# Find parametric equations for the line of slope 5 that passes through the point (7, 4). (Use t as...

## Question:

Find parametric equations for the line of slope 5 that passes through the point (7, 4). (Use t as your parametrized variable).

## Parametric Equation of the Line:

The procedure for finding the parametric equation of a line is to first find two points that allow you to establish its direction vector, which is a vector that is parallel to the line. With the vector direction and any point that passes through the line, we can replace them in the parametric equation formula of the line.

## Answer and Explanation:

{eq}\eqalign{ & {\text{The equation of the line in point - slope form is given by:}} \cr & \,\,\,\,\,\,\,y = m\left( {x - {x_0}} \right) + {y_0} \cr & {\text{where }}\,m\,\left[ {{\text{slope}}\,{\text{of}}\,{\text{the}}\,{\text{line}}} \right]\,{\text{ and }}\,P\left( {{x_0},{y_0}} \right)\,\left[ {{\text{point on the line}}} \right] \cr & {\text{In this case we have that the line that passes through the point }} \cr & P\left( {{x_0},{y_0}} \right) = P\left( {7,4} \right)\,{\text{ and has the slope }}m = 5.\; \cr & {\text{So point - slope form of the equation of the line is determined by:}} \cr & \,\,\,\,\,\,\,y - {y_0} = m\left( {x - {x_0}} \right) \cr & {\text{Thus}}{\text{, substituting the given values we have:}} \cr & \,\,\,\,\,\,\,y = 5\left( {x - 7} \right) + 4 \cr & \,\,\,\,\,\,\,y = 5x - 31 \cr & {\text{The parametric equation of the line with vector direction }}\,\vec v = \left\langle {{v_1},{v_2}} \right\rangle \,{\text{ }} \cr & {\text{and passing through point }}\,P\left( {{x_0},{y_0}} \right)\,{\text{ is given by:}} \cr & \,\,\ell \left( t \right)\,:\,\,\,\,\,\,\,\,x = {x_0} + t{v_1},\,\,\,\,y = {y_0} + t{v_2} \cr & {\text{In this particular we have the line }}\,\,y = 5x - 31,{\text{ then:}} \cr & {\text{for }}\,{x_1} = 0\,\,\, \Rightarrow {y_1} = 5\left( 0 \right) + 2 = - 31\,\,\,\, \Rightarrow Q\left( {0, - 31} \right) \cr & {\text{So the direction vector }}\,\vec v = \overrightarrow {PQ} \,\,\,{\text{of the line is given by:}} \cr & \,\,\,\vec v = \overrightarrow {PQ} = \left\langle {0 - 7, - 31 - 4} \right\rangle = \left\langle { - 7, - 35} \right\rangle \cr & {\text{Thus}}{\text{, using the vector }}\,\vec v = \left\langle {{v_1},{v_2}} \right\rangle {\text{ = }}\left\langle { - 7, - 35} \right\rangle \,{\text{ and the point}} \cr & P\left( {{x_0},{y_0}} \right){\text{ = }}\left( {7,4} \right){\text{ we can write the parametric equation:}} \cr & {\text{ }}\,\,\,\,\,\,\ell \left( t \right)\,:\,\,\,x = {x_0} + t{v_1},\,\,\,\,y = {y_0} + t{v_2} \cr & \Rightarrow \,\,\,\boxed{{\text{ }}\ell \left( t \right)\,:\,\,\,x = 7 - 7t,\,\,\,\,y = 4 - 35t} \cr} {/eq}