Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2$3y - x = 2$ and x + y = 2$x + y = 2$ , 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

### Related Formula The orthocentre P$P$ of a triangle is the point of intersection of its altitudes. Area of a triangle with a horizontal base lying on the x-axis is: textArea = frac12 times textbase times textheight = frac12 times |x_C - x_B| times |y_P|$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times |x_C - x_B| \times |y_P|$$ ### Core Logic Find vertex A$A$ by solving the line equations AB$AB$ and AC$AC$: 3y - x = 2 implies x = 3y - 2$$3y - x = 2 \implies x = 3y - 2$$ Substitute into x + y = 2 implies (3y - 2) + y = 2 implies 4y = 4 implies y = 1$x + y = 2 \implies (3y - 2) + y = 2 \implies 4y = 4 \implies y = 1$. Then x = 3(1) - 2 = 1$x = 3(1) - 2 = 1$. So vertex A$A$ is (1, 1)$(1, 1)$. Find vertices B$B$ and C$C$ where the lines cross the x-axis (y = 0$y = 0$): - For B$B$ (on line AB$AB$): 3(0) - x = 2 implies x = -2 implies B(-2, 0)$3(0) - x = 2 \implies x = -2 \implies B(-2, 0)$ - For C$C$ (on line AC$AC$): x + 0 = 2 implies x = 2 implies C(2, 0)$x + 0 = 2 \implies x = 2 \implies C(2, 0)$ Base length BC = |2 - (-2)| = 4$BC = |2 - (-2)| = 4$. ### Step 1: Find Equations of Altitudes

1. **Altitude from A to BC**: Since BC$BC$ lies along the x-axis, the altitude from A(1, 1)$A(1, 1)$ must be a vertical line: textEquation of Altitude 1: x = 1$$\text{Equation of Altitude 1}: x = 1$$ 2. **Altitude from B to AC**: Slope of line AC$AC$ (x + y = 2$x + y = 2$) is m_AC = -1$m_{AC} = -1$. Therefore, the slope of the altitude perpendicular to AC$AC$ is m_2 = -frac1-1 = 1$m_2 = -\frac{1}{-1} = 1$. Passing through B(-2, 0)$B(-2, 0)$: y - 0 = 1(x - (-2)) implies y = x + 2 implies x - y + 2 = 0$$y - 0 = 1(x - (-2)) \implies y = x + 2 \implies x - y + 2 = 0$$ ### Step 2: Solve for Orthocentre coordinates P Intersect the altitude equations: x = 1$x = 1$ and y = x + 2$y = x + 2$: y = 1 + 2 = 3$y = 1 + 2 = 3$ Hence, the orthocentre is P(1, 3)$P(1, 3)$. ### Step 3: Compute Area of Triangle PBC Triangle PBC$PBC$ has base BC = 4$BC = 4$ on the x-axis, and vertex P(1, 3)$P(1, 3)$ gives a height of 3$3$. textArea = frac12 times 4 times 3 = 6$$\text{Area} = \frac{1}{2} \times 4 \times 3 = 6$$ ### Pattern Recognition When a triangle has its base sitting entirely on the coordinate axis, the altitude dropped from the opposite vertex is simplified to a pure horizontal or vertical line path, heavily reducing your linear derivation equation workload. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines