Math Problem Statement

The straight line joining the points P(5, 6) to the point Q(q, 2) is perpendicular to the straight line joining the point Q to the point R(9, -1). Calculate the values of q.

Solution

We are given two lines:

  1. The line joining points P(5,6)P(5, 6) and Q(q,2)Q(q, 2).
  2. The line joining points Q(q,2)Q(q, 2) and R(9,1)R(9, -1).

Since the lines are perpendicular, the product of their slopes must equal 1-1.

Step 1: Find the slopes of both lines.

The formula for the slope between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

Slope of line PQPQ:

Using points P(5,6)P(5, 6) and Q(q,2)Q(q, 2): mPQ=26q5=4q5m_{PQ} = \frac{2 - 6}{q - 5} = \frac{-4}{q - 5}

Slope of line QRQR:

Using points Q(q,2)Q(q, 2) and R(9,1)R(9, -1): mQR=129q=39qm_{QR} = \frac{-1 - 2}{9 - q} = \frac{-3}{9 - q}

Step 2: Use the condition that the lines are perpendicular.

For perpendicular lines, the product of their slopes must be 1-1: mPQ×mQR=1m_{PQ} \times m_{QR} = -1 Substitute the slopes: 4q5×39q=1\frac{-4}{q - 5} \times \frac{-3}{9 - q} = -1 Simplify: 12(q5)(9q)=1\frac{12}{(q - 5)(9 - q)} = -1

Step 3: Solve for qq.

Multiply both sides by (q5)(9q)(q - 5)(9 - q) to eliminate the fraction: 12=(q5)(9q)12 = -(q - 5)(9 - q) Expand the right side: 12=(q(9q)5(9q))12 = -(q(9 - q) - 5(9 - q)) 12=(9qq245+5q)12 = -(9q - q^2 - 45 + 5q) Simplify the equation: 12=(q2+14q45)12 = -(-q^2 + 14q - 45) 12=q214q+4512 = q^2 - 14q + 45 Move all terms to one side: q214q+33=0q^2 - 14q + 33 = 0

Step 4: Solve the quadratic equation.

Use the quadratic formula: q=(14)±(14)24(1)(33)2(1)q = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(33)}}{2(1)} q=14±1961322q = \frac{14 \pm \sqrt{196 - 132}}{2} q=14±642q = \frac{14 \pm \sqrt{64}}{2} q=14±82q = \frac{14 \pm 8}{2}

So, the two possible values for qq are: q=14+82=11orq=1482=3q = \frac{14 + 8}{2} = 11 \quad \text{or} \quad q = \frac{14 - 8}{2} = 3

Final answer:

The values of qq are q=11q = 11 and q=3q = 3.


Would you like further details on any of the steps? Here are some related questions to explore:

  1. How can we verify if two lines are perpendicular using their slopes?
  2. Can the perpendicular condition apply to horizontal and vertical lines?
  3. What happens if the lines are parallel instead of perpendicular?
  4. How do we solve a quadratic equation using factorization instead of the quadratic formula?
  5. Can this problem be solved geometrically rather than algebraically?

Tip: Always double-check the condition (perpendicular, parallel, etc.) to ensure you're using the correct relationship between slopes.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Slopes of Lines
Perpendicular Lines
Quadratic Equations

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Perpendicular slopes condition: m1 * m2 = -1
Quadratic formula: q = (-b ± √(b² - 4ac)) / 2a

Theorems

Perpendicular lines theorem

Suitable Grade Level

Grades 9-12