Math Problem Statement

The coordinates of the vertices of a triangle are P(-2, 6), Q(6, 2), and R(-8, a). Find the gradient of line PQ, determine the gradient of PR if angle RPQ is a right angle, use this to find the value of a, calculate the distance PQ, and find the midpoint of QR. Also, for a line joining A(s, 8) to B(-2, t) with a midpoint M(2, 3), calculate the values of s and t, and find the equation of the line perpendicular to AB passing through M.

Solution

Let's solve the problems given in the uploaded image:

Problem 4

a. Find the gradient of the line PQ.

Given points P(2,6)P(-2, 6) and Q(6,2)Q(6, 2). The gradient mm is calculated by: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} m=266(2)=48=12m = \frac{2 - 6}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2}

b. If angle RPQ is a right angle, what is the gradient of the line PR?

Since PQPQ and PRPR are perpendicular, the product of their gradients is 1-1. Given mPQ=12m_{PQ} = -\frac{1}{2}, mPRmPQ=1mPR(12)=1mPR=2m_{PR} \cdot m_{PQ} = -1 \Rightarrow m_{PR} \cdot \left(-\frac{1}{2}\right) = -1 \Rightarrow m_{PR} = 2

c. Use your answer from (b), or otherwise, to find the value of aa.

For PRPR, given R(8,a)R(8, a) and P(2,6)P(-2, 6), using the gradient formula: mPR=a68(2)=a610m_{PR} = \frac{a - 6}{8 - (-2)} = \frac{a - 6}{10} Setting mPR=2m_{PR} = 2, a610=2a6=20a=26\frac{a - 6}{10} = 2 \Rightarrow a - 6 = 20 \Rightarrow a = 26

d. Calculate the distance PQ.

Using the distance formula, d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} d=(6(2))2+(26)2=82+(4)2=64+16=80=45d = \sqrt{(6 - (-2))^2 + (2 - 6)^2} = \sqrt{8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}

e. Calculate the midpoint of QR.

Given Q(6,2)Q(6, 2) and R(8,a)R(8, a), with a=26a = 26, M=(6+82,2+262)=(7,14)M = \left(\frac{6 + 8}{2}, \frac{2 + 26}{2}\right) = (7, 14)

Problem 5

a. Calculate the values of ss and tt.

Given A(s,8)A(s, 8) to B(2,t)B(-2, t) with midpoint M(2,3)M(2, 3). Using the midpoint formula, s+(2)2=2s2=4s=6\frac{s + (-2)}{2} = 2 \Rightarrow s - 2 = 4 \Rightarrow s = 6 8+t2=38+t=6t=2\frac{8 + t}{2} = 3 \Rightarrow 8 + t = 6 \Rightarrow t = -2

b. Find the equation of the straight line perpendicular to AB, passing through the point M.

Gradient of ABAB: mAB=t82s=2826=108=54m_{AB} = \frac{t - 8}{-2 - s} = \frac{-2 - 8}{-2 - 6} = \frac{-10}{-8} = \frac{5}{4} The perpendicular gradient is 45-\frac{4}{5}. Using point-slope form, y3=45(x2)y=45x+85+3=45x+235y - 3 = -\frac{4}{5}(x - 2) \Rightarrow y = -\frac{4}{5}x + \frac{8}{5} + 3 = -\frac{4}{5}x + \frac{23}{5} y=45x+235y = -\frac{4}{5}x + \frac{23}{5}

Would you like more details on any step, or do you have any other questions? Here are some related questions to consider:

  1. How do you verify the right angle in a triangle using slopes?
  2. What is the equation of the circle passing through points P, Q, and R?
  3. How do you determine if three points are collinear?
  4. Can you find the area of triangle PQR?
  5. How does the distance formula derive from the Pythagorean theorem?

Tip: Remember to verify your calculations by plugging values back into the original equations to ensure consistency and correctness.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Gradients
Midpoints
Distance Formula
Perpendicular Lines
Right Angle Theorem

Formulas

Gradient Formula: m = (y2 - y1) / (x2 - x1)
Distance Formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Midpoint Formula: M = ((x1 + x2)/2, (y1 + y2)/2)
Perpendicular Gradient: m1 * m2 = -1

Theorems

Perpendicular Gradient Theorem
Pythagoras' Theorem (for distance)

Suitable Grade Level

Grade 10-12