Question

Find the distance from the point \(( - 2,3)\) to the line \(3x - 4y + 5 = 0\).

Short answer

Distance between Point\(( - 2,3)\)and line \(3x - 4y + 5 = 0\)

is\(\frac{{13}}{5}\).

Step-by-step solution

Formula used to find the distance from a point to the line

The expression to find the distance between point\(({x_1}{y_1})\)and line\((ax + by + c)\):

Distance\( = \frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}\) …… (1)

Calculation to find the distance

Given:

Point\(( - 2,3)\)and line

In equation (1), substitute \(3\) for \(a\), \( - 4\) for \(b\), \(5\) for \(c\), \( - 2\) for \({x_1}\) and 3 for \({y_1}\).

Distance:

\(\begin{aligned}{l} &= \frac{{|(3)( - 2) + ( - 4)(3) + (5)|}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }}\\ &= \frac{{| - 6 - 12 + 5|}}{{\sqrt {9 + 16} }}\\ &= \frac{{| - 13|}}{{\sqrt {25} }}\\ &= \frac{{13}}{5}\end{aligned}\)

Thus, the required distance is \(\frac{{13}}{5}\).