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 the x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to:
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Step-by-Step Solution
Step 1: Find the coordinates of vertices A, B, and C
First, we find the coordinates of the vertices of triangle ABC. Vertex A is the intersection of lines AB (x−3y=−2) and AC (x+y=2). Solving these two equations simultaneously gives x=1 and y=1, so A=(1,1). Since B and C lie on the x -axis, their y -coordinates are 0. For B, substitute y=0 into the equation of line AB (3y−x=2) to get x=−2, so B=(−2,0). For C, substitute y=0 into the equation of line AC (x+y=2) to get x=2, so C=(2,0).
Step 2: Calculate the slopes of the sides
Next, we calculate the slopes of the sides AB and AC. The equation of line AB is 3y−x=2, which can be rewritten as y=31x+32. So, the slope of AB, mAB, is 31. The equation of line AC is x+y=2, which can be rewritten as y=−x+2. So, the slope of AC, mAC, is −1.
Step 3: Find the equations of altitudes from B and C
The orthocentre P is the intersection of the altitudes. The altitude from B to AC is perpendicular to AC. Since mAC=−1, the slope of the altitude from B is mB=−−11=1. Using point-slope form with B(−2,0), the equation of this altitude is y−0=1(x−(−2)), which simplifies to y=x+2. The altitude from C to AB is perpendicular to AB. Since mAB=31, the slope of the altitude from C is mC=−1/31=−3. Using point-slope form with C(2,0), the equation of this altitude is y−0=−3(x−2), which simplifies to y=−3x+6.
Step 4: Find the coordinates of the orthocentre P
The orthocentre P is the intersection of the two altitudes. We solve the system of equations for the altitudes: y=x+2 and y=−3x+6. Substituting the first into the second gives x+2=−3x+6. This simplifies to 4x=4, so x=1. Substituting x=1 into y=x+2 gives y=1+2=3. Thus, the orthocentre is P(1,3).
Step 5: Calculate the area of triangle PBC
Finally, we calculate the area of triangle PBC. The base BC lies on the x -axis. The length of the base BC is the distance between B(−2,0) and C(2,0), which is ∣2−(−2)∣=4 units. The height of the triangle PBC with respect to base BC is the absolute value of the y -coordinate of P, which is ∣3∣=3 units. Using the formula for the area of a triangle, Area =21×base×height, we get Area =21×4×3=6 square units.