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Given one point and the gradient determine the equation of the line.

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The gradient intercept form of the line that passes through the point ({point_x},{point_y}) with a gradient of $\\simplify{{rise}/{run}}$ is $y=$ [[0]].

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There are two common ways to approach these questions:

\n
    \n
  1. Use the one point gradient formula $y-y_1=m(x-x_1)$. Or,

  2. \n
  3. Substitute the $x$ and $y$ values of the point and the value for $m$ into the gradient intercept form of a line $y=mx+b$ to determine $b$. Now you have $m$ and $b$ so you can write the equation of the line in the form $y=mx+b$.
  4. \n
\n

\n
\n

\n

For example, suppose we had to find the equation of the line through $(2,3)$ with a gradient of $\\frac{1}{4}$. The following are examples of using the above approaches:

\n
    \n
  1. Substitute $m=\\frac{1}{4}$ and $(x_1,y_1)=(2,3)$ into the equation $y-y_1=m(x-x_1)$. This gives $y-3=\\frac{1}{4}(x-2)$. Expanding the brackets we have $y-3=\\frac{x}{4}-\\frac{1}{2}$. Making $y$ the subject gives $y=\\frac{x}{4}+\\frac{5}{2}$.

  2. \n
  3. Take the point $(2,3)$ as $(x,y)$ and substitute this and the gradient into the equation $y=mx+b$. This gives $3=\\frac{1}{4}(2)+b$. Solving for $b$ gives $b=\\frac{5}{2}$. Therefore the equation of the line is $y=\\frac{x}{4}+\\frac{5}{2}$.
  4. \n
\n

\n
\n

\n

Recall $\\frac{x}{4}$ is the same as $\\frac{1}{4}x$.

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The gradient intercept form of the line that passes through the point ({bpoint_x},{bpoint_y}) with a gradient of $\\simplify{{brise}/{brun}}$ is $y=$ [[0]].

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There are two common ways to approach these questions:

\n
    \n
  1. Use the one point gradient formula $y-y_1=m(x-x_1)$. Or,

  2. \n
  3. Substitute the $x$ and $y$ values of the point and the value for $m$ into the gradient intercept form of a line $y=mx+b$ to determine $b$. Now you have $m$ and $b$ so you can write the equation of the line in the form $y=mx+b$.
  4. \n
\n

\n
\n

\n

For example, suppose we had to find the equation of the line through $(2,3)$ with a gradient of $-\\frac{1}{4}$. The following are examples of using the above approaches:

\n
    \n
  1. Substitute $m=-\\frac{1}{4}$ and $(x_1,y_1)=(2,3)$ into the equation $y-y_1=m(x-x_1)$. This gives $y-3=-\\frac{1}{4}(x-2)$. Expanding the brackets we have $y-3=-\\frac{x}{4}+\\frac{1}{2}$. Making $y$ the subject gives $y=-\\frac{x}{4}+\\frac{7}{2}$.

  2. \n
  3. Take the point $(2,3)$ as $(x,y)$ and substitute this and the gradient into the equation $y=mx+b$. This gives $3=-\\frac{1}{4}(2)+b$. Solving for $b$ gives $b=\\frac{7}{2}$. Therefore the equation of the line is $y=-\\frac{x}{4}+\\frac{7}{2}$.
  4. \n
\n

\n
\n

\n

Recall $-\\frac{x}{4}$ is the same as $-\\frac{1}{4}x$.

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