Question

Line 1 passes through the points A(-15,-8) and B(-3,0). Line 2 has equation 3x + 2y - 35 = 0. Line 3 is parallel to line 1 and intersects line 2 when x = 5. Determine the equation of line 3. Express your answer in standard form: Ax + By + C = 0,

Answer 1

Given that;

Line 1 passes through the points A(-15,-8) and B(-3,0)

To find the slope, m, of the line, the formula is

`\begin{gathered} m=(y_2-y_1)/(x_2-x_1) \\ (x_1,y_1)=(-15,-8) \\ (x_2,y_2)=(-3,0) \end{gathered}`

Substituting the coordinates to find the slope of line 1

`\begin{gathered} m=(0-(-8))/(-3-(-15))=(8)/(-3+15)=(8)/(12)=(2)/(3) \\ m=(2)/(3) \end{gathered}`

Since, line 3 is parallel to line 1, then, they will have the same slope,

Thus, the slope pf line 3 is 2/3

Line 2 has equation 3x + 2y - 35 = 0.

Where line 3 intersects line 2 at x = 5, substitute for x into the equation of line 2 to find the value of y

`\begin{gathered} 3x+2y-35=0 \\ 3(5)+2y-35=0 \\ 15+2y-35=0 \\ -20=-2y \\ y=(-10)/(-2)=5 \\ y=5 \end{gathered}`

Where

The slope and the coordinates on line 3 are

`\begin{gathered} m=(2)/(3) \\ (x_1,y_1)=(5,10) \end{gathered}`

Applying the point-slope formula to find the equation of line 2

`\begin{gathered} y-y_1=m(x-x_1) \\ y-10=(2)/(3)(x-5) \\ 3(y-10)=2(x-5) \\ 3y-30=2x-10 \\ 2x-10-(3y-30)=0 \\ 2x-10-3y+30=0 \\ 2x-3y+20=0 \end{gathered}`

Hence, the equation of line 3 is

`2x-3y+20=0`