cheme":"- Point $A$ is $(-1, 6)$, gradient is $\\frac{1}{3}$ M1\n- $y - 6 = \\frac{1}{3}(x + 1)$ M1\n- $y = \\frac{1}{3}x + \\frac{19}{3}$ A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Identifies the coordinates of point $A$ or $B$ using $k=4$.","code":"M"},{"id":"p2","marks":1,"description":"Uses the gradient $\\frac{1}{3}$ and a point to form the equation.","code":"M"},{"id":"p3","marks":1,"description":"Provides the final equation $y = \\frac{1}{3}x + \\frac{19}{3}$.","code":"A"}],"finalAnswers":["y = 1/3x + 19/3"],"generatedAt":"2026-05-12T02:14:06.602Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"point_slope","label":"Use point–slope form","steps":[{"type":"text","order":0,"title":"Identify point A","content":"Using $k=4$ from the previous part, point $A$ is $(-1,6)$. Together with the given gradient $\frac13$, this earns the first M1.","highlights":[{"text":"Given the gradient of $AB$ is $\\frac{1}{3}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"y-y_1=\\htmlData{anchor=form}{m(x-x_1)}"},{"latex":"y-6=\\frac13\\htmlData{anchor=substitution}{\\bigl(x-(-1)\\bigr)}"},{"latex":"y-6=\\frac13\\htmlData{anchor=simplified}{(x+1)}"}],"highlights":[{"color":"teal","style":"text-color","anchor":"simplified"}],"connections":[{"to":"substitution","from":"form","color":"sky","label":"substitute"},{"to":"simplified","from":"substitution","color":"teal","label":"simplify x minus negative 1"}]},"order":1,"title":"Use the point","description":"Substitute the gradient and coordinates of point A into point-slope form. This gives the required point-slope equation and earns the second M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"y-6=\\frac{1}{3}\\htmlData{anchor=expand_start}{\\bigl(x+1\\bigr)}"},{"latex":"y-6=\\htmlData{anchor=expand_end}{\\frac{1}{3}x+\\frac{1}{3}}"},{"latex":"y=\\frac{1}{3}x+\\htmlData{anchor=add_six}{\\frac{1}{3}+6}"},{"latex":"y=\\frac{1}{3}x+\\htmlData{anchor=final_constant}{\\frac{19}{3}}"}],"highlights":[{"color":"green","style":"text-color-with-background","anchor":"final_constant"}],"connections":[{"to":"expand_end","from":"expand_start","color":"sky","label":"expand the bracket","style":"solid"},{"to":"add_six","from":"expand_end","color":"orange","label":"add 6 to both sides","style":"dashed"},{"to":"final_constant","from":"add_six","color":"teal","label":"simplify the constants","style":"dashed"}]},"order":2,"title":"Rearrange the equation","description":"Expand the bracket, then add 6 to both sides. The correct final equation earns A1."}]}],"generatedAt":"2026-07-21T04:08:08.165Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]},{"id":"5daee371-dbf3-4209-9aa2-7cd0bc8ea235","label":"c","content":"Find the equation of the perpendicular bisector of $AB$ and $B$.\nGive your answer in the form $ax + by + c = 0$","markscheme":"- Midpoint of $AB = (\\frac{-1+5}{2}, \\frac{6+8}{2}) = (2, 7)$ M1\n- Perpendicular gradient $= -3$ M1\n- $y - 7 = -3(x - 2)$ M1\n- $3x + y - 13 = 0$ A1","marks":4,"order":2,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Calculates the midpoint of $AB$.","code":"M"},{"id":"p2","marks":1,"description":"Determines the perpendicular gradient.","code":"M"},{"id":"p3","marks":1,"description":"Uses the midpoint and perpendicular gradient to form an equation.","code":"M"},{"id":"p4","marks":1,"description":"Rearranges to the form $3x + y - 13 = 0$.","code":"A"}],"finalAnswers":["3x + y - 13 = 0"],"generatedAt":"2026-05-12T02:14:06.602Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"midpoint_gradient","label":"Midpoint and perpendicular gradient","steps":[{"type":"text","order":0,"title":"Find k first","content":"First use the given gradient to find $k$, so the coordinates are fully determined for this part. The gradient formula gives\n\n$$\\frac{8-(k+2)}{(2k-3)-(-1)}=\\frac13.$$\n

Straight Line Graphs

Question 2

Straight Line Graphs Questions

Straight Line Graphs

Question 2

Straight Line Graphs Questions