s of the first road, $(4, 7)$ and $(8, 9)$:\n\n$$m = \\frac{9 - 7}{8 - 4} = \\frac{2}{4} = 0.5$$\n\nSince parallel lines have the same gradient, the parallel road will also have a gradient of $m = 0.5$. This step earns you the first mark M1.","highlights":[{"text":"(4, 7)","location":"specification"},{"text":"(8, 9)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Point-slope formula: } \\htmlData{anchor=eq}{\\color{@sky}{y}} - \\htmlData{anchor=y1}{\\color{@pink}{y_1}} = \\htmlData{anchor=m-val}{\\color{@amber}{m}}(\\htmlData{anchor=x-val}{\\color{@sky}{x}} - \\htmlData{anchor=x1}{\\color{@purple}{x_1}})$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=point-coord}{\\color{@pink}{2}} \\text{ for } y_1, \\; \\htmlData{anchor=grad-val}{\\color{@amber}{0.5}} \\text{ for } m, \\text{ and } \\htmlData{anchor=x-coord}{\\color{@purple}{10}} \\text{ for } x_1:$"},{"latex":"$$y - \\color{@pink}{2} = \\color{@amber}{0.5}(x - \\color{@purple}{10})$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"y1"},{"color":"amber","style":"text-color","anchor":"m-val"},{"color":"purple","style":"text-color","anchor":"x1"}],"connections":[{"to":"point-coord","from":"y1","color":"pink","label":null,"style":"solid"},{"to":"grad-val","from":"m-val","color":"amber","label":null,"style":"solid"},{"to":"x-coord","from":"x1","color":"purple","label":null,"style":"solid"}]},"order":1,"title":"Apply point-slope form","description":"We can write the equation of the parallel road directly using the point-slope formula: $$y - y_1 = m(x - x_1)$$ We substitute the point $(10, 2)$ as $(x_1, y_1)$ and the gradient $m = 0.5$."},{"type":"text","order":2,"title":"Rearrange the equation","content":"Now, expand and simplify the equation to express it in slope-intercept form ($y = mx + c$):\n\n$$\\begin{aligned} y - 2 &= 0.5(x - 10) \\\\ y - 2 &= 0.5x - 5 \\end{aligned}$$\n\nAdd $2$ to both sides to solve for $y$:\n\n$$y = 0.5x - 3$$\n\nThis algebraic rearrangement and substitution earns you the second method mark M1."},{"type":"text","order":3,"title":"State the final answer","content":"The final equation of the parallel road is:\n\n$$y = 0.5x - 3$$\n\nThis final line equation earns you the accuracy mark A1."}]}],"generatedAt":"2026-06-12T21:08:04.952Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"aa0f185c-eb42-4ff3-8633-5f87a2d46b4d","revisionId":"89923d41-719a-4096-b459-e0919b9fc080","questionNumber":null,"questionType":"LA","specification":"","difficulty":4,"difficultyBand":"medium","totalMarks":4,"sourceTitle":null,"calculatorUse":"non_calc","parts":[{"id":"abc9ca6b-9f30-46c2-8e2f-31596507c55e","label":"a","content":"Line P passes through the points $(-3, 1)$ and $(1, 3)$. Line Q passes through the points $(-1, -2)$ and $(7, 2)$. Show that Line P and Line Q are parallel.","markscheme":"- Gradient of P: $m_P = \\frac{3 - 1}{1 - (-3)} = \\frac{2}{4} = \\frac{1}{2}$ M1 A1\n- Gradient of Q: $m_Q = \\frac{2 - (-2)}{7 - (-1)} = \\frac{4}{8} = \\frac{1}{2}$ M1\n- Conclusion: Since $m_P = m_Q$, the lines are parallel. C1","marks":4,"order":0,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Correct substitution into the gradient formula for Line P.","code":"M"},{"id":"p2","marks":1,"description":"Correct evaluation of Line P's gradient as 0.5 or 1/2.","code":"A"},{"id":"p3","marks":1,"description":"Correct calculation for Line Q's gradient, showing it is also 0.5 or 1/2.","code":"M"},{"id":"p4","marks":1,"description":"Conclusion stating $m_P = m_Q$ implies the lines are parallel.","code":"C"}],"finalAnswers":["Gradients are equal (1/2 or 0.5)","Lines are parallel"],"generatedAt":"2026-05-06T14:01:16.677Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"compare_gradients","label":"Compare the gradients","steps":[{"type":"text","order":0,"title":"Recall the gradient test","content":"For two points $(x_1

Parallel and Perpendicular Lines

Question 9

Parallel and Perpendicular Lines Questions

Parallel and Perpendicular Lines

Question 9

Parallel and Perpendicular Lines Questions