Given points P(1,1), Q(0,−2), and the equations of lines PR (y = −x + 2) and QR (x + 3y + 6 = 0), find the…

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Given points P(1,1), Q(0,−2), and the equations of lines PR (y = −x + 2) and QR (x + 3y + 6 = 0), find the gradient of PQ, prove that angle PQR is 90 degrees, and determine the coordinates of R.

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Step 1: Identify the coordinates of P and Q. P is (1, 1) and Q is (0, −2). Step 2: Apply the gradient formula m = (y₂ − y₁)/(x₂ − x₁). mQP = (1 − (−2))/(1 − 0); mQP = (1 + 2)/(1); mQP = 3/1; mQP = 3 The gradient of QP is 3. 6.2 Prove that ∠ PQR = 90°. Step 1: Determine the gradient of PQ. From 6.1, mPQ = 3. Step 2: Determine the gradient of QR. The equation of line QR is x + 3y + 6 = 0. Rearrange the equation into the form y = mx + c: 3y = −x − 6; y = −1/3x − 2; The gradient of QR is mQR = −1/3. Step 3: Calculate the product of the gradients of PQ and QR. mPQ × mQR = 3 × (−1/3); mPQ × mQR = −1 Since the product of the gradients of PQ and QR is −1, the lines PQ and QR are perpendicular. Therefore, ∠ PQR = 90°. 6.3 Determine the coordinates of R. Step 1: Identify the equations of lines PR and QR. Equation of PR: y = −x + 2 (1); Equation of QR: x + 3y + 6 = 0 (2) R is the intersection point of these two lines. Step 2: Substitute equation (1) into equation (2). Substitute y = −x + 2 into x + 3y + 6 = 0: x + 3(−x + 2) + 6 = 0; x − 3x + 6 + 6 = 0; −2x + 12 = 0; −2x = −12; x = (−12)/(−2); x = 6 Step 3: Substitute the value of x back into equation (1) to find y. y = −x + 2; y = −(6) + 2; y = −6 + 2; y = −4 The coordinates of R are (6, −4) . 6.4 Calculate the length of PR. Leave your answer in surd form. Step 1: Identify the coordinates of P and R. P is (1, 1) and R is (6, −4). Step 2: Apply the distance formula d = √((x₂ − x₁)² + (y₂ − y₁)²). PR = √((6 − 1)² + (−4 − 1)²); PR = √((5)² + (−5)²); PR = √(25 + 25); PR = √(50) Step 3: Simplify the surd. PR = √(25 × 2); PR = 5√(2) The length of PR is 5√(2) units. 6.5 Determine the equation of the line (s) in the form y = mx + c. (s) is perpendicular to line y = −x + 2 and passes through the point P. Step 1: Determine the gradient of line (s). The given line is y = −x + 2, so its gradient is m₁ = −1. Line (s) is perpendicular to this line. If m_s is the gradient of line (s), then m_s × m₁ = −1. m_s × (−1) = −1; m_s = 1; Step 2: Use the point-slope form y − y₁ = m(x − x₁) with point P(1, 1) and gradient m_s = 1. y − 1 = 1(x − 1); y − 1 = x − 1; y = x − 1 + 1; y = x; The equation of line (s) is y = x. 6.6 Calculate the size of θ. Step 1: Identify the gradients of the lines forming angle θ = ∠ PRQ. The angle θ is formed by lines PR and QR. From the equation of PR, y = −x + 2, the gradient is mPR = −1. From the equation of QR, x + 3y + 6 = 0 ⇒ y = −1/3x − 2, the gradient is mQR = −1/3. Step 2: Use the formula for the angle between two lines, tan θ = | (m₁ − m₂)/(1 + m₁ m₂) |. Let

Given points P(1,1), Q(0,-2), and the equations of lines PR (y = -x + 2) and QR (x + 3y + 6 = 0), find the gradient of PQ, prove that angle PQR is 90 degrees, and determine the coordinates of R.

Answers

Step-by-step solution

Part 6.2· 3 steps

Determine the gradient of PQ.

Determine the gradient of QR.

Calculate the product of the gradients of PQ and QR.

Part 6.3· 3 steps(6, -4)

Identify the equations of lines PR and QR.

Substitute equation (1) into equation (2).

Substitute the value of $x$ back into equation (1) to find $y$ .

Part 6.4· 3 steps
$5\sqrt{2} \text{ units}$

Identify the coordinates of P and R.

Apply the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

Simplify the surd.

Part 6.5· 2 steps
$y = x$

Determine the gradient of line (s).

Use the point-slope form $y - y_1 = m(x - x_1)$ with point P(1, 1) and gradient $m_s = 1$ .

Part 6.6· 2 steps

Identify the gradients of the lines forming angle $\theta = \angle PRQ$ .

Use the formula for the angle between two lines, $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ .

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