Question

Find the perpendicular distance from the point (3, -4) to the line 3x- 4y + 10 =0.​

Answer 1

Answer:

$distance = | \frac{3(3) - 4( - 4) + 10}{ \sqrt{( {3}^{2} + { ( - 4)}^{2} } } | \\ = | \frac{9 + 16 + 10}{ \sqrt{(9 + 16)} } | \\ = | \frac{35}{5} | \\ = |7| \\ distance = 7units$

Answer 2

ANSWER:

To Find:

• Perpendicular distance of Point A(3,-4) to the line 3x - 4y + 10 = 0

Solution:

$\text{\sf{We know that,}} \\\\\text{\sf{The perpendicular distance of a point(p,q) to a line Ax + By + C =0 is:}}\\\\:\longrightarrow\sf{d = \dfrac{\left|\,Ap+Bq+C\,\right|}{\sqrt{A^2+B^2\,\,}}}\\\\\text{\sf{Here, A = 3, B = -4, C = 10, p = 3 and q = -4}}\\\\\text{\sf{So,}}\\\\:\implies\sf{d = \dfrac{\left|\,Ap+Bq+C\,\right|}{\sqrt{A^2+B^2\,\,}}}\\\\:\implies\sf{d = \dfrac{\left|\,(3)(3)+(-4)(-4)+10\,\right|}{\sqrt{(3)^2+(-4)^2\,\,}}}\\\\:\implies\sf{d = \dfrac{\left|\,9+16+10\,\right|}{\sqrt{9+16\,\,}}}$

$:\implies\sf{d = \dfrac{\left|\,35\,\right|}{\sqrt{25\,\,}}}\\\\:\implies\sf{d = \dfrac{\,35\,}{\sqrt{5^2\,\,}}}\\\\:\implies\sf{d = \dfrac{35\!\!\!\!/^{\,\,\,7}}{5\!\!\!/_{\,\,\,1}}}\\\\:\implies\bf{d = 7units}$

Formula Used:

$\:\:\:\:\bullet\:\:\:\:\sf{d = \dfrac{\left|\,Ap+Bq+C\,\right|}{\sqrt{A^2+B^2\,\,}}}$