This page 10th grade math solution exercise 5.5 part1 is going to provide you solution for every problems that you find in the exercise no 5.5

(1) Find the slope of the straight line

(i) 3x + 4y -6 = 0

Solution:

Slope (m) = - coefficient of x/coefficient of y

= -3/4

(ii) y = 7x + 6

Solution:

y = 7 x + 6

The given equation is in the form y = mx + b. By comparing the given equation with this form,we get m = 7

(iii) 4 x = 5y + 3

Solution:

4 x - 5 y - 3 = 0

Slope (m) = - coefficient of x/coefficient of y

= -4/(-5)

= 4/5

(2) Show that the straight lines x + 2 y + 1 = 0 and 3x + 6y + 2 = 0 are parallel

Solution:

If two lines are parallel then slopes of those lines are equal.

m1 = m2

Slope of the first line x + 2 y + 1 = 0

Slope (m) = - coefficient of x/coefficient of y

m1 = -1/2

Slope of the second line 3x + 6y + 2 = 0

m2 = -3/6

= -1/2

Since both slopes are equal.They are parallel.

(3) Show that the straight lines 3x – 5y + 7 = 0 and 15 x + 9y + 4 = 0 are perpendicular.

Solution:

If two lines are perpendicular,then the product of their slopes will be equal to -1

m1 x m2 = -1

Slope of the first line 3x - 5 y + 7 = 0

Slope (m) = - coefficient of x/coefficient of y

m1 = -3/(-5)

= 3/5

Slope of the second line 15x + 9y + 4 = 0

m2 = -15/9

= -5/3

m1 x m2 = (3/5) x (-5/3) = -1

Since the product of those slopes is -1 .They are perpendicular.

(4) If the straight lines y/2 = x – p and ax + 5 = 3y are parallel,then find a.

Solution:

Since the given lines are parallel,slopes of their will be equal

y/2 = x – p

y = 2(x - p)

y = 2x - 2p

m1 = 2

ax + 5 = 3y

y = (a x + 5)/3

y = (a/3)x + (5/3)

m2 = a/3

m1 = m2

2 = a/3

2x3 = a

a = 6

In the page10th grade math solution exercise 5.5 part1 we are going to see the solution of next problem

(5) Find the value of a if the straight lines 5x – 2y – 9 = 0 and ay + 2x – 11 = 0 are perpendicular to each other.

Solution:

If two lines are perpendicular,then product of their slopes will be equal to -1

slope of the first line 5x – 2y – 9 = 0

m1 = -5/(-2)

m1 = 5/2

Slope of the second line ay + 2x – 11 = 0

m2 = -a/2

m1 x m2 = -1

(5/2) x (-a/2) = -1

-5a/4 = -1

5 a = 4

a = 4/5

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