Course Content
Class 10th Maths
0/86
Online Class For 10th Standard Students (CBSE) (English Medium)
About Lesson

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.

Exercise Files
No Attachment Found
No Attachment Found
Wisdom Academy