find coordinates point which divides the lines (-7,4) and (-6,-5) in ratio 7:2 .
Answers
Step-by-step explanation:
Lets call the point (-3,5 ) ‘A’.
Lets call the point (4,2) ‘B’.
Let the line AB intersect the y-axis at ‘C’.
Now we need to calculate the distances ‘AC’ and ‘CB’.
For that we need to find the coordinates of the point ‘C’.
The slope of ‘AB’ is (y2 - y1) / (x2 - x1) = (2–5) / (4- (-3) ) = -3 / 7.
The equation of a line can be represented as
y = mx + c
y = (-3/7)x + c. Lets put (4,2) in this equation.
(Since (4,2) lies on the line, it must satisfy the equation of the line.
=> 2 = -12/7 + c OR c = 26/7.
Hence the equation of the line is
y = (-3/7)x + 26/7.
OR 7y + 3x = 26. ……………. (1)
Check this for (4,2) .
LHS : 14 + 12 = 26 = RHS
Check this for (-3, 5)
LHS : 35 - 9 = 26 = RHS.
Hence the equation is correct.
Also, if we put x=0 in the equation (1) above, we can find the point of intersection of the line with the y-axis.
When we do that, we get y = 26/7.
Hence, A : (-3,5) ;; B : (4,2) ;; C : (0,26/7)
The distance between two points can be found out by the distance formula
Dist = sqrt [(y2 - y1)^2 + (x2 -x1)^2]
AB = sqrt [(7^2) + (-3)^2 ] = 7.615
AC = sqrt [(3)^2 + (5–26/7)^2 ] = 3.264
Hence, CB = 7.615 - 3.264 = 4.351
AC/AB = 3.264/7.615 = 0.4286
CB/AB = 1–0.4286 = 0.5714
OR, the y-axis divides the given line in the ratio of 42.86 : 57.14 of their lengths.