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

For any right-angled triangle, the orthocentre lies exactly at the vertex containing the 90°$90^{\circ}$ right angle. Distance formula:

Core Logic

Analyze slopes of lines forming Triangle 1: L₁: 4x - 7y + 10 = 0 m₁ = (4)/(7)$L_1: 4x - 7y + 10 = 0 \implies m_1 = \frac{4}{7}$ L₂: 7x + 4y - 15 = 0 m₂ = -(7)/(4)$L_2: 7x + 4y - 15 = 0 \implies m_2 = -\frac{7}{4}$ Notice m₁ · m₂ = ((4)/(7))(-(7)/(4)) = -1$m_1 \cdot m_2 = \left(\frac{4}{7}\right)\left(-\frac{7}{4}\right) = -1$.

Thus, Triangle 1 is a right-angled triangle. Its orthocentre B$B$ is the intersection point of L₁$L_1$ and L₂$L_2$: Solving 4x - 7y = -10$4x - 7y = -10$ and 7x + 4y = 15$7x + 4y = 15$: Multiplying first by 4, second by 7, and adding yields x = 1, y = 2 B(1, 2)$x = 1, y = 2 \implies B(1, 2)$.

Step 1: Locate Second Orthocentre

Triangle 2 is formed by x = 0$x = 0$, y = 0$y = 0$, and x + y = 1$x + y = 1$. This is a right triangle with vertices at (0,0), (1,0), (0,1)$(0,0), (1,0), (0,1)$. The right-angled vertex is at the origin P(0, 0)$P(0, 0)$. Thus, its orthocentre is P(0, 0)$P(0, 0)$.

Step 2: Distance Computation

Find the distance between B(1,2)$B(1,2)$ and P(0,0)$P(0,0)$:

Pattern Recognition

Always check for mutually perpendicular side orientations (m₁ · m₂ = -1$m_1 \cdot m_2 = -1$) when finding orthocentres in competitive math papers. This completely cuts out lengthy altitude equation derivation tracks.

Chapter Mix

Class 11 Mathematics: Straight Lines

Related Formula

The product of slopes of two mutually perpendicular lines is always equal to -1$-1$:

Core Logic

Let the vertices of the triangle be P$P$, Q$Q$, and R$R$. The lines given are:

• y = x + 1$y = x + 1$
• y = 4x - 8$y = 4x - 8$
• y = mx + c$y = mx + c$

Solving lines y = x + 1$y = x + 1$ and y = 4x - 8$y = 4x - 8$ gives the vertex P(3, 4)$P(3, 4)$.

The orthocentre is given as H(3, -1)$H(3, -1)$. Notice that the x$x$-coordinate of P$P$ and H$H$ are identical (x = 3$x = 3$). This implies that the altitude from vertex P$P$ to the base line y = mx + c$y = mx + c$ is a vertical line along x = 3$x = 3$.

Step 1: Determine the Slopes

Since the altitude from P$P$ is vertical, the side opposite to it (which lies on y = mx + c$y = mx + c$) must be a horizontal line.

Therefore, the slope of the line y = mx + c$y = mx + c$ must be zero:

m = 0$m = 0$

Step 2: Solve for c

Let's find point Q$Q$ by intersecting y = x + 1$y = x + 1$ and y = mx + c$y = mx + c$. Since m = 0$m = 0$, y = c$y = c$, we get Q(c-1, c)$Q(c-1, c)$.

Using the property that the line segment connecting Q$Q$ to the opposite side's altitude is perpendicular to line PR$PR$ (y = 4x - 8$y = 4x - 8$):

Step 3: Evaluate m - c

Substituting the values of m$m$ and c$c$:

Pattern Recognition

When the x$x$-coordinate of a vertex matches the x$x$-coordinate of the orthocentre, the altitude is vertical, forcing the opposite base to be purely horizontal (m=0$m=0$). This observation cuts down calculation time completely.

Chapter Mix

Class 11 Mathematics: Straight Lines

Related Formula

The midpoint between a point and its reflection image must lie exactly on the line mirror equation.

Core Logic

Find the midpoint M$M$ between point P(1, 2)$P(1, 2)$ and its given reflection image Q((57)/(13), (-40)/(13))$Q\left(\frac{57}{13}, \frac{-40}{13}\right)$:

Since M$M$ lies on the reflecting line 2x - 3y + λ = 0$2x - 3y + \lambda = 0$:

Step 1: Apply Concurrency Determinant

For three straight lines to intersect at a single concurrent point, the determinant of their linear coefficients must equal zero:

Expand the determinant along the first row:

Step 2: Compute Final Product Target

Multiply the absolute values of the determined parameters together:

Pattern Recognition

Using the midpoint property to evaluate unknown line parameters from reflection images is often much faster than using full distance formulas.

Chapter Mix

Class 11 Mathematics: Straight Lines

Related Formula

Angle θ$\theta$ between two lines with slopes m₁$m_1$ and m₂$m_2$:

Core Logic

Given lines for the equal sides:

• -x + 2y = 4 y = (1)/(2)x + 2 m₁ = (1)/(2)$-x + 2y = 4 \implies y = \frac{1}{2}x + 2 \implies m_1 = \frac{1}{2}$
• x + y = 4 y = -x + 4 m₂ = -1$x + y = 4 \implies y = -x + 4 \implies m_2 = -1$

In an isosceles triangle, the third side makes equal angles θ$\theta$ with both equal sides. Let the slope of the third side be m$m$:

Step 1: Solve the Slope Equation

Case 1 (Same sign):

Case 2 (Opposite sign):

Step 2: Sum of Roots

The quadratic equation for m$m$ is m² - 6m - 1 = 0$m^2 - 6m - 1 = 0$. The sum of possible distinct values of m$m$ is given by the sum of roots of this quadratic:

Pattern Recognition

Instead of solving for the explicit values of the slopes (which involve radicals), using Vieta's relations directly on the quadratic equation m² - 6m - 1 = 0$m^2 - 6m - 1 = 0$ gives the final answer instantly.

Chapter Mix

Class 11 Mathematics: Straight Lines

Related Formula

Core Logic

First, find the vertices A, B, C$A, B, C$ by solving the lines pairwise. Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and x + 2y - 31 = 0$x + 2y - 31 = 0$ gives A(9,11)$A(9,11)$. Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives B(3,4)$B(3,4)$. Solving x + 2y - 31 = 0$x + 2y - 31 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives C(5,13)$C(5,13)$.

Step 1: Determine the Centroid

Step 2: Find the Image (h, k)

Using the line 3x + 6y - 53 = 0$3x + 6y - 53 = 0$:

Solving for h$h$ and k$k$ yields:

Step 3: Compute final algebraic target

Pattern Recognition

Instead of solving fractions endlessly, substitute potential integer coordinates early into the slope relationship (k - yG)/(h - xG) = -1/m$(k - y_G)/(h - x_G) = -1/m$ to accelerate competitive solving time.

Chapter Mix

Class 11 Mathematics: Straight Lines