Question

derive the formula for the perpendicular distance of a point (x1,y1)from the line Ax+By+c=0.(plz no spam only give the correct answer) ​

Answer 1

Answer:

Step-by-step explanation:

We will learn how to find the perpendicular distance of a point from a straight line.

Prove that the length of the perpendicular from a point (x1, y1) to a line ax + by + c = 0 is |ax1+by1+c|a2+b2√

Let AB be the given straight line whose equation is ax + by + c = 0 ………………… (i) and P (x1, y1) be the given point.

To find the length of the perpendicular drawn from P upon the line (i).

Firstly, we assume that the line ax + by + c = 0 meets x-axis at y = 0.

Therefore, putting y = 0 in ax + by + c = 0 we get ax + c = 0 ⇒ x = -ca.

Therefore, the coordinate of the point A where the line ax + by + c = 0 intersect at x-axis are (-ca, 0).

Answer 2

Given:

• Line segment: Ax+By+C = 0
• Points: ($x_1,y_1$)

To Find:

• The formula for the perpendicular distance of a given point from the given line segement.

Solution:

• Consider a line segment L on the graph passing x and y axes.
• Now draw another line that is perpendicular to line L.
• We get two points for x and y axes respectively:  Q(-C/A,0) and R(0,-C/B)
• This forms a triangle, hence by using the area formula we can find the perpendicular distance.
• Area of ΔPQR = $\frac{1}{2}*b*h$ , b - base, h -height

• ΔPQR = $\frac{1}{2}*QR*PM$
• Here we are finding PM as we need the perpendicular distance.
• PM = 2(area of ΔPQR)/QR  (1)
• Area of ΔPQR = $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
• $(x_1,y_1) = P , (x_2,y_2) = (-C/A,0), (x_3,y_3) = (0,-C/B)$  
• Area of ΔPQR =  $\frac{1}{2}(x_1(0+\frac{C}{B})-\frac{C}{A}(-\frac{C}{B}-y_1)+0)$ = $\frac{1}{2}(\frac{Cx_1}{B}+\frac{Cy_1}{A}+\frac{C^2}{AB})$
• 2*Area of ΔPQR = $|\frac{C}{AB}|Ax_1+By_1+C|$  (2)
• And QR = $\sqrt{(0+\frac{C}{A})^2+(\frac{C}{B}-0)^2 }$  = $\frac{C^2}{A^2}+\frac{C^2}{B^2} = |\frac{C}{AB}|\sqrt{A^2+B^2}$   (3)
• substitute (2) and (3) in (1) we get,
• $PM = \frac{|\frac{C}{AB}||Ax_1+By_1+C| }{|\frac{C}{AB}| \sqrt{A^2+B^2} }$  

∴ d = $\frac{|Ax_1+By_1+C| }{\sqrt{ A^2+B^2} }$ , perpendicular distance of a point from a given line segment.