# Lines and Angles Worksheets - Page 3

Lines and Angles Worksheets
• Page 3
21.
What are the $x$ and $y$ - intercepts of the line 5$x$ + 6$y$ - 30 = 0?
 a. - 6, 5 b. - 6, - 5 c. 6, - 5 d. 6, 5

#### Solution:

Given line is 5x + 6y - 30 = 0

5x + 6(0) = 30 gives x = 6
[Substitute y = 0 for x - intercept.]

So, x - intercept of the line is 6

5(0) + 6y - 30 = 0 gives y = 5
[Substitute y = 0 for x - intercept.]

So, y - intercept is 5

22.
Find the slope and $y$ - intercept of the line 11$x$ - 10$y$ + 20 = 0.
 a. $\frac{11}{10}$, 2 b. $\frac{11}{10}$, - 2 c. $\frac{10}{11}$, 2 d. $\frac{1}{10}$, 2

#### Solution:

Given line is 11x - 10y + 20 = 0

y = (11 / 10)x + 2
[Solve for y.]

This equation is in the slope - intercept form, y = mx + c

So the slope of the above line = m = 1110 and y - intercept of the above line = c = 2

23.
Write the equation of the line having slope 5 and $x$- intercept 4.
 a. 5 $x$ - $y$ - 20 = 0 b. 5 $x$ + $y$ + 20 = 0 c. 5 $x$ + $y$ - 20 = 0 d. 5 $x$ - $y$ + 20 = 0

#### Solution:

Let L be the line with slope m = 5 and with x - intercept = 4

Since the x - intercept of the line is 4, it passes through the point (4, 0).

The equation of the line L is y - y1 = m(x - x1)
[Slope- point form.]

y - 0 = 5 (x - 4)

y = 5x - 20

5x - y - 20 = 0.

24.
Find the equation of line passing through (10, 8) and parallel to line 8$x$ + 9$y$ - 46 = 0.
 a. $x$ + 9$y$ - 152 = 0 b. 8$x$ - 9$y$ - 152 = 0 c. 8$x$ + 9$y$ - 152 = 0 d. 8$x$ + 9$y$ + 152 = 0

#### Solution:

Given line is 8x + 9y - 46 = 0 , whose slope = -89

Let L be the required line parallel to the above line.

Since the slopes of parallel lines are equal, Slope of line L = m = slope of given line = -89

Since line L is passing through A(10, 8), the equation of line L is y - y1 = m(x - x1)
[Slope -point form of line.]

y - 8 = - 89 (x - 10)

9y - 72 = - 8x + 80

8x + 9y - 152 = 0

25.
Write the equation of the line passing through (6, 3) and perpendicular to the line 5$x$ + 6$y$ - 38 = 0.
 a. 6$x$ + 5$y$ + 21 = 0 b. 6$x$ + 5$y$ - 21 = 0 c. 6$x$ - 5$y$ + 21 = 0 d. 6$x$ - 5$y$ - 21 = 0

#### Solution:

Given line is 5x + 6y - 38 = 0.

y = (- 56)x + 19 / 3
[Solve for y.]

On comparing the above line with the slope intercept form y = mx + c we get, m = (- 56), c = 19 / 3

So, slope of given line is m = (- 56).

Let L be the required line perpendicular to the given line.

Since product of slopes of perpendicular lines = -1

( slope of line L) × ( slope of line 5x + 6y - 38 = 0) = - 1

Slope of line L = m1 = -1(-56) = 65

Since line L is passing through A(6, 3), the equation of line L is y - y1 = m1 (x - x1)
[Slope - point form of line.]

y - 3 = 65 (x - 6)

5y - 15 = 6x - 36

6x - 5y - 21 = 0