# 10th GRADE MATH SOLUTION EXERCISE 5.5 part2

This page 10th grade math solution exercise 5.5 part2 is going to provide you solution for every problems that you find in the exercise no 5.5

## 10th grade math solution exercise 5.5 part2

(6) Find the values of p for which the straight
lines 8px + (2-3p)y + 1 = 0 and px + 8y + 7 = 0 are perpendicular to each
other.

Solution:

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

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

Slope of the first line 8px + (2-3p)y + 1 = 0

m1 = -8p/(2-3p)

Slope of the second line px + 8y + 7 = 0

m2 = -p/8

m1 x m2 = -1

[-8p/(2-3p)] x [-p/8] = -1

[(8 p²)/8(2-3p)] = -1

p²/(2-3p) = -1

p² = -1(2-3 p)

p² = -2 + 3 p

p²- 3 p + 2 = 0

(p - 2) (p - 1) = 0

p - 2 = 0 p - 1 = 0

p = 2 p = 1

(7) If the straight lines passing through the
points (h,3) and (4,1) intersects the line 7x – 9y – 19 = 0 at right
angle,then find the value of h.

Solution:

the straight lines passing through the
points (h,3) and (4,1) intersects the line 7x – 9y – 19 = 0 at right
angle

then the product of their slopes will be equal to -1

from this first we are going to find the slope of the line passing through the points (h,3) and (4,1)

m = (y2-y1)/(x2-x1)

= (1-3)/(4-h)

m1 = -2/(4-h)

slope of the second line 7x – 9y – 19 = 0

m = -7/(-9)

= 7/9

m1 x m2 = -1

[-2/(4-h)] x (7/9) = -1

[2/(4-h)] x (7/9) = 1

14/9(4-h) = 1

14/(36-9h) = 1

14 = 36 - 9 h

9h = 36-14

9h = 22

h = 22/9

(8) Find the equation of the straight line parallel to the line
3x – y + 7 = 0 and passing through the point (1,-2).

Solution:

Here,the required line is parallel to the given line 3x - y + 7 = 0.Then their slopes will be equal.

m = -3/(-1)

m = 3

Equation of the line

(y - y1) = m (x - x1)

(y - (-2)) = 3(x - 1)

y + 2 = 3 x - 3

3x - y - 3 -2 = 0

3x - y - 5 = 0

In the page10th grade math solution exercise 5.5 part2 we are going to see the solution of next problem

(9) Find the equation of the straight line perpendicular to
the straight line x – 2y + 3 = 0 and passing through the point (1,-2).

Solution:

Here the required line is perpendicular to the given line x – 2y + 3 = 0

m = -1/(-2)

= 1/2

slope of the required line = -1/m

= -1/(1/2)

= -2

Equation of the line

(y - y1) = m (x - x1)

(y - (-2)) = -2(x - 1)

y + 2 = -2 x + 2

2x + y + 2 - 2 = 0

2x + y = 0

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