Frequently Asked Questions on Chapter 7 Coordinate Geometry
Q1.(Coordinate Geometry)
Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3
Let P(x,y) be the required point.
By using the section formula,
x = (2×4 + 3×(-1))/(2+3)
= (8-3)/5
= 1
y = (2×(-3) + 3×7)/(2+3)
= (-6+21)/5
= 3
Hence, the point is (1,3).
Q2.(Coordinate Geometry)
Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6)
Consider the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k :1.
-1 = ( 6k-3)/(k+1)
–k – 1 = 6k -3
7k = 2
k = 2/7
Hence, the required ratio is 2:7.
Q3.(Coordinate Geometry)
In the given points (7, -2), (5, 1), (3, -k) find the value of ‘k’, for which the points are collinear.
For collinear points, the area of the triangle formed by them is always zero.
So, let the points (7, -2) (5, 1), and (3, k) be the vertices of a triangle.
Area of triangle = 1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0
7 – 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4
Hence, the value of k is 4.