Question

from Find the length of the perpendicular a point is given against to the following Straight lines.
i) x-3y-4=0 ; (0,0)

please I need answer ​

Answer 1

Step-by-step explanation:

3x+4y=10

We know that standard form of straight line ax+by+c=0, perpendicular distance is =

∣

∣

∣

∣

∣

a

2

+b

2

ax+by+c

∣

∣

∣

∣

∣

Given point is (4,5)

Therefore perpendicular distance =

∣

∣

∣

∣

∣

3

2

+4

2

4×3+4×5−10

∣

∣

∣

∣

∣

⇒

5

22

Answer 2

Answer:

x2 + 8x + 15 = x2 + 5x + 3x + 15 = (x + 3) (x + 5)

x2 + 3x – 10 = x2 + 5x – 2x – 10 = (x – 2) (x + 5)

Clearly, the common factor is x + 5.

Polynomials Class 9 Extra Questions Short Answer Type 1

Question 1.

Expand :

(i) (y – √3)2

(ii) (x – 2y – 3z)2

Solution: (i)

(y – √3)2 = y2 -2 × y × √3 + (√3)2 = y2 – 2√3 y + 3 (x – 2y – 3z)2

= x2 + 1 – 2y)2 + (-3z)2 + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x

= x2 + 4y2 + 9z2 – 4xy + 12yz – 6zx

Question 2.

If x + = 1x = 7, then find the value of x3 + 1x3

Solution:

We have x + 1x = 7

Cubing both sides, we have

Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers 2

Question 3.

Show that p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.

Solution:

Let f(p) = p10 + p8 + p6 – p4 – p2 – 1

Put p = 1, we obtain

f(1) = 110 + 18 + 16 – 14 – 12 – 1

= 1 + 1 + 1 – 1 – 1 – 1 = 0

Hence, p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.

Question 4.

If 3x + 2y = 12 and xy = 6, find the value of 27x3 + 8y3

Solution