Question

the distance of the point 1,2 from the line x+y+5=0 measured along the line parallel to 3x-y=7 is equal to

Answer 1

Solution:

As you can see from the graph lines , x+y+5=0 and 3 x -y-7=0 are intersecting lines.

Substituting , y= -x-5 in  3 x -y -7 =0

→ 3 x -(-x -5) -7=0

→3 x +x+5-7=0

→4 x -2=0

→4 x =2

→x =2÷4

→x=1/2

y= -1/2-5=-11/2

The point of intersection of two lines are (1/2,-11/2).

Distance of point (a,b) from (p,q) is given by=$\sqrt{(a-p)^{2} +(b-q)^{2}}$

Distance of point (1,2) from (1/2,-11/2) is given by=$\sqrt{(1-1/2)^{2}+(2+11/2)^{2}}= \sqrt{(\frac{1}{2})^{2} + (\frac{15}{2})^{2}}  =\sqrt{\frac{1}{4} +\frac{225}{4}}= \sqrt{\frac{226}{4}}= \sqrt{56.5}$=7.51(approx)