Question

find the distance between the line 4x + 3y + 4 = 0,and a point (i) (- 2,4) (ii) (7, - 3)

Answer 1

there is one formula to solve this problem
=ax1+by2+c/√a²+b²
then in this equation a=4 ,b=3,c=4
and the point x1=-2 y1=4 and you can put the value same as second point

Answer 2

Given:

The line 4x+3y+4=0 and the point (- 2,4), (7, - 3)

To find:

Distance between the line and point.

Solution:

Part(i):

Distance between the lineax+by+c=0 and the point($x_{1} ,y_{1}$) is ±(a$x_{1}$+b$y_{1}$+c)/$\sqrt{a^{2}+b^{2} }$

Distance between the line4x+3y+4=0 and the point(-2,4) is ±{4(-2)+3(4)+4}/$\sqrt{4^{2}+3^{2} }$

⇒distance=±(-8+12+4)/$\sqrt{25}$

Hence, Distance=8/5(∵distance is non-negative)

Part (ii):

Distance between the line4x+3y+4=0 and the point(7,-3) is ±{4(7)+3(-3)+4}/$\sqrt{4^{2}+3^{2} }$

⇒distance=±(28-9+4)/$\sqrt{25}$

Hence, Distance=23/5(∵distance is non-negative)