Solution:
Given line x – y = 5 ..(i)
2x + y + 6 = 0 ..(ii)
Slope of above line is m = 1
Parallel lines have same slope.
Equation line parallel to passing through (2, 3) and slope = 1 is
y – 3 = 1(x – 2)
=> y = x – 2 + 3
=> y = x + 1 ..(iii)
Solving (iii) and (ii) we get the intersection points of line (ii) and (iii)
2x + y + 6 = 0
x – y + 1 = 0
+> 3x = -7
=> x = -7/3
=> y = x+1
= -7/3 + 1
= -4/3
So the intersection point is (-7/3, -4/3).
Distance between (2, 3) and (-7/3, -4/3) can be calculated using distance formula.
Distance = √((-7/3)-2)2 + (-4/3 – 3)2)
= √((132/9) + (132/9))
= √(169 + 169)/9
= 13√2/3 units.
Tags:
Maths