k (**M1**).","highlights":[{"text":"passes through $(2, 7)$","location":"specification"}]},{"type":"text","order":3,"title":"State final equation","content":"Finally, we substitute the value of $c = 4$ back into our equation to get the full equation of the line:\n\n$$y = \\frac{3}{2}x + 4$$\n\nThis final correct equation earns you the third mark (**A1**)."}]},{"id":"748ac42d-1426-4a13-9b6a-22504135cc91:point_slope","label":"Point-Slope Method","steps":[{"type":"text","order":0,"title":"Find the gradient of line A","content":"To find the gradient ($m$) of Line A, we use the gradient formula for two points $(x_1, y_1)$ and $(x_2, y_2)$:\n\n$$m = \\frac{y_2 - y_1}{x_2 - x_1}$$\n\nSubstituting the given points $(-2, 1)$ and $(4, 10)$:\n\n$$m = \\frac{10 - 1}{4 - (-2)} = \\frac{9}{6} = \\frac{3}{2}$$\n\nThis earns the first mark M1.","highlights":[{"text":" $(-2, 1)$ and $(4, 10)$","location":"specification"}]},{"type":"text","order":1,"title":"Determine the parallel gradient","content":"Since the line we want to find is **parallel** to Line A, it must have the same gradient.\n\nTherefore, the gradient of our new line is also:\n\n$$m = \\frac{3}{2}$$","highlights":[{"text":"parallel to A","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"Now, we use the point-slope formula with the point $(x_1, y_1) = (2, 7)$ and the gradient $m = \\frac{3}{2}$:"},{"latex":"$$\\htmlData{anchor=y-var}{y} - \\htmlData{anchor=y1-val}{\\color{@sky}{7}} = \\htmlData{anchor=m-val}{\\color{@amber}{\\frac{3}{2}}}\\left(\\htmlData{anchor=x-var}{x} - \\htmlData{anchor=x1-val}{\\color{@purple}{2}}\\right)$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y1-val"},{"color":"amber","style":"text-color","anchor":"m-val"},{"color":"purple","style":"text-color","anchor":"x1-val"}],"connections":[]},"order":2,"title":"Apply point-slope formula","description":"We substitute our gradient and point directly into the point-slope template $y - y_1 = m(x - x_1)$."},{"type":"text","order":3,"title":"Simplify to final form","content":"Expand and simplify the equation to find the final line equation:\n\n$$\\begin{aligned} y - 7 &= \\frac{3}{2}(x - 2) \\\\[1ex] y - 7 &= \\frac{3}{2}x - 3 \\\\[1ex] y &= \\frac{3}{2}x + 4 \\end{aligned}$$\n\nThis step earns the final two marks M1 A1.","highlights":[{"text":"passes through (2, 7)","location":"specification"}]}]}],"generatedAt":"2026-06-12T15:50:38.063Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"803659de-f970-4e6f-8303-8ec103b02f49","revisionId":"89fbc148-d469-4a76-9f38-a4239d5ac96e","questionNumber":"15","questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":3,"sourceTitle":"Parallel and Perpendicular Lines","calculatorUse":"non_calc","parts":[{"id":"2fe1db51-6e18-4d76-a3c4-790bd6e4ebb1","label":"","content":"Line A passes through the points $(2, -5)$ and $(10, -1)$. Find the equation of the line perpendicular to A that passes through $(4, 3)$.","markscheme":"- Calculates gradient of A: $m = \\frac{-1 - (-5)}{10 - 2} = \\frac{4}{8} = \\frac{1}{2}$ M1\n- Identifies perpendicular gradient $m = -2$ and substitutes $(4, 3)$ into $y = -2x + c$: $3 = -2(4) + c \\Rightarrow 3 = -8 + c \\Rightarrow c = 11$ M1\n- Final equation $y = -2x + 11$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Calculates the gradient of Line A as $\\frac{1}{2}$ and identifies the perpendicular gradient as $-2$.","code":"M"},{"id":"p2","marks":1,"description":"Substitutes $(4, 3)$ into $y = -2x + c$ to find $c = 11$ and states the final equation.","code":"A"}],"finalAnswers":["y=-2x+11"],"generatedAt":"2026-05-05T15:38:06.273Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"gradient_and_intercept","label":"Gradient and intercept","steps":[{"type":"text","order":0,"title":"Find gradient A","content":"Use $m=\\dfrac{\\text{change in }y}{\\text{change i

Parallel and Perpendicular Lines

Question 17

Parallel and Perpendicular Lines Questions

Parallel and Perpendicular Lines

Question 17

Parallel and Perpendicular Lines Questions