CO-ORDINATE GEOMETRY Part 1
( Conformed 3 mark Question)
1. Show that the points (0, – 1), (2, 1), (0, 3) and (– 2, 1) are the corners of a square.
idKwau ik ibMdU (0, – 1), (2, 1), (0, 3) Aqy(– 2, 1) ie`k vrg dy isKr hn [
2. Prove that the points (6, 9), (0, 1) and (– 6, –7) are collinear.
is`D kro ibMdU (6, 9), (0, 1) Aqy (– 6, – 7) smryKI hn [
3. Find the value of k, if the point P (0, 2) is equidistant from (3, k) and (k, 5).
Kdw mu`l pqw kro jykr ibMdU P (0, 2) ibMdUAW(3, k) Aqy (k, 5) qoN brwbr durI qy hovy [
4. Prove that the points A (0, 1), B (1, 4), C (4, 3), D (3, 0) are the vertices of a square.
idKwau ik ibMdU A (0, 1), B (1, 4), C (4, 3), D (3, 0) ie`k vrg ABCD dy isKr hn [
5. Find the point on x-axis which is equidistant from (– 2, 5) and (2, – 3).
xDury qy auh ibMdU pqw kro jo ibMdUAW (3,4) Aqy (2,-3) qoN brwbr dUrI qy hY [
6. Find the point on the y-axis which is equidistant from (– 5, – 2) and (3, 2).
yDury qy auh ibMdU pqw kro jo ibMdUAW (-5,-2) Aqy (3,2) qoN brwbr dUrI qy hY [
7. Show that the points (1, 0), (0, 1) and (– 3, 4) lie on a straight line.
is`D kro ibMdU (1,0), (0, 1) Aqy (– 3, 4) smryKI hn [
8. Find a point on the x-axis which is equidistant from the points (5, 4) and (– 2, 3).
xDury qy auh ibMdU pqw kro jo ibMdUAW (5,4) Aqy (-2,3) qoN brwbr dUrI qy hY [
9. Find the co-ordinates of the point which divides the line joining the points (3, 5), (4, 2) internally in the ratio 3 : 2.
aus ibMdU dy inrdyS AMk pqw kro jo (3, 5), (4, 2) nMU imlwaux vwly ryKw KMf nMU AMdrUnI 3:2 iv`c vMfdw hY [
10. Find the middle point of the line joining (– 3, – 6) and (1, – 2).
ibMdU (-3,-6), (1,-2) nMU imlwaux vwly ryKw KMf dy m`D ibMdU pqw kro[
11. Find the co-ordinates of the points which divides internally the line joining A(– 1, 2) to B (4, – 5) in the ratio 2 : 3.
aus ibMdU dy inrdyS AMk pqw kro jo A(– 1, 2) qoN B (4, – 5) nMU imlwaux vwly ryKw KMf nMU AMdrUnI 2:3 iv`c vMfdw hY [
12. In what ratio is the line joining the points A(4, 4) and B(7, 7) divided by P (–1, – 1)?
ibMdUAW A(4, 4) , B (7, 7 ) nMU imlwaux vwly ryKw KMf nMU ibMdU P (– 1, – 1) iks Anupwq iv`c vMfdw hY [