Question

Q 2: AB is a line segment of length 8 cm. Locate a

point C on AB such that AC = 1

/3

CB.​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Answer:

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MATHS

Draw a line segment AB of length 10 cm.Mark a point P on AB such That AP = 4 cm. Draw a line through P perpendicular to AB.

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ANSWER

Step I: We draw line L and take a point A on it.

Step II: Using a ruler and a compass, we mark a point B, 10 cm from A, on the line L. AB is the required line segment of 10 cm.

Step III: Again, we mark a point P, which is 4 cm from A, in the direction of B.

Step IV: With P as centre, take a radius of 4 cm and construct an arc intersecting the line L at two points A and E.

Step V: With A and E as centres, take a radius of 6 cm and construct two arcs intersecting each other at R.