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STRAIGHT LINES IN LINEAR EQUATIONS

There are two forms of line equations, namely:

The gradient of a line is a number that indicates the degree of slope of a line. The horizontal line (parallel to the X-axis) has a gradient of 0, and the vertical line (parallel to the Y-axis) has a gradient of
 
Note the following image:
If a line y = mx + c passes through point A(x 1 ,y 1 ) and point B(x 2 ,y 2 ) then the slope of the line is defined as:
So that the gradient of a line is the rate of change of the value of y to the value of x on that line.
If the gradient is positive, the change in the value of y to the value of x always increases (m > 0), for example by the g line above
If the gradient is negative then the change of the value of y to the value of x always decreases (m> 0), exemplified by the line h above
 
The equation of a line passing through point A (x 1 , y 1 ) with a gradient m is formulated:
 
y – y1 = m(x – x1 )
 
The equations of the lines passing through points A (x 1 , y 1  ) and B (x 2 , y 2  ) are formulated:

For more details, follow the example problem below:
01. Determine the value of the gradient of the line 2x + 6y = 5.
Answer 02. A line passes through point A(–2, –3) and point B(6, –5). Determine the value of the gradient of the line. Answer 03. Find the equation of the line with a gradient of 3 and through the point P(2, –4). Answer y – y 1 = m(x – x 1 ) y – (–4) = 3(x – 2)

y + 4 = 3x – 6
y = 3x – 10
04. Find the equation of the line that passes through point P(–2, 3) and point Q(2, –5)
Answer 05. A new car is tested for roadworthiness by driving it for 10 hours. In the first 4 hours the car has covered a distance of 242 km and after 6 hours the car has covered 362 km. If the car is always fixed, then determine the equation of the line that describes the speed of the car. Answer . Let x be the time the car travels and y be the distance traveled by the car, then: For x = 4 , y = 242, the point (4, 242) is obtained. For x = 6, y = 362 is obtained the point (6, 362) We get the equation of the line:



To draw the equation of a line on a Cartesius graph, at least two test points are required. For more details, follow the following example question:
07. Draw the line y = 2x – 6 on the Cartesius graph
Answer


08. Draw the line 3x + 2y = 12 on the Cartesian graph
Answer 09. Determine the line equation in the following figure is Answer Lines through (6, 0) and (0, 4), then There are two kinds of position of two lines, namely:







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Analytically, the conditions for the position of the two lines, namely y = m 1 x + c 1  and y = m 2 x + c 2 are:
The two lines are parallel if the slopes are the same (m 1  = m 2 ).
The two lines intersect if the gradients are not the same (m 1  m 2 ).
For more details, follow the example questions below:
01. What is the position of the lines 2x + 3y = –5 and 3x + 4y = –6.
Answer 02. A line g passes through the point A(4, –2). If the line g is parallel to the line 3x + 2y = 6 then find the equation of the line g. Answer



03. A line y = 2x + p intersects a line y = px – 4q at the point (3, 5). Determine the value of p + q = Two lines g and h will intersect perpendicularly if the product of the two gradients is -1. In other words: For more details, follow the example problem below: 04. A line passes through the point A(–3, 4). If the line is perpendicular to the line 2x – 5y = 8 then find the equation of the line. Answer 06. A line ax + by = c. If the line is perpendicular to 3x – 2y = 8 and passes through the point (6, –2) then determine the equation of the line. Answer





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