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High School Mathematics
9.15 Problems on Slope

Example - I:
The slope of a line passing through (5,3) and (6,a) is 8/3. Find the value of a.
Solution:
Given that,
The line is passing through (5,3) and (6,a)
Therefore, the slope of these two points is
m = (y2 - y1) / (x2 - x1)
m = (a - 3) / (6 - 5)
m = a - 3
Given that, m = 8/3
a - 3 = 8/3
a = 8/3 - 3
a = (8 - 9)/3
a = -1/3

Example - II:
The line 4x + 3y - 5 = 0 is perpendicular to 3x - ay + 7 = 0, find the value of a.
Solution:
Let the slope of 4x + 3y - 5 = 0, be m1,
Then m1 = -x coefficient / y coefficient.
i.e., m1 = -4/3
If m2 is the slope of 3x - ay + 7 = 0, then
m2 = -3 / -a
m2 = 3/a
Given that the lines are perpendicular, we know that the product of the perpendicular lines slope is equal to -1.
i.e., m1m2 = -1
-4/3 * 3/a = -1
-4/a = -1
a = 4


Direction: Solve the following problems. Also write at least 10 examples of your own.
Q 1: The slope of a line passing through (4,5) and (7,a) is 4. Find the value of a.
15
13
17
14

Q 2: The slope of a line passing through (4,6) and (8,a) is 6. Find the value of a.
20
30
15
25

Q 3: The line 15x - 7y + 21 = 0 is perpendicular to 4x - ay + 5 = 0, find the value of a.
-60/7
7
60/7
60

Q 4: The line 4x + 7y - 5 = 0 is perpendicular to 3x - ay - 7 = 0, find the value of a.
7
12
7/12
12/7

Q 5: The slope of a line passing through (3,8) and (5,a) is 7/2. Find the value of a.
15
12
14
13

Q 6: The line 3x + 5y - 15 = 0 is perpendicular to ax + 3y - 5 = 0, find the value of a.
5
-3
3
-5

Question 7: This question is available to subscribers only!

Question 8: This question is available to subscribers only!


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