Find the equation of a straight line parallel to the line joining the points (7,5) and (1,3) and passing through the point (–3,4).

Line joining the points (7,5) and (1,3) and passing through the point \$(–3,4)\$.

Slope of the straight line joining the points \$(7,5)\$ and \$(1,3)\$ is

\begin{align} \require{cancel} m_1 &= \frac{y_2 - y_1}{x_2 - x_1} \\ m_1 &= \frac{3 - 5}{1 - 7} \\ &= \frac{-2}{-6} \\ &= \frac{1}{3} \\ \end{align}

If two straight lines are parallel then their slopes are equal.

\$∴\$ Slope of the required line is

\$m_2 = \frac{1}{3}\$

The equation of a straight line parallel to the line joining the points (7,5) and (1,3) and passing through the point (–3,4) is

\begin{align} \require{cancel} && y - y_1 &= m_2(x – x_1) \\ \Rightarrow && y - 4 &= \frac{1}{3}(x - (-3) ) \\ \Rightarrow && 3y - 12 &= x + 3 \\ \Rightarrow && x - 3y + 15 &= 0 \\ \end{align}

Answer

The equation of a straight line parallel to the line joining the points (7,5) and (1,3) and passing through the point (–3,4) is \$ x - 3y + 15 = 0\$