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$c$.","mcqOptions":[{"text":"gradient","correct":true,"explanation":"In $y=mx+c$, $m$ measures the change in $y$ for each unit change in $x$, so it represents the line’s gradient."},{"text":"y-intercept","correct":false,"explanation":"The y-intercept is represented by $c$, not $m$. The coefficient $m$ gives the gradient, which measures the line’s rate of change."},{"text":"x-intercept","correct":false,"explanation":"The x-intercept is found by setting $y=0$ and is not generally represented by $m$. In $y=mx+c$, $m$ represents the gradient."},{"text":"constant","correct":false,"explanation":"The constant is $c$, the term independent of $x$. The coefficient $m$ represents the gradient because it determines the line’s steepness and direction."}]},{"id":"17741f8b-ef6c-46ef-aae2-76c9a02d8153","cardType":"cloze","points":null,"front":"In the straight-line equation $y=mx+c$, the gradient is represented by $m$ and the [ ] is represented by $c$.","back":"In the straight-line equation $y=mx+c$, the gradient is represented by $m$ and the y-intercept is represented by $c$.","mcqOptions":[{"text":"y-intercept","correct":true,"explanation":"In $y=mx+c$, setting $x=0$ gives $y=c$. Therefore, $c$ represents the y-intercept, where the line crosses the y-axis."},{"text":"gradient","correct":false,"explanation":"The gradient is represented by $m$, not $c$. The value $c$ is the y-intercept because it is the value of $y$ when $x=0$."},{"text":"x-intercept","correct":false,"explanation":"The x-intercept is found by setting $y=0$ and solving for $x$; it is not represented directly by $c$. Here, $c$ is the y-intercept."},{"text":"origin","correct":false,"explanation":"The origin is the point $(0,0)$ and is not represented by $c$ unless $c=0$. In this equation, $c$ represents the y-intercept."}]},{"id":"79d29c10-a9fc-4baa-adfe-dcedbcb2b136","cardType":"basic","points":2,"front":"How do you find the gradient of a line given the equation $2x+5y=10$?","back":"Rearrange the equation to make $y$ the subject ($y=mx+c$) before reading the coefficient of $x$.","mcqOptions":[{"text":"Rearrange the equation to make $y$ the subject ($y=mx+c$) before reading the coefficient of $x$.","correct":true,"explanation":"Rearranging into $y=mx+c$ isolates $y$, making the gradient visible as the coefficient of $x$. Here, division by 5 gives a gradient of $-2/5$."},{"text":"Read the coefficient of $x$ directly from the equation, so the gradient is $2$.","correct":false,"explanation":"Reading 2 directly ignores the coefficient of $y$ and the rearrangement needed. Making $y$ the subject gives $y=-2x/5+2$, so the gradient is $-2/5$."},{"text":"Divide the coefficient of $x$ by $5$, so the gradient is $2/5$.","correct":false,"explanation":"Dividing the coefficient of $x$ by 5 misses the negative sign introduced when $2x$ moves across the equals sign. The correct gradient is $-2/5$."},{"text":"Set $x=0$ and solve for $y$ before reading the resulting value as the gradient.","correct":false,"explanation":"Setting $x=0$ finds the y-intercept rather than the gradient. Instead, rearrange to $y=-2x/5+2$ and read the gradient as $-2/5$."}]},{"id":"c236471c-3f8b-4f5f-8763-59fe0651d8fd","cardType":"basic","points":2,"front":"What is the gradient and y-intercept of the line $6y=3x-12$?","back":"

Gradient: $1/2$
y-intercept: $-2$

","mcqOptions":[{"text":"

Gradient: $1/2$
y-intercept: $-2$

","correct":true,"explanation":"Dividing every term by 6 gives $y=(3/6)x-12/6=(1/2)x-2$. Thus the coefficient of $x$ is $1/2$ and the y-intercept is $-2$."},{"text":"

Gradient: $3$
y-intercept: $-12$

","correct":false,"explanation":"This reads the coefficients on the right without dividing them by 6. Dividing the entire equation by 6 gives a gradient of $1/2$ and a y-intercept of $-2$."},{"text":"

Gradient: $1/2$
y-intercept: $2$

","correct":false,"explanation":"The gradient has been divided correctly, but the intercept’s sign has been lost. Since $-12/6=-2$, the gradient is $1/2$ and the y-intercept is $-2$."},{"text":"

Gradient: $2$
y-intercept: $-2$

","correct":false,"explanation":"This reverses the gradient fraction when dividing by 6. Since $3/6=1/2$ and $-12/6=-2$, the correct values are $1/2$ and $-2$."}]},{"id":"ad5ddfd5-f1ab-438b-a6f1-002667e247f9","cardType":"basic","points":1,"front":"What is the relationship between the gradients of two parallel lines?","back":"They have the same gradient.","mcqOptions":[{"text":"They have the same gradient.","correct":true,"explanation":"Parallel lines have identical steepness and direction, so their gradients are equal. Their y-intercepts may differ, allowing them to remain distinct without meeting."},{"text":"They have negative reciprocal gradients.","correct":false,"explanation":"Negative reciprocal gradients describe perpendicular lines, not parallel lines. Parallel lines point in the same direction and therefore have the same gradient."},{"text":"They have opposite gradients.","correct":false,"explanation":"Opposite gradients generally produce lines with different directions that intersect. Parallel lines must have equal gradients, although their y-intercepts are different for distinct lines."},{"text":"They have reciprocal gradients.","correct":false,"explanation":"Reciprocal gradients are not generally equal and do not ensure that lines remain the same distance apart. Parallel lines must have the same gradient."}]},{"id":"ce501f26-e240-4ed1-8fec-c585f4a4fa49","cardType":"basic","points":1,"front":"A line is parallel to $y=3x+1$ and has no specific point given. Give one possible equation.","back":"Any equation in the form $y=3x+c$ where $c \\neq 1$ (e.g. $y=3x-4$).","mcqOptions":[{"text":"Any equation in the form $y=3x+c$ where $c \\neq 1$ (e.g. $y=3x-4$).","correct":true,"explanation":"A parallel line must retain the gradient 3 while having a different y-intercept from the original line. Therefore, any $y=3x+c$ with $c\\neq1$ is suitable."},{"text":"Any equation in the form $y=-\\frac{1}{3}x+c$ where $c \\neq 1$ (e.g. $y=-\\frac{1}{3}x-4$).","correct":false,"explanation":"Changing the gradient to $-1/3$ produces a line perpendicular to the original, not parallel. A distinct parallel line has form $y=3x+c$ with $c\\neq1$."},{"text":"Any equation in the form $y=\\frac{1}{3}x+c$ where $c \\neq 1$ (e.g. $y=\\frac{1}{3}x-4$).","correct":false,"explanation":"Using $1/3$ confuses the gradient with its reciprocal, so the resulting line is not parallel. The gradient must stay 3, giving $y=3x+c$ with $c\\neq1$."},{"text":"Any equation in the form $y=-3x+c$ where $c \\neq 1$ (e.g. $y=-3x-4$).","correct":false,"explanation":"Changing the sign of the gradient changes the line’s direction, so it will intersect the original. A parallel line requires gradient 3 and a different intercept."}]},{"id":"e537bc58-079c-491a-8d9c-f03cd3041d31","cardType":"basic","points":1,"front":"What is the relationship between the gradients $m_1$ and $m_2$ of two perpendicular lines?","back":"They are negative reciprocals ($m_1 \\times m_2 = -1$).","mcqOptions":[{"text":"They are negative reciprocals ($m_1 \\times m_2 = -1$).","correct":true,"explanation":"Perpendicular non-vertical lines have gradients whose product is $-1$. Consequently, each gradient is the negative reciprocal of the other."},{"text":"They are equal gradients ($m_1 \\times m_2 = 1$).","correct":false,"explanation":"Equal gradients make two distinct lines parallel, not perpendicular. For perpendicular lines, the gradients are negative reciprocals and satisfy $m_1\\times m_2=-1$."},{"text":"They are positive reciprocals ($m_1 \\times m_2 = 1$).","correct":false,"explanation":"Taking only the reciprocal omits the required sign change. Perpendicular gradients are negative reciprocals, so their product must be $-1$, not $1$."},{"text":"They are opposite gradients ($m_1 + m_2 = 0$).","correct":false,"explanation":"Simply changing the sign does not usually create a right angle. Perpendicular gradients require both reversing the fraction and changing its sign, giving product $-1$."}]},{"id":"c0a31420-9946-4b8e-8400-6da5e2c48f37","cardType":"basic","points":1,"front":"What is the negative reciprocal of 4?","back":"$-1/4$","mcqOptions":[{"text":"$-1/4$","correct":true,"explanation":"Taking a negative reciprocal means changing the sign and inverting the number. Since $4=4/1$, its negative reciprocal is $-1/4$."},{"text":"$1/4$","correct":false,"explanation":"This takes the reciprocal but misses the required sign change. Writing $4$ as $4/1$, then inverting and negating, gives $-1/4$."},{"text":"$-4$","correct":false,"explanation":"This changes the sign but does not take the reciprocal. The fraction $4/1$ must also be inverted, making the negative reciprocal $-1/4$."},{"text":"$-1/16$","correct":false,"explanation":"Dividing by 4 again confuses the reciprocal with squaring the denominator. Inverting $4/1$ gives $1/4$, then changing the sign gives $-1/4$."}]},{"id":"0a1f86a9-b079-4682-837f-12c5d5f6c26b","cardType":"basic","points":1,"front":"What is the negative reciprocal of $-2/3$?","back":"$3/2$","mcqOptions":[{"text":"$3/2$","correct":true,"explanation":"Taking the reciprocal of $-2/3$ gives $-3/2$, and changing its sign gives $3/2$. Its product with the original gradient is $-1$, as required."},{"text":"$2/3$","correct":false,"explanation":"This changes the sign but does not take the reciprocal. The fraction must first become $-3/2$ and then change sign, giving the negative reciprocal $3/2$."},{"text":"$-3/2$","correct":false,"explanation":"This takes the reciprocal but fails to change the sign. A negative reciprocal requires both operations, so the correct result is $3/2$, not $-3/2$."},{"text":"$-2/3$","correct":false,"explanation":"This leaves the original value unchanged. The reciprocal of $-2/3$ is $-3/2$, and reversing its sign gives the required answer $3/2$."}]},{"id":"fcab2999-9f6d-4583-85ad-b80ae37eba05","cardType":"cloze","points":null,"front":"To find the gradient of a line perpendicular to $y=mx+c$, you must [ ] and change the sign.","back":"To find the gradient of a line perpendicular to $y=mx+c$, you must flip the fraction and change the sign.","mcqOptions":[{"text":"flip the fraction","correct":true,"explanation":"Perpendicular non-vertical gradients multiply to $-1$, so the original gradient’s numerator and denominator must be interchanged before its sign is reversed."},{"text":"square the fraction","correct":false,"explanation":"Squaring the gradient does not produce a perpendicular gradient. You must flip the fraction and then change its sign so that the two gradients multiply to $-1$."},{"text":"simplify the fraction","correct":false,"explanation":"Simplifying changes only the fraction’s form, not its value, so it cannot create the required perpendicular gradient. The correct operation is to flip the fraction before changing the sign."},{"text":"double the fraction","correct":false,"explanation":"Doubling the gradient is not the negative-reciprocal operation and will not generally make the product of the gradients $-1$. You must flip the fraction and change its sign."}]},{"id":"c620a9ee-bc09-46b4-a567-e29c3ee5f168","cardType":"cloze","points":null,"front":"To find the gradient of a line perpendicular to $y=mx+c$, you must flip the fraction and [ ].","back":"To find the gradient of a line perpendicular to $y=mx+c$, you must flip the fraction and change the sign.","mcqOptions":[{"text":"change the sign","correct":true,"explanation":"After reversing the numerator and denominator, changing the sign makes the new gradient the negative reciprocal. The original and perpendicular gradients therefore multiply to $-1$."},{"text":"keep the sign","correct":false,"explanation":"Keeping the sign produces the reciprocal rather than the negative reciprocal, so the lines are not generally perpendicular. After flipping the fraction, you must change the sign."},{"text":"double the denominator","correct":false,"explanation":"Changing the denominator alone does not reliably form the negative reciprocal. After flipping the whole fraction, the required final operation is to change the sign."},{"text":"square the result","correct":false,"explanation":"Squaring the gradient does not create a negative reciprocal and will not make the gradients multiply to $-1$. The correct final step is to change the sign."}]},{"id":"b6d639cb-176d-4add-b46e-3ec9bbbbb1c4","cardType":"basic","points":1,"front":"What type of line is perpendicular to the horizontal line $y=4$?","back":"A vertical line (e.g. $x=-1$).","mcqOptions":[{"text":"A vertical line (e.g. $x=-1$).","correct":true,"explanation":"Every point on $y=4$ has the same $y$-coordinate, so the line is horizontal. A line meeting it at a right angle must therefore be vertical."},{"text":"A horizontal line (e.g. $y=-1$).","correct":false,"explanation":"Another horizontal line would be parallel to $y=4$, not perpendicular to it. A line perpendicular to a horizontal line must be vertical, such as $x=-1$."},{"text":"A diagonal line (e.g. $y=x$).","correct":false,"explanation":"A line with positive gradient is slanted and does not generally meet a horizontal line at $90°$. The perpendicular line must be vertical, such as $x=-1$."},{"text":"A parallel line (e.g. $y=-1$).","correct":false,"explanation":"A line parallel to $y=4$ is also horizontal, so the two lines never meet. Its perpendicular must instead be vertical, such as $x=-1$."}]},{"id":"f74227d0-5abc-4a4d-a86b-7635a7c734cb","cardType":"basic","points":1,"front":"State the formula for the gradient ($m$) between two points $(x_1, y_1)$ and $(x_2, y_2)$.","back":"$$m = \\frac{y_2 - y_1}{x_2 - x_1}$$","mcqOptions":[{"text":"$$m = \\frac{y_2 - y_1}{x_2 - x_1}$$","correct":true,"explanation":"Gradient is the change in vertical coordinate divided by the corresponding change in horizontal coordinate. Subtracting both coordinate pairs in the same order gives this formula."},{"text":"$$m = \\frac{x_2 - x_1}{y_2 - y_1}$$","correct":false,"explanation":"This reverses rise and run, giving the reciprocal of the gradient. Gradient must be change in $y$ divided by change in $x$, using a consistent subtraction order."},{"text":"$$m = \\frac{y_2 + y_1}{x_2 + x_1}$$","correct":false,"explanation":"Adding the coordinates does not measure the change between the two points. Gradient uses their differences: $y_2-y_1$ divided by $x_2-x_1$."},{"text":"$$m = \\frac{y_2 - y_1}{x_1 - x_2}$$","correct":false,"explanation":"The numerator and denominator use opposite subtraction orders here, incorrectly reversing the gradient’s sign. Both differences must follow the same point order."}]},{"id":"00bfa38c-b2bb-461a-a28a-7647319cfd8b","cardType":"cloze","points":null,"front":"When calculating gradient, the [ ] must be the numerator and the change in $x$ must be the denominator.","back":"When calculating gradient, the change in $y$ must be the numerator and the change in $x$ must be the denominator.","mcqOptions":[{"text":"change in $y$","correct":true,"explanation":"Gradient measures rise divided by run, so the vertical change, or change in $y$, belongs in the numerator above the horizontal change in $x$."},{"text":"change in $x$","correct":false,"explanation":"Putting the change in $x$ in the numerator reverses rise over run and gives the reciprocal gradient. The numerator must be the change in $y$."},{"text":"sum of the $y$-values","correct":false,"explanation":"The sum of the $y$-coordinates does not measure vertical displacement between the points. The numerator must be their difference, which is the change in $y$."},{"text":"sum of the $x$-values","correct":false,"explanation":"The sum of the $x$-coordinates is unrelated to the line’s vertical rise. The gradient’s numerator is the change in $y$, with change in $x$ below it."}]},{"id":"c86ac74a-c2aa-4579-98a1-3d03a9302322","cardType":"cloze","points":null,"front":"When calculating gradient, the change in $y$ must be the numerator and the [ ] must be the denominator.","back":"When calculating gradient, the change in $y$ must be the numerator and the change in $x$ must be the denominator.","mcqOptions":[{"text":"change in $x$","correct":true,"explanation":"Gradient is rise over run, so the horizontal change, or change in $x$, forms the denominator beneath the vertical change in $y$."},{"text":"change in $y$","correct":false,"explanation":"Using the change in $y$ in both positions does not calculate rise over run and would usually give $1$. The denominator must be the change in $x$."},{"text":"sum of the $x$-values","correct":false,"explanation":"Adding the $x$-coordinates does not measure horizontal displacement between the points. The denominator must be their difference, which is the change in $x$."},{"text":"sum of the $y$-values","correct":false,"explanation":"The sum of the $y$-coordinates is not the line’s horizontal run. Gradient requires the change in $x$ as the denominator beneath the change in $y$."}]},{"id":"a15b324a-91c4-4f30-a67d-f80af30942bd","cardType":"basic","points":2,"front":"How do you find the y-intercept ($c$) when you know the gradient ($m$) and a point $(x, y)$?","back":"Substitute the values of $m$, $x$, and $y$ into $y=mx+c$ and solve for $c$.","mcqOptions":[{"text":"Substitute the values of $m$, $x$, and $y$ into $y=mx+c$ and solve for $c$.","correct":true,"explanation":"The known point satisfies the line’s equation, so inserting its $x$- and $y$-coordinates and the known gradient leaves $c$ as the only unknown to solve."},{"text":"Multiply the values of $m$, $x$, and $y$ together and use the result as $c$.","correct":false,"explanation":"Multiplying all three known values does not use the equation of the line and cannot isolate the intercept. Substitute them into $y=mx+c$ and solve for $c$."},{"text":"Substitute the values of $m$ and $x$ into $y=mx$ and use the result as $c$.","correct":false,"explanation":"The product $mx$ gives the line’s change from its intercept, not the intercept itself. Substitute into $y=mx+c$ and rearrange to obtain $c=y-mx$."},{"text":"Divide the value of $y$ by $x$ and subtract the value of $m$ to find $c$.","correct":false,"explanation":"Dividing $y$ by $x$ attempts to find a gradient from the origin and ignores the known gradient. Substitute into $y=mx+c$ and solve for $c$."}]},{"id":"a965a45f-c4fd-4e16-ad8a-bc1b1f930799","cardType":"basic","points":2,"front":"Find the equation of a line parallel to $y=-3x+8$ passing through $(0, 5)$.","back":"$y=-3x+5$","mcqOptions":[{"text":"$y=-3x+5$","correct":true,"explanation":"Parallel lines have the same gradient, so the gradient remains $-3$. Passing through $(0,5)$ means the $y$-intercept is $5$, giving $y=-3x+5$."},{"text":"$y=3x+5$","correct":false,"explanation":"Changing the sign of the gradient produces a non-parallel line. The gradient must remain $-3$, while the point $(0,5)$ gives an intercept of $5$, so the answer is $y=-3x+5$."},{"text":"$y=-3x+8$","correct":false,"explanation":"Keeping the original equation ignores the new point: when $x=0$, this line gives $y=8$, not $5$. The required parallel line is $y=-3x+5$."},{"text":"$y=-1/3x+5$","correct":false,"explanation":"Taking the reciprocal of the gradient is not the rule for parallel lines. Parallel gradients are equal, so the gradient is $-3$ and the equation is $y=-3x+5$."}]},{"id":"e937b6a5-0f34-44ff-a4db-7fcdf6f20d43","cardType":"basic","points":2,"front":"Find the equation of a line perpendicular to $y=1/4x-2$ passing through $(0, -6)$.","back":"$y=-4x-6$","mcqOptions":[{"text":"$y=-4x-6$","correct":true,"explanation":"A perpendicular gradient is the negative reciprocal of $1/4$, which is $-4$. Since the line passes through $(0,-6)$, its intercept is $-6$."},{"text":"$y=4x-6$","correct":false,"explanation":"This takes the reciprocal but forgets to change its sign. The negative reciprocal of $1/4$ is $-4$, so the required equation is $y=-4x-6$."},{"text":"$y=-1/4x-6$","correct":false,"explanation":"Merely changing the gradient’s sign does not make lines perpendicular. The negative reciprocal of $1/4$ is $-4$, giving the equation $y=-4x-6$."},{"text":"$y=-4x-2$","correct":false,"explanation":"The gradient $-4$ is correct, but retaining the original intercept ignores the given point. Passing through $(0,-6)$ requires an intercept of $-6$, so the equation is $y=-4x-6$."}]},{"id":"11d45c36-c126-4564-b46e-c739ca5c08f2","cardType":"basic","points":2,"front":"Line A has gradient 3. Line B passes through $(-2, 1)$ and $(0, 7)$. Are they parallel?","back":"Yes. Line B gradient is $\\frac{7-1}{0-(-2)} = 3$. Gradients are equal.","mcqOptions":[{"text":"Yes. Line B gradient is $\\frac{7-1}{0-(-2)} = 3$. Gradients are equal.","correct":true,"explanation":"Line B’s gradient is the change in $y$ divided by the change in $x$: $6/2=3$. This equals Line A’s gradient, so the lines are parallel."},{"text":"No. Line B gradient is $\\frac{7-1}{0-(-2)} = 6$. Gradients are not equal.","correct":false,"explanation":"This divides incorrectly: $\\frac{7-1}{0-(-2)}=\\frac{6}{2}=3$, not $6$. Both lines therefore have gradient $3$, so they are parallel."},{"text":"No. Line B gradient is $\\frac{0-(-2)}{7-1} = 1/3$. Gradients are not equal.","correct":false,"explanation":"This reverses the gradient formula by calculating change in $x$ over change in $y$. Gradient is change in $y$ over change in $x$, giving $3$, so the lines are parallel."},{"text":"No. Line B gradient is $\\frac{1-7}{0-(-2)} = -3$. Gradients are not equal.","correct":false,"explanation":"The $y$-differences and $x$-differences must be taken in the same direction. Using consistent subtraction gives $\\frac{7-1}{0-(-2)}=3$, so the lines are parallel."}]},{"id":"f53a0b9b-6d15-4132-91e3-c6d3eb75bacb","cardType":"basic","points":2,"front":"How do you prove two lines are perpendicular using their gradients?","back":"Show that the product of the two gradients is $-1$.","mcqOptions":[{"text":"Show that the product of the two gradients is $-1$.","correct":true,"explanation":"Perpendicular non-vertical lines have gradients that are negative reciprocals. Multiplying a gradient by its negative reciprocal gives $-1$, which proves the lines are perpendicular."},{"text":"Show that the product of the two gradients is $1$.","correct":false,"explanation":"A product of $1$ means the gradients are reciprocals with the same sign, not negative reciprocals. For perpendicular lines, their gradient product must be $-1$."},{"text":"Show that the sum of the two gradients is $-1$.","correct":false,"explanation":"Adding the gradients does not test perpendicularity. The gradients must be negative reciprocals, which is shown by their product being $-1$, not their sum."},{"text":"Show that the difference of the two gradients is $-1$.","correct":false,"explanation":"The difference between gradients has no general perpendicularity rule. Perpendicular gradients are negative reciprocals, so multiplying them must produce $-1$."}]},{"id":"a7f0b032-b9fe-41c7-8710-962f3ae14c63","cardType":"basic","points":1,"front":"What is the gradient of a line perpendicular to $y=3/5x-2$?","back":"$-5/3$","mcqOptions":[{"text":"$-5/3$","correct":true,"explanation":"The perpendicular gradient is the negative reciprocal of $3/5$. Swapping numerator and denominator and changing the sign gives $-5/3$."},{"text":"$5/3$","correct":false,"explanation":"This takes the reciprocal but fails to change the sign. A perpendicular gradient is the negative reciprocal of $3/5$, so it is $-5/3$."},{"text":"$-3/5$","correct":false,"explanation":"This changes only the sign rather than taking the negative reciprocal. Swapping $3$ and $5$ as well as changing the sign gives $-5/3$."},{"text":"$3/5$","correct":false,"explanation":"Keeping the same gradient would make the lines parallel, not perpendicular. The negative reciprocal of $3/5$ is $-5/3$."}]},{"id":"4d52c9b5-1444-4cad-8c61-eb7f0b67c1a3","cardType":"basic","points":2,"front":"Rearrange $2x+3y=9$ into the form $y=mx+c$.","back":"$y = -2/3x + 3$","mcqOptions":[{"text":"$y = -2/3x + 3$","correct":true,"explanation":"Subtracting $2x$ gives $3y=-2x+9$. Dividing every term by $3$ then gives $y=-2/3x+3$."},{"text":"$y = 2/3x + 3$","correct":false,"explanation":"Moving $2x$ to the other side changes its sign. From $3y=-2x+9$, dividing by $3$ gives $y=-2/3x+3$."},{"text":"$y = -2/3x + 9$","correct":false,"explanation":"This divides the $x$ coefficient by $3$ but not the constant. Every term in $3y=-2x+9$ must be divided by $3$, giving $y=-2/3x+3$."},{"text":"$y = -3/2x + 3$","correct":false,"explanation":"This incorrectly reverses the coefficient when dividing by $3$. Since $3y=-2x+9$, division gives an $x$ coefficient of $-2/3$, so $y=-2/3x+3$."}]},{"id":"f1826821-941b-4143-bbe8-1d0739c645aa","cardType":"basic","points":2,"front":"Line A passes through $(-1, 2)$ and $(3, 4)$. What is its gradient?","back":"$1/2$ (Calculated as $\\frac{4-2}{3-(-1)}$)","mcqOptions":[{"text":"$1/2$ (Calculated as $\\frac{4-2}{3-(-1)}$)","correct":true,"explanation":"The change in $y$ is $4-2=2$ and the change in $x$ is $3-(-1)=4$. Therefore the gradient is $2/4=1/2$."},{"text":"$2$ (Calculated as $\\frac{3-(-1)}{4-2}$)","correct":false,"explanation":"This reverses the gradient formula by dividing change in $x$ by change in $y$. Gradient is change in $y$ over change in $x$, so it is $2/4=1/2$."},{"text":"$1/4$ (Calculated as $\\frac{4-2}{3-(-1)}$)","correct":false,"explanation":"The differences are $2$ and $4$, but $2/4$ simplifies to $1/2$, not $1/4$. Therefore the line’s gradient is $1/2$."},{"text":"$-1/2$ (Calculated as $\\frac{2-4}{3-(-1)}$)","correct":false,"explanation":"Subtracting the $y$-coordinates in one direction while subtracting the $x$-coordinates in the opposite direction creates an incorrect negative sign. Consistent subtraction gives a gradient of $1/2$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Parallel and Perpendicular Lines","title":"Parallel and Perpendicular Lines Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"8ac75b46-6b9d-45dd-86c7-369c2e941b54","walled":"$L39"}],"$L3a","$L3b"]