Question

Let ABC be the triangle such that the equations of lines AB and AC be 3y−x=2 and x+y=2 , respectively, and the points B and C lie on x -axis. If P is the orthocentre of the triangle ABC , then the area of the triangle PBC is equal to

Answer 1

6

Answer 2

1. Find the coordinates of vertices A, B, and C:

• Point A (Intersection of AB and AC):
  Given equations:
  Line AB: $3y - x = 2 \implies x - 3y = -2$ (1)
  Line AC: $x + y = 2$ (2)

  Subtract equation (1) from equation (2):
  $(x + y) - (x - 3y) = 2 - (-2)$
  $4y = 4 \implies y = 1$

  Substitute $y = 1$ into equation (2):
  $x + 1 = 2 \implies x = 1$
  So, the coordinates of A are $(1, 1)$.

• Point B (Intersection of AB and x-axis, i.e., y=0):
  Substitute $y = 0$ into the equation of AB:
  $x - 3(0) = -2 \implies x = -2$
  So, the coordinates of B are $(-2, 0)$.

• Point C (Intersection of AC and x-axis, i.e., y=0):
  Substitute $y = 0$ into the equation of AC:
  $x + 0 = 2 \implies x = 2$
  So, the coordinates of C are $(2, 0)$.

2. Find the coordinates of the orthocenter P:
The orthocenter is the intersection of the altitudes of the triangle.

• Altitude from A to BC:
  The side BC lies on the x-axis (y=0). Therefore, the altitude from A to BC will be a vertical line passing through A.
  The equation of this altitude is $x = x_A \implies x = 1$.

• Altitude from B to AC:
  First, find the slope of AC ($m_{AC}$):
  $m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{0 - 1}{2 - 1} = \frac{-1}{1} = -1$.
  The altitude from B to AC is perpendicular to AC. Its slope ($m_{BE}$) will be the negative reciprocal of $m_{AC}$:
  $m_{BE} = -\frac{1}{m_{AC}} = -\frac{1}{-1} = 1$.
  Now, use the point-slope form to find the equation of the altitude passing through B(-2, 0) with slope 1:
  $y - y_B = m_{BE}(x - x_B)$
  $y - 0 = 1(x - (-2))$
  $y = x + 2$

• Coordinates of P:
  The orthocenter P is the intersection of the two altitudes found.
  Substitute $x = 1$ (from the first altitude equation) into the second altitude equation ($y = x + 2$):
  $y = 1 + 2 \implies y = 3$
  So, the coordinates of the orthocenter P are $(1, 3)$.

3. Calculate the area of triangle PBC:
The vertices of triangle PBC are P(1, 3), B(-2, 0), and C(2, 0).
The base BC lies on the x-axis.

• Length of base BC:
  $BC = |x_C - x_B| = |2 - (-2)| = |2 + 2| = 4$ units.

• Height of the triangle (perpendicular distance from P to BC):
  Since BC lies on the x-axis, the height is the absolute value of the y-coordinate of P.
  Height $h = |y_P| = |3| = 3$ units.

• Area of triangle PBC:
  Area $= \frac{1}{2} \times \text{base} \times \text{height}$
  Area $= \frac{1}{2} \times 4 \times 3$
  Area $= 2 \times 3 = 6$ square units.

The final answer is 6.

Explanation of the solution:

1. Found vertices A, B, C by solving line equations and using the fact that B and C lie on the x-axis. A(1,1), B(-2,0), C(2,0).
2. Determined orthocenter P by finding equations of two altitudes:
   • Altitude from A to BC (x-axis) is $x=1$.
   • Altitude from B to AC (slope 1) is $y=x+2$.
   • Intersection of these lines gives P(1,3).
3. Calculated area of triangle PBC using base BC (on x-axis) and height (y-coordinate of P).
   • Base BC = $2 - (-2) = 4$.
   • Height = $3$.
   • Area = $\frac{1}{2} \times 4 \times 3 = 6$.