﻿ Lines and Angles Worksheets - Page 2 | Problems & Solutions
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Lines and Angles Worksheets
• Page 2
11.
Find the slope of the line passing through the points A (7, 9) and B (9, 13). a. - $\frac{1}{2}$ b. - 2 c. $\frac{1}{2}$ d. 2

#### Solution:

The slope of the line passing through the points A and B is m = ΔyΔx = y2-y1x2-x1

= 13 - 99 - 7 = 2
[Use the slope formula.]

Correct answer : (4)
12.
If $t$ > 1and $m$ is the slope of the line joining the points ($\frac{2}{{t}^{2}}$ , $\frac{4}{t}$) and (2$t$2 , 4$t$), then which of the following is correct? a. $m$ =1 b. $m$ > 1 c. $m$ < 1 d. $m$ =2

#### Solution:

The points are P(2t2 , 4t)and Q (2t2 , 4t)

Slope of line joining P,Q = m = ΔyΔx = (4t-4t)(2t2-2t2)
[Use the slope formula.]

= 4(t-1t)2(t-1t)(t+1t)

So, m = 2t +1t
[Cancel the common factor.]

2t +1t < 1
[ For t >1, t + 1t > 2.]

So, m < 1.

Correct answer : (3)
13.
Find the values of $k$ such that the points ($k$ + 2, 2), (2$k$ + 2, 4) and (2$k$ + 3, 2$k$ + 1) are collinear. a. - $\frac{1}{2}$ , 2 b. $\frac{1}{2}$ , 2 c. $\frac{1}{2}$ , -2 d. - $\frac{1}{2}$ , -2

#### Solution:

The points are A(k + 2, 2), B(2k + 2, 4) and C(2k + 3, 2k + 1)

Slope of line joining A, B = slope of line joining B,C .
[The three points A, B, C are collinear.]

4 - 22k + 2 - k - 2 = 2k + 1 - 42k + 3 - 2k - 2
[Use the slope formula.]

2k = 2k - 31

2k2 - 3k - 2 = 0

(2k + 1)(k - 2) = 0
[Factor.]

k = - 12 , 2.

Correct answer : (1)
14.
What is the slope of the line 4$x$ + 9$y$ + 13 = 0? a. $\frac{9}{4}$ b. $\frac{4}{9}$ c. - $\frac{4}{9}$ d. - $\frac{9}{4}$

#### Solution:

Given line is 4x + 9y + 13 = 0

y = (- 49)x + (- 13 / 9)
[Solve for y.]

As the above equation is in the slope intercept form y = mx + c, slope of the line = m = - 4 / 9

Correct answer : (3)
15.
What is the equation of the line passing through the points (2, 5) and (5, 2)? a. $x$ + $y$ + 3 = 0 b. $x$ + $y$ - 7 = 0 c. $x$ - $y$ + 7 = 0 d. $x$ - $y$ + 3 = 0

#### Solution:

The points are A(x1 , y1) = (2, 5) and B(x2, y2) = (5, 2)

The slope of the line passing through the points A and B is m = ΔyΔx = y2-y1x2-x1
[Use the slope formula.]

= 2 - 55 - 2 = - 1

The equation of the line passing through A(x1, y1) = (2, 5) having slope m = - 1 is y - y1 = m(x - x1)
[Slope point form.]

y - 5 = - 1(x - 2)

x + y - 7 = 0

Correct answer : (2)
16.
What is the equation of the line passing through (- 3, - 7) with the slope 5? a. 5$x$ - $y$ + 8 = 0 b. 5$x$ - $y$ + 22 = 0 c. $x$ - 5$y$ - 8 = 0 d. $x$ - 5$y$ - 22 = 0

#### Solution:

Given point is A(x1, y1) = (- 3, - 7)

Let L be the line with the slope m = 5

If the line L passes through A then its equation is y - y1 = m(x - x1)
[Slope point form.]

y + 7 = 5(x + 3)

y + 7 = 5x + 15

5x - y + 8 = 0

Correct answer : (1)
17.
Find the equation of the line with slope 7 and $y$ - intercept 11. a. 7$x$ - $y$ - 11 = 0 b. $x$ - 7$y$ - 11 = 0 c. 7$x$ - $y$ + 11 = 0 d. $x$ - 7$y$ + 11 = 0

#### Solution:

Let L be the line with the slope, m = 7 and having y - intercept, c = 11

So, the equation of line L is y = mx + c.
[Slope - intercept form.]

y = 7x + 11

7x - y + 11 = 0

Correct answer : (3)
18.
What is the equation of the line which has 7 and - 8 as the $x$, $y$ - intercepts? a. 8$x$ + 7 $y$ - 56 = 0 b. 8$x$ + 7 $y$ + 56 = 0 c. 8$x$ - 7 $y$ - 56 = 0 d. 8$x$ - 7 $y$ + 56 = 0

#### Solution:

Let L be the line whose x-intercept = 7, y-intercept = - 8

Let the equation of line L be y = mx + c
[Slope -intercept form.]

Substitute x = 0 in y = mx + c we get
[For y-intercept.]

y = c = - 8

Substitute y = 0 in y = mx + c we get
[For x-intercept. ]

mx + c = 0

x = - (cm) = 7
[x-intercept = 7.]

- (- 8m) = 7 and hence m = 87
[c = - 8.]

So, the equation of the line L is y = ( 8 / 7) x + (- 8)

8x - 7y - 56 = 0

Correct answer : (3)
19.
What is the equation of the line passing through (6, 12) and has an angle of inclination $\frac{3\pi }{4}$? a. $x$ + $y$ - 18 = 0 b. 6 $x$ + $y$ = 18 c. $x$ - $y$ + 18 = 0 d. 6 $x$ - $y$ = 0

#### Solution:

Let L be the line passing through A(x1, y1) = (6, 12), whose angle of inclination θ = 3π4.

The slope of the line L = m = tan 3π4 = -1
[Use the slope definition.]

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

y - 12 = -1(x - 6)

y - 12 = - x + 6

x + y - 18 = 0

Correct answer : (1)
20.
What is the equation of a vertical line whose $x$ - intercept is 3? a. $x$ = 3 b. $x$ + $y$ = 3 c. $y$ = 3 d. $x$ - $y$ = 3

#### Solution:

Let L be the required vertical line.

Equation of line L is x = k where k is a constant.
[Equation of a vertical line.]

Since x - intercept of line L is 3, it passes through (3, 0).

Since the line L passes through (3, 0), 3 = k
[Substitute 3 for x into x = k.]

So, the equation of the line L is x = 3

Correct answer : (1)

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