Question

(8,1), (K-4), (2,-5)
Given points are collinear, then area of triangle is 0 sq.units
Sol: Given, (x,y)=(8,1),(xx,y)=(K,-4),(xx,y;)= (2,-5)
28
=03
5 1
-32-K-5K+8+2+40]=0=6K-18=0= 6k=18 = k=3
2
124​

Answer 1

Answer:

Draw the x and y axes. Mark the units along the x and y axes with a suitable scale.

(i) To plot the point A (3 , 5)

Here, the x-coordinate of A is 3 and the y-coordinate of A is 5. Both are positive. Hence the point A (3 , 5) lies in the quadrant I. Start at the Origin. Move three units to the right along the x-axis. Then turn and move 5 units up parallel to Y-axis and mark the point A (3 , 5).

(ii) To plot the point B (2 , 7)

Here, the x-coordinate of B is -2 which is negative and the y-coordinate of B is 7 which is positive. Hence the point B (-2 , 7) lies in the quadrant II. Start at the Origin. Move 2 units to the left along the x-axis. Then turn and move 7 units up parallel to y-axis and mark the point B (-2 , 7).

(iii) To plot the point C (-3 , -5) Here, the x-coordinate of C is -3 and the y-coordinate of C is -5. Both are negative. Hence the point C (-3 , -5) lies in the quadrant III. Start at the Origin. Move 3 units to the left along the x-axis. Then turn and move 5 units down parallel to y-axis. and mark the point C (-3, -5).

(iv) To plot the point D (2 , 7)

Here, the x-coordinate of the point D is 2 which is positive and the y-coordinate of D is -7 which is negative. Hence the point D (2 , -7) lies in the quadrant IV. Start at the Origin. Move 2 units to the right along the x-axis. Then turn and move 7 units down parallel to y-axis and mark the point D (2 , -7).

(v) To plot the point O (0, 0)

This is the origin. Both the x and y coordinates are zeros. It is the point of intersection of the axes x and y. Mark the point O (0,0).