3, \frac{9}{2})$

Right bisector of $\overline{A B}$ :

Perpendicular to $\overline{A B}$ (slope $\frac{- 4}{3}$ ), so slope $= \frac{3}{4}$ . Through $M_{1} (\frac{11}{2}, 4)$ :

$y - 4 = \frac{3}{4} (x - \frac{11}{2})$ $4 y - 16 = 3 x - \frac{33}{2}$ $8 y - 32 = 6 x - 33$ $\Rightarrow 6 x - 8 y - 1 = 0$

Right bisector of $\overline{B C}$ :

Perpendicular to $\overline{B C}$ (slope $\frac{- 1}{5}$ ), so slope $= 5$ . Through $M_{2} (\frac{9}{2}, \frac{5}{2})$ :

$y - \frac{5}{2} = 5 (x - \frac{9}{2})$ $2 y - 5 = 10 x - 45$ $\Rightarrow 5 x - y - 20 = 0$

Right bisector of $\overline{C A}$ :

Perpendicular to $\overline{C A}$ (slope $\frac{3}{2}$ ), so slope $= \frac{- 2}{3}$ . Through $M_{3} (3, \frac{9}{2})$ :

$y - \frac{9}{2} = \frac{- 2}{3} (x - 3)$

Multiplying by 6: $6 y - 27 = - 4 x + 12$ $\Rightarrow 4 x + 6 y - 39 = 0$

Proving Concurrency:

$654 - 8 - 1 6 - 1 - 20 - 39$

$= 6 - 1 6 - 20 - 39 + 8 54 - 20 - 39 - 1 54 - 1 6$

$= 6 (39 + 120) + 8 (- 195 + 80) - 1 (30 + 4)$ $= 6 (159) + 8 (- 115) - 34$ $= 954 - 920 - 34 = 0$

Thus, the right bisectors are concurrent.

(c) Medians

Using the same midpoints $M_{1}$ , $M_{2}$ , $M_{3}$ calculated above:

Median $\overline{A M_{2}}$ from $A (4, 6)$ to $M_{2} (\frac{9}{2}, \frac{5}{2})$ :

Using two-point form: $\frac{x - 4}{\frac{9}{2} - 4} = \frac{y - 6}{\frac{5}{2} - 6}$ $\frac{x - 4}{\frac{1}{2}} = \frac{y - 6}{\frac{- 7}{2}}$ $\Rightarrow y - 6 = - 7 (x - 4)$ $\Rightarrow 7 x + y - 34 = 0$

Median $\overline{B M_{3}}$ from $B (7, 2)$ to $M_{3} (3, \frac{9}{2})$ :

$\frac{x - 7}{3 - 7} = \frac{y - 2}{\frac{9}{2} - 2}$ $\frac{x - 7}{- 4} = \frac{2 ( y - 2 )}{5}$ $5 (x - 7) = - 8 (y - 2)$ $5 x - 35 = - 8 y + 16$ $\Rightarrow 5 x + 8 y - 51 = 0$

Median $\overline{C M_{1}}$ from $C (2, 3)$ to $M_{1} (\frac{11}{2}, 4)$ :

$\frac{x - 2}{\frac{11}{2} - 2} = \frac{y - 3}{4 - 3}$ $\frac{2 ( x - 2 )}{7} = y - 3$ $2 x - 4 = 7 y - 21$ $\Rightarrow 2 x - 7 y + 17 = 0$

Proving Concurrency:

$752 18 - 7 - 34 - 51 17$

$= 7 8 - 7 - 51 17 - 1 52 - 51 17 - 34 52 8 - 7$

$= 7 (136 - 357) - 1 (85 + 102) - 34 (- 35 - 16)$ $= 7 (- 221) - 187 - 34 (- 51)$ $= - 1547 - 187 + 1734 = 0$

Therefore, the medians are concurrent.

Part (ii) — Triangle $P QR$ with $P (- 4, 0)$ , $Q (2, 0)$ , $R (0, 3)$

(a) Altitudes

Calculate side slopes:

$Slope of \overline{P Q} = \frac{0 - 0}{2 - ( - 4 )} = 0$ $Slope of \overline{QR} = \frac{3 - 0}{0 - 2} = \frac{- 3}{2}$ $Slope of \overline{R P} = \frac{3 - 0}{0 - ( - 4 )} = \frac{3}{4}$

Altitude $\overline{P A}$ from $P$ to $\overline{QR}$ :

$Slope = \frac{- 1}{\frac{- 3}{2}} = \frac{2}{3}$

Equation through $P (- 4, 0)$ : $y = \frac{2}{3} (x + 4)$ $\Rightarrow 2 x - 3 y + 8 = 0$

Altitude $\overline{QB}$ from $Q$ to $\overline{R P}$ :

$Slope = \frac{- 1}{\frac{3}{4}} = \frac{- 4}{3}$

Equation through $Q (2, 0)$ : $y = \frac{- 4}{3} (x - 2)$ $3 y = - 4 x + 8$ $\Rightarrow 4 x + 3 y - 8 = 0$

Altitude $\overline{R C}$ from $R$ to $\overline{P Q}$ :

Since $\overline{P Q}$ is horizontal (slope 0), the altitude is vertical: $\Rightarrow x = 0$

Proving Concurrency:

$241 - 3 30 8 - 8 0$

$= 2 (0 - 0) + 3 (0 + 8) + 8 (0 - 3)$ $= 0 + 24 - 24 = 0$

The altitudes are concurrent.

(b) Right Bisectors

Midpoints: $M_{1} = (\frac{- 4 + 2}{2}, 0) = (- 1, 0)$ $M_{2} = (\frac{2 + 0}{2}, \frac{0 + 3}{2}) = (1, \frac{3}{2})$ $M_{3} = (\frac{- 4 + 0}{2}, \frac{3}{2}) = (- 2, \frac{3}{2})$

Right bisector of $\overline{P Q}$ :

Perpendicular to horizontal line $\overline{P Q}$ is vertical through $M_{1}$ : $\Rightarrow x + 1 = 0$

Right bisector of $\overline{QR}$ :

Slope $= \frac{2}{3}$ through $M_{2} (1, \frac{3}{2})$ : $y - \frac{3}{2} = \frac{2}{3} (x - 1)$ $6 y - 9 = 4 x - 4$ $\Rightarrow 4 x - 6 y + 5 = 0$

Right bisector of $\overline{R P}$ :

Slope $= \frac{- 4}{3}$ through $M_{3} (- 2, \frac{3}{2})$ : $y - \frac{3}{2} = \frac{- 4}{3} (x + 2)$ $6 y - 9 = - 8 x - 16$ $\Rightarrow 8 x + 6 y + 7 = 0$

Proving Concurrency:

$148 0 - 6 6 157$

$= 1 (- 42 - 30) - 0 + 1 (24 + 48)$ $= - 72 + 72 = 0$

The right bisectors are concurrent.

(c) Medians

Median $\overline{P M_{2}}$ from $P (- 4, 0)$ to $M_{2} (1, \frac{3}{2})$ :

$\frac{x + 4}{5} = \frac{y}{\frac{3}{2}} = \frac{2 y}{3}$ $3 x + 12 = 10 y$ $\Rightarrow 3 x - 10 y + 12 = 0$

Median $\overline{Q M_{3}}$ from $Q (2, 0)$ to $M_{3} (- 2, \frac{3}{2})$ :

$\frac{x - 2}{- 4} = \frac{y}{\frac{3}{2}} = \frac{2 y}{3}$ $3 x - 6 = - 8 y$ $\Rightarrow 3 x + 8 y - 6 = 0$

Median $\overline{R M_{1}}$ from $R (0, 3)$ to $M_{1} (- 1, 0)$ :

$\frac{x}{- 1} = \frac{y - 3}{- 3}$ $3 x = y - 3$ $\Rightarrow 3 x - y + 3 = 0$

Proving Concurrency:

$333 - 10 8 - 1 12 - 6 3$

$= 3 (24 - 6) + 10 (9 + 18) + 12 (- 3 - 24)$ $= 3 (18) + 10 (27) + 12 (- 27)$ $= 54 + 270 - 324 = 0$

All medians are concurrent.

Key Formulas or Methods Used

Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Perpendicularity Condition: $m_{⊥} = - \frac{1}{m}$ (negative reciprocal)
Point-Slope Form: $y - y_{1} = m (x - x_{1})$
Two-Point Form: $\frac{y - y _{1}}{y _{2} - y _{1}} = \frac{x - x _{1}}{x _{2} - x _{1}}$
Midpoint Formula: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Concurrency Condition: For lines $a_{i} x + b_{i} y + c_{i} = 0$ , the determinant $a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = 0$

Summary of Steps

Calculate side slopes using the slope formula for all three sides of the triangle
Determine perpendicular slopes (negative reciprocals) for altitudes and right bisectors
Find midpoints of each side for right bisectors and medians
Derive line equations using point-slope form (through vertices for altitudes/medians, through midpoints for right bisectors)
Arrange equations in standard form $a x + b y + c = 0$ to extract coefficients
Evaluate the determinant of the $3 \times 3$ coefficient matrix—if zero, the lines are concurrent

Question Statement

Prove that (a) altitudes, (b) right bisectors, and (c) medians of the following triangles are concurrent:

(i) $A (4, 6)$ , $B (7, 2)$ , $C (2, 3)$

(ii) $P (- 4, 0)$ , $Q (2, 0)$ , $R (0, 3)$

Background and Explanation

Solution

Part (i) — Triangle $A B C$ with $A (4, 6)$ , $B (7, 2)$ , $C (2, 3)$

(a) Altitudes

First, we calculate the slopes of the sides of $△ A B C$ :

$Slope of \overline{A B} = \frac{6 - 2}{4 - 7} = \frac{- 4}{3}$

$Slope of \overline{B C} = \frac{2 - 3}{7 - 2} = \frac{- 1}{5}$

$Slope of \overline{C A} = \frac{6 - 3}{4 - 2} = \frac{3}{2}$

Altitude $\overline{A E}$ from $A$ to $\overline{B C}$ :

Since $\overline{A E} ⊥ \overline{B C}$ :

$Slope of \overline{A E} = \frac{- 1}{Slope of B C} = \frac{- 1}{\frac{- 1}{5}} = 5$

Using point-slope form through $A (4, 6)$ :

$y - 6 = 5 (x - 4)$ $y - 6 = 5 x - 20$ $\Rightarrow 5 x - y - 14 = 0$

Altitude $\overline{B F}$ from $B$ to $\overline{C A}$ :

Since $\overline{B F} ⊥ \overline{C A}$ :

$Slope of \overline{B F} = \frac{- 1}{\frac{3}{2}} = \frac{- 2}{3}$

Equation through $B (7, 2)$ :

$y - 2 = - \frac{2}{3} (x - 7)$ $3 y - 6 = - 2 x + 14$ $\Rightarrow 2 x + 3 y - 20 = 0$

Altitude $\overline{C D}$ from $C$ to $\overline{A B}$ :

Since $\overline{C D} ⊥ \overline{A B}$ :

$Slope of \overline{C D} = \frac{- 1}{\frac{- 4}{3}} = \frac{3}{4}$

Equation through $C (2, 3)$ :

$y - 3 = \frac{3}{4} (x - 2)$ $4 y - 12 = 3 x - 6$ $\Rightarrow 3 x - 4 y + 6 = 0$

Proving Concurrency:

To verify that the three altitudes meet at one point, we evaluate the determinant of their coefficients:

$523 - 1 3 - 4 - 14 - 20 6$

Expanding along the first row:

$= 5 3 - 4 - 20 6 + 1 23 - 20 6 - 14 23 3 - 4$

$= 5 (18 - 80) + 1 (12 + 60) - 14 (- 8 - 9)$ $= 5 (- 62) + 72 - 14 (- 17)$ $= - 310 + 72 + 238 = 0$

Since the determinant equals zero, the altitudes are concurrent.

(b) Right Bisectors

Let $M_{1}$ , $M_{2}$ , $M_{3}$ be the midpoints of sides $\overline{A B}$ , $\overline{B C}$ , and $\overline{C A}$ respectively:

$M_{1} = (\frac{4 + 7}{2}, \frac{6 + 2}{2}) = (\frac{11}{2}, 4)$

$M_{2} = (\frac{2 + 7}{2}, \frac{3 + 2}{2}) = (\frac{9}{2}, \frac{5}{2})$

$M_{3} = (\frac{2 + 4}{2}, \frac{3 + 6}{2}) = (3, \frac{9}{2})$

Right bisector of $\overline{A B}$ :

Perpendicular to $\overline{A B}$ (slope $\frac{- 4}{3}$ ), so slope $= \frac{3}{4}$ . Through $M_{1} (\frac{11}{2}, 4)$ :

$y - 4 = \frac{3}{4} (x - \frac{11}{2})$ $4 y - 16 = 3 x - \frac{33}{2}$ $8 y - 32 = 6 x - 33$ $\Rightarrow 6 x - 8 y - 1 = 0$

Right bisector of $\overline{B C}$ :

Perpendicular to $\overline{B C}$ (slope $\frac{- 1}{5}$ ), so slope $= 5$ . Through $M_{2} (\frac{9}{2}, \frac{5}{2})$ :

$y - \frac{5}{2} = 5 (x - \frac{9}{2})$ $2 y - 5 = 10 x - 45$ $\Rightarrow 5 x - y - 20 = 0$

Right bisector of $\overline{C A}$ :

Perpendicular to $\overline{C A}$ (slope $\frac{3}{2}$ ), so slope $= \frac{- 2}{3}$ . Through $M_{3} (3, \frac{9}{2})$ :

$y - \frac{9}{2} = \frac{- 2}{3} (x - 3)$

Multiplying by 6: $6 y - 27 = - 4 x + 12$ $\Rightarrow 4 x + 6 y - 39 = 0$

Proving Concurrency:

$654 - 8 - 1 6 - 1 - 20 - 39$

$= 6 - 1 6 - 20 - 39 + 8 54 - 20 - 39 - 1 54 - 1 6$

$= 6 (39 + 120) + 8 (- 195 + 80) - 1 (30 + 4)$ $= 6 (159) + 8 (- 115) - 34$ $= 954 - 920 - 34 = 0$

Thus, the right bisectors are concurrent.

(c) Medians

Using the same midpoints $M_{1}$ , $M_{2}$ , $M_{3}$ calculated above:

Median $\overline{A M_{2}}$ from $A (4, 6)$ to $M_{2} (\frac{9}{2}, \frac{5}{2})$ :

Median $\overline{B M_{3}}$ from $B (7, 2)$ to $M_{3} (3, \frac{9}{2})$ :

$\frac{x - 7}{3 - 7} = \frac{y - 2}{\frac{9}{2} - 2}$ $\frac{x - 7}{- 4} = \frac{2 ( y - 2 )}{5}$ $5 (x - 7) = - 8 (y - 2)$ $5 x - 35 = - 8 y + 16$ $\Rightarrow 5 x + 8 y - 51 = 0$

Median $\overline{C M_{1}}$ from $C (2, 3)$ to $M_{1} (\frac{11}{2}, 4)$ :

$\frac{x - 2}{\frac{11}{2} - 2} = \frac{y - 3}{4 - 3}$ $\frac{2 ( x - 2 )}{7} = y - 3$ $2 x - 4 = 7 y - 21$ $\Rightarrow 2 x - 7 y + 17 = 0$

Proving Concurrency:

$752 18 - 7 - 34 - 51 17$

$= 7 8 - 7 - 51 17 - 1 52 - 51 17 - 34 52 8 - 7$

$= 7 (136 - 357) - 1 (85 + 102) - 34 (- 35 - 16)$ $= 7 (- 221) - 187 - 34 (- 51)$ $= - 1547 - 187 + 1734 = 0$

Therefore, the medians are concurrent.

Part (ii) — Triangle $P QR$ with $P (- 4, 0)$ , $Q (2, 0)$ , $R (0, 3)$

(a) Altitudes

Calculate side slopes:

$Slope of \overline{P Q} = \frac{0 - 0}{2 - ( - 4 )} = 0$ $Slope of \overline{QR} = \frac{3 - 0}{0 - 2} = \frac{- 3}{2}$ $Slope of \overline{R P} = \frac{3 - 0}{0 - ( - 4 )} = \frac{3}{4}$

Altitude $\overline{P A}$ from $P$ to $\overline{QR}$ :

$Slope = \frac{- 1}{\frac{- 3}{2}} = \frac{2}{3}$

Equation through $P (- 4, 0)$ : $y = \frac{2}{3} (x + 4)$ $\Rightarrow 2 x - 3 y + 8 = 0$

Altitude $\overline{QB}$ from $Q$ to $\overline{R P}$ :

$Slope = \frac{- 1}{\frac{3}{4}} = \frac{- 4}{3}$

Equation through $Q (2, 0)$ : $y = \frac{- 4}{3} (x - 2)$ $3 y = - 4 x + 8$ $\Rightarrow 4 x + 3 y - 8 = 0$

Altitude $\overline{R C}$ from $R$ to $\overline{P Q}$ :

Since $\overline{P Q}$ is horizontal (slope 0), the altitude is vertical: $\Rightarrow x = 0$

Proving Concurrency:

$241 - 3 30 8 - 8 0$

$= 2 (0 - 0) + 3 (0 + 8) + 8 (0 - 3)$ $= 0 + 24 - 24 = 0$

The altitudes are concurrent.

(b) Right Bisectors

Midpoints: $M_{1} = (\frac{- 4 + 2}{2}, 0) = (- 1, 0)$ $M_{2} = (\frac{2 + 0}{2}, \frac{0 + 3}{2}) = (1, \frac{3}{2})$ $M_{3} = (\frac{- 4 + 0}{2}, \frac{3}{2}) = (- 2, \frac{3}{2})$

Right bisector of $\overline{P Q}$ :

Perpendicular to horizontal line $\overline{P Q}$ is vertical through $M_{1}$ : $\Rightarrow x + 1 = 0$

Right bisector of $\overline{QR}$ :

Slope $= \frac{2}{3}$ through $M_{2} (1, \frac{3}{2})$ : $y - \frac{3}{2} = \frac{2}{3} (x - 1)$ $6 y - 9 = 4 x - 4$ $\Rightarrow 4 x - 6 y + 5 = 0$

Right bisector of $\overline{R P}$ :

Slope $= \frac{- 4}{3}$ through $M_{3} (- 2, \frac{3}{2})$ : $y - \frac{3}{2} = \frac{- 4}{3} (x + 2)$ $6 y - 9 = - 8 x - 16$ $\Rightarrow 8 x + 6 y + 7 = 0$

Proving Concurrency:

$148 0 - 6 6 157$

$= 1 (- 42 - 30) - 0 + 1 (24 + 48)$ $= - 72 + 72 = 0$

The right bisectors are concurrent.

(c) Medians

Median $\overline{P M_{2}}$ from $P (- 4, 0)$ to $M_{2} (1, \frac{3}{2})$ :

$\frac{x + 4}{5} = \frac{y}{\frac{3}{2}} = \frac{2 y}{3}$ $3 x + 12 = 10 y$ $\Rightarrow 3 x - 10 y + 12 = 0$

Median $\overline{Q M_{3}}$ from $Q (2, 0)$ to $M_{3} (- 2, \frac{3}{2})$ :

$\frac{x - 2}{- 4} = \frac{y}{\frac{3}{2}} = \frac{2 y}{3}$ $3 x - 6 = - 8 y$ $\Rightarrow 3 x + 8 y - 6 = 0$

Median $\overline{R M_{1}}$ from $R (0, 3)$ to $M_{1} (- 1, 0)$ :

$\frac{x}{- 1} = \frac{y - 3}{- 3}$ $3 x = y - 3$ $\Rightarrow 3 x - y + 3 = 0$

Proving Concurrency:

$333 - 10 8 - 1 12 - 6 3$

$= 3 (24 - 6) + 10 (9 + 18) + 12 (- 3 - 24)$ $= 3 (18) + 10 (27) + 12 (- 27)$ $= 54 + 270 - 324 = 0$

All medians are concurrent.

Key Formulas or Methods Used

Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Perpendicularity Condition: $m_{⊥} = - \frac{1}{m}$ (negative reciprocal)
Point-Slope Form: $y - y_{1} = m (x - x_{1})$
Two-Point Form: $\frac{y - y _{1}}{y _{2} - y _{1}} = \frac{x - x _{1}}{x _{2} - x _{1}}$
Midpoint Formula: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Concurrency Condition: For lines $a_{i} x + b_{i} y + c_{i} = 0$ , the determinant $a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = 0$

Summary of Steps

Calculate side slopes using the slope formula for all three sides of the triangle
Determine perpendicular slopes (negative reciprocals) for altitudes and right bisectors
Find midpoints of each side for right bisectors and medians
Derive line equations using point-slope form (through vertices for altitudes/medians, through midpoints for right bisectors)
Arrange equations in standard form $a x + b y + c = 0$ to extract coefficients
Evaluate the determinant of the $3 \times 3$ coefficient matrix—if zero, the lines are concurrent

6.1 Q-8

Question Statement

Background and Explanation

Solution

Part (i) — Triangle $A B C$ with $A (4, 6)$ , $B (7, 2)$ , $C (2, 3)$

(a) Altitudes

(b) Right Bisectors

(c) Medians

Part (ii) — Triangle $P QR$ with $P (- 4, 0)$ , $Q (2, 0)$ , $R (0, 3)$

(a) Altitudes

(b) Right Bisectors

(c) Medians

Key Formulas or Methods Used

Summary of Steps

6.1 Q-8

Question Statement

Background and Explanation

Solution

Part (i) — Triangle $A B C$ with $A (4, 6)$ , $B (7, 2)$ , $C (2, 3)$

(a) Altitudes

(b) Right Bisectors

(c) Medians

Part (ii) — Triangle $P QR$ with $P (- 4, 0)$ , $Q (2, 0)$ , $R (0, 3)$

(a) Altitudes

(b) Right Bisectors

(c) Medians

Key Formulas or Methods Used

Summary of Steps